Why can `(*)` be omitted from a function pointer inside a function parameter list?

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8
down vote

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Compiling with gcc8:



#include <stdio.h>
void some_func(void f1(void), void (*f2)(void))

printf("%dn", f1);
printf("%dn", f2);



Gives (only) the following warnings:



<source>:11:14: warning: format '%d' expects argument of type 'int', but argument 2 has type 'void (*)(void)' [-Wformat=]
printf("%dn", f1);
<source>:12:14: warning: format '%d' expects argument of type 'int', but argument 2 has type 'void (*)(void)' [-Wformat=]
printf("%dn", f2);


Why is the type of f1 the same as f2? Only f2 is declared as a function pointer. I would expect f1 not to compile at all, as it names a function type, not a function pointer. What is the rule that says, that a function type inside a function parameter list changes to a pointer to that function type?










share|improve this question



















  • 1




    Answered this on Quora once: quora.com/…
    – PSkocik
    2 hours ago










  • Regarding the warnings, please use %p format specifier when targeting pointers.
    – paddy
    2 hours ago






  • 2




    @paddy I know, and not really, %p is only for void*, so I would need to cast it. The warnings serve only to show to me the deduced by the compiler variable types, not to print the pointers addresses.
    – Kamil Cuk
    2 hours ago










  • As a note, just stick to one style, be constant, I prefer the second one but, there is not a clear rule about that.
    – Stargateur
    1 hour ago










  • Why? Because the language standard says so.
    – AnT
    1 hour ago














up vote
8
down vote

favorite












Compiling with gcc8:



#include <stdio.h>
void some_func(void f1(void), void (*f2)(void))

printf("%dn", f1);
printf("%dn", f2);



Gives (only) the following warnings:



<source>:11:14: warning: format '%d' expects argument of type 'int', but argument 2 has type 'void (*)(void)' [-Wformat=]
printf("%dn", f1);
<source>:12:14: warning: format '%d' expects argument of type 'int', but argument 2 has type 'void (*)(void)' [-Wformat=]
printf("%dn", f2);


Why is the type of f1 the same as f2? Only f2 is declared as a function pointer. I would expect f1 not to compile at all, as it names a function type, not a function pointer. What is the rule that says, that a function type inside a function parameter list changes to a pointer to that function type?










share|improve this question



















  • 1




    Answered this on Quora once: quora.com/…
    – PSkocik
    2 hours ago










  • Regarding the warnings, please use %p format specifier when targeting pointers.
    – paddy
    2 hours ago






  • 2




    @paddy I know, and not really, %p is only for void*, so I would need to cast it. The warnings serve only to show to me the deduced by the compiler variable types, not to print the pointers addresses.
    – Kamil Cuk
    2 hours ago










  • As a note, just stick to one style, be constant, I prefer the second one but, there is not a clear rule about that.
    – Stargateur
    1 hour ago










  • Why? Because the language standard says so.
    – AnT
    1 hour ago












up vote
8
down vote

favorite









up vote
8
down vote

favorite











Compiling with gcc8:



#include <stdio.h>
void some_func(void f1(void), void (*f2)(void))

printf("%dn", f1);
printf("%dn", f2);



Gives (only) the following warnings:



<source>:11:14: warning: format '%d' expects argument of type 'int', but argument 2 has type 'void (*)(void)' [-Wformat=]
printf("%dn", f1);
<source>:12:14: warning: format '%d' expects argument of type 'int', but argument 2 has type 'void (*)(void)' [-Wformat=]
printf("%dn", f2);


Why is the type of f1 the same as f2? Only f2 is declared as a function pointer. I would expect f1 not to compile at all, as it names a function type, not a function pointer. What is the rule that says, that a function type inside a function parameter list changes to a pointer to that function type?










share|improve this question















Compiling with gcc8:



#include <stdio.h>
void some_func(void f1(void), void (*f2)(void))

printf("%dn", f1);
printf("%dn", f2);



Gives (only) the following warnings:



<source>:11:14: warning: format '%d' expects argument of type 'int', but argument 2 has type 'void (*)(void)' [-Wformat=]
printf("%dn", f1);
<source>:12:14: warning: format '%d' expects argument of type 'int', but argument 2 has type 'void (*)(void)' [-Wformat=]
printf("%dn", f2);


Why is the type of f1 the same as f2? Only f2 is declared as a function pointer. I would expect f1 not to compile at all, as it names a function type, not a function pointer. What is the rule that says, that a function type inside a function parameter list changes to a pointer to that function type?







c






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









Boann

36.1k1286118




36.1k1286118










asked 2 hours ago









Kamil Cuk

5,8111219




5,8111219







  • 1




    Answered this on Quora once: quora.com/…
    – PSkocik
    2 hours ago










  • Regarding the warnings, please use %p format specifier when targeting pointers.
    – paddy
    2 hours ago






  • 2




    @paddy I know, and not really, %p is only for void*, so I would need to cast it. The warnings serve only to show to me the deduced by the compiler variable types, not to print the pointers addresses.
    – Kamil Cuk
    2 hours ago










  • As a note, just stick to one style, be constant, I prefer the second one but, there is not a clear rule about that.
    – Stargateur
    1 hour ago










  • Why? Because the language standard says so.
    – AnT
    1 hour ago












  • 1




    Answered this on Quora once: quora.com/…
    – PSkocik
    2 hours ago










  • Regarding the warnings, please use %p format specifier when targeting pointers.
    – paddy
    2 hours ago






  • 2




    @paddy I know, and not really, %p is only for void*, so I would need to cast it. The warnings serve only to show to me the deduced by the compiler variable types, not to print the pointers addresses.
    – Kamil Cuk
    2 hours ago










  • As a note, just stick to one style, be constant, I prefer the second one but, there is not a clear rule about that.
    – Stargateur
    1 hour ago










  • Why? Because the language standard says so.
    – AnT
    1 hour ago







1




1




Answered this on Quora once: quora.com/…
– PSkocik
2 hours ago




Answered this on Quora once: quora.com/…
– PSkocik
2 hours ago












Regarding the warnings, please use %p format specifier when targeting pointers.
– paddy
2 hours ago




Regarding the warnings, please use %p format specifier when targeting pointers.
– paddy
2 hours ago




2




2




@paddy I know, and not really, %p is only for void*, so I would need to cast it. The warnings serve only to show to me the deduced by the compiler variable types, not to print the pointers addresses.
– Kamil Cuk
2 hours ago




@paddy I know, and not really, %p is only for void*, so I would need to cast it. The warnings serve only to show to me the deduced by the compiler variable types, not to print the pointers addresses.
– Kamil Cuk
2 hours ago












As a note, just stick to one style, be constant, I prefer the second one but, there is not a clear rule about that.
– Stargateur
1 hour ago




As a note, just stick to one style, be constant, I prefer the second one but, there is not a clear rule about that.
– Stargateur
1 hour ago












Why? Because the language standard says so.
– AnT
1 hour ago




Why? Because the language standard says so.
– AnT
1 hour ago












2 Answers
2






active

oldest

votes

















up vote
11
down vote













Because the standard (6.7.6.3p8) says that




A declaration of a parameter as ''function returning type'' shall be
adjusted to ''pointer to function returning type'', as in 6.3.2.1.




It's similar to how arrays parameters are adjusted to pointers (6.7.63.p7) , if you think about it.



void some_func(void (void));
void some_func(void (*)(void));


are compatible declarations, just like:



void other_func(char string);
void other_func(char *string);


are.




Note that the adjustment doesn't make void some_func(void (*)(void) compatible with void some_other_func(void (**)(void) or void yet_another_func(void (*****)(void) and that as far as functions are concerned, declarations don't really reflect use anymore, (despite that being the intention of the language's original author). In standardized C, due to how function identifiers decay to pointers and due to how it doesn't matter whether you use a function type or a function pointer type to make a call, you can call any function with arbitrarily many *:



#include <stdio.h>
int main()

(*puts)("hello world");
(******puts)("hello world");
(***&*&*puts)("hello world"); //& cancels a * as per 6.5.3.2p3

int (*p)(char const*) = puts;
int (**pp)(char const*) = &p;
int (***ppp)(char const*) = &pp;

(**ppp)("hello world"); //at least two asterisks required here






share|improve this answer





























    up vote
    0
    down vote













    Because in C, in that case, the function name itself is a function pointer. See this answer: Why is using the function name as a function pointer equivalent to applying the address-of operator to the function name?






    share|improve this answer






















    • Then why does void some_func(void (*****f3)(void)) here f3 names a different type then f1 and f2.
      – Kamil Cuk
      2 hours ago










    • I don't see why it should be the same type as f1 or f2. There is an historical and practical reason why the function name is a pointer to the function itself, like explained in that answer, and I don't see these reasons can be applied to any arbitrary pointer to pointer (...) to function.
      – ad3angel1s
      2 hours ago






    • 4




      In C, a function name itself is not a function pointer. The function name is an identifier. When used as an expression, it has function type. Per C 6.3.2.1 4, an expression that has function type is called a function designator. When it is not used as the operand of sizeof or unary &, then a function designator is converted to a pointer (again by 6.3.2.1 4). I understand you may think of this conversion as so automatic and ubiquitous that one can think of a function name as a pointer, but we are creating a permanent record of technical answers explaining the C standard here.
      – Eric Postpischil
      1 hour ago










    • @EricPostpischil does this mean that you can only call only function pointers and not functions? Because () operator is not sizeof or &.
      – Ajay Brahmakshatriya
      1 hour ago






    • 1




      @AjayBrahmakshatriya: In a function call expression, such as SomeExpression(), if the SomeExpression is a function designator (any expression that has function type), then it is automatically converted to a pointer to the function. In effect, a “proper” way to write a function call such as f1(3) is (&f1)(3). When you write f1(3), the compiler automatically converts it to (&f1)(3) for you.
      – Eric Postpischil
      1 hour ago










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    11
    down vote













    Because the standard (6.7.6.3p8) says that




    A declaration of a parameter as ''function returning type'' shall be
    adjusted to ''pointer to function returning type'', as in 6.3.2.1.




    It's similar to how arrays parameters are adjusted to pointers (6.7.63.p7) , if you think about it.



    void some_func(void (void));
    void some_func(void (*)(void));


    are compatible declarations, just like:



    void other_func(char string);
    void other_func(char *string);


    are.




    Note that the adjustment doesn't make void some_func(void (*)(void) compatible with void some_other_func(void (**)(void) or void yet_another_func(void (*****)(void) and that as far as functions are concerned, declarations don't really reflect use anymore, (despite that being the intention of the language's original author). In standardized C, due to how function identifiers decay to pointers and due to how it doesn't matter whether you use a function type or a function pointer type to make a call, you can call any function with arbitrarily many *:



    #include <stdio.h>
    int main()

    (*puts)("hello world");
    (******puts)("hello world");
    (***&*&*puts)("hello world"); //& cancels a * as per 6.5.3.2p3

    int (*p)(char const*) = puts;
    int (**pp)(char const*) = &p;
    int (***ppp)(char const*) = &pp;

    (**ppp)("hello world"); //at least two asterisks required here






    share|improve this answer


























      up vote
      11
      down vote













      Because the standard (6.7.6.3p8) says that




      A declaration of a parameter as ''function returning type'' shall be
      adjusted to ''pointer to function returning type'', as in 6.3.2.1.




      It's similar to how arrays parameters are adjusted to pointers (6.7.63.p7) , if you think about it.



      void some_func(void (void));
      void some_func(void (*)(void));


      are compatible declarations, just like:



      void other_func(char string);
      void other_func(char *string);


      are.




      Note that the adjustment doesn't make void some_func(void (*)(void) compatible with void some_other_func(void (**)(void) or void yet_another_func(void (*****)(void) and that as far as functions are concerned, declarations don't really reflect use anymore, (despite that being the intention of the language's original author). In standardized C, due to how function identifiers decay to pointers and due to how it doesn't matter whether you use a function type or a function pointer type to make a call, you can call any function with arbitrarily many *:



      #include <stdio.h>
      int main()

      (*puts)("hello world");
      (******puts)("hello world");
      (***&*&*puts)("hello world"); //& cancels a * as per 6.5.3.2p3

      int (*p)(char const*) = puts;
      int (**pp)(char const*) = &p;
      int (***ppp)(char const*) = &pp;

      (**ppp)("hello world"); //at least two asterisks required here






      share|improve this answer
























        up vote
        11
        down vote










        up vote
        11
        down vote









        Because the standard (6.7.6.3p8) says that




        A declaration of a parameter as ''function returning type'' shall be
        adjusted to ''pointer to function returning type'', as in 6.3.2.1.




        It's similar to how arrays parameters are adjusted to pointers (6.7.63.p7) , if you think about it.



        void some_func(void (void));
        void some_func(void (*)(void));


        are compatible declarations, just like:



        void other_func(char string);
        void other_func(char *string);


        are.




        Note that the adjustment doesn't make void some_func(void (*)(void) compatible with void some_other_func(void (**)(void) or void yet_another_func(void (*****)(void) and that as far as functions are concerned, declarations don't really reflect use anymore, (despite that being the intention of the language's original author). In standardized C, due to how function identifiers decay to pointers and due to how it doesn't matter whether you use a function type or a function pointer type to make a call, you can call any function with arbitrarily many *:



        #include <stdio.h>
        int main()

        (*puts)("hello world");
        (******puts)("hello world");
        (***&*&*puts)("hello world"); //& cancels a * as per 6.5.3.2p3

        int (*p)(char const*) = puts;
        int (**pp)(char const*) = &p;
        int (***ppp)(char const*) = &pp;

        (**ppp)("hello world"); //at least two asterisks required here






        share|improve this answer














        Because the standard (6.7.6.3p8) says that




        A declaration of a parameter as ''function returning type'' shall be
        adjusted to ''pointer to function returning type'', as in 6.3.2.1.




        It's similar to how arrays parameters are adjusted to pointers (6.7.63.p7) , if you think about it.



        void some_func(void (void));
        void some_func(void (*)(void));


        are compatible declarations, just like:



        void other_func(char string);
        void other_func(char *string);


        are.




        Note that the adjustment doesn't make void some_func(void (*)(void) compatible with void some_other_func(void (**)(void) or void yet_another_func(void (*****)(void) and that as far as functions are concerned, declarations don't really reflect use anymore, (despite that being the intention of the language's original author). In standardized C, due to how function identifiers decay to pointers and due to how it doesn't matter whether you use a function type or a function pointer type to make a call, you can call any function with arbitrarily many *:



        #include <stdio.h>
        int main()

        (*puts)("hello world");
        (******puts)("hello world");
        (***&*&*puts)("hello world"); //& cancels a * as per 6.5.3.2p3

        int (*p)(char const*) = puts;
        int (**pp)(char const*) = &p;
        int (***ppp)(char const*) = &pp;

        (**ppp)("hello world"); //at least two asterisks required here







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 1 hour ago

























        answered 2 hours ago









        PSkocik

        28.9k43965




        28.9k43965






















            up vote
            0
            down vote













            Because in C, in that case, the function name itself is a function pointer. See this answer: Why is using the function name as a function pointer equivalent to applying the address-of operator to the function name?






            share|improve this answer






















            • Then why does void some_func(void (*****f3)(void)) here f3 names a different type then f1 and f2.
              – Kamil Cuk
              2 hours ago










            • I don't see why it should be the same type as f1 or f2. There is an historical and practical reason why the function name is a pointer to the function itself, like explained in that answer, and I don't see these reasons can be applied to any arbitrary pointer to pointer (...) to function.
              – ad3angel1s
              2 hours ago






            • 4




              In C, a function name itself is not a function pointer. The function name is an identifier. When used as an expression, it has function type. Per C 6.3.2.1 4, an expression that has function type is called a function designator. When it is not used as the operand of sizeof or unary &, then a function designator is converted to a pointer (again by 6.3.2.1 4). I understand you may think of this conversion as so automatic and ubiquitous that one can think of a function name as a pointer, but we are creating a permanent record of technical answers explaining the C standard here.
              – Eric Postpischil
              1 hour ago










            • @EricPostpischil does this mean that you can only call only function pointers and not functions? Because () operator is not sizeof or &.
              – Ajay Brahmakshatriya
              1 hour ago






            • 1




              @AjayBrahmakshatriya: In a function call expression, such as SomeExpression(), if the SomeExpression is a function designator (any expression that has function type), then it is automatically converted to a pointer to the function. In effect, a “proper” way to write a function call such as f1(3) is (&f1)(3). When you write f1(3), the compiler automatically converts it to (&f1)(3) for you.
              – Eric Postpischil
              1 hour ago














            up vote
            0
            down vote













            Because in C, in that case, the function name itself is a function pointer. See this answer: Why is using the function name as a function pointer equivalent to applying the address-of operator to the function name?






            share|improve this answer






















            • Then why does void some_func(void (*****f3)(void)) here f3 names a different type then f1 and f2.
              – Kamil Cuk
              2 hours ago










            • I don't see why it should be the same type as f1 or f2. There is an historical and practical reason why the function name is a pointer to the function itself, like explained in that answer, and I don't see these reasons can be applied to any arbitrary pointer to pointer (...) to function.
              – ad3angel1s
              2 hours ago






            • 4




              In C, a function name itself is not a function pointer. The function name is an identifier. When used as an expression, it has function type. Per C 6.3.2.1 4, an expression that has function type is called a function designator. When it is not used as the operand of sizeof or unary &, then a function designator is converted to a pointer (again by 6.3.2.1 4). I understand you may think of this conversion as so automatic and ubiquitous that one can think of a function name as a pointer, but we are creating a permanent record of technical answers explaining the C standard here.
              – Eric Postpischil
              1 hour ago










            • @EricPostpischil does this mean that you can only call only function pointers and not functions? Because () operator is not sizeof or &.
              – Ajay Brahmakshatriya
              1 hour ago






            • 1




              @AjayBrahmakshatriya: In a function call expression, such as SomeExpression(), if the SomeExpression is a function designator (any expression that has function type), then it is automatically converted to a pointer to the function. In effect, a “proper” way to write a function call such as f1(3) is (&f1)(3). When you write f1(3), the compiler automatically converts it to (&f1)(3) for you.
              – Eric Postpischil
              1 hour ago












            up vote
            0
            down vote










            up vote
            0
            down vote









            Because in C, in that case, the function name itself is a function pointer. See this answer: Why is using the function name as a function pointer equivalent to applying the address-of operator to the function name?






            share|improve this answer














            Because in C, in that case, the function name itself is a function pointer. See this answer: Why is using the function name as a function pointer equivalent to applying the address-of operator to the function name?







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 1 hour ago

























            answered 2 hours ago









            ad3angel1s

            314112




            314112











            • Then why does void some_func(void (*****f3)(void)) here f3 names a different type then f1 and f2.
              – Kamil Cuk
              2 hours ago










            • I don't see why it should be the same type as f1 or f2. There is an historical and practical reason why the function name is a pointer to the function itself, like explained in that answer, and I don't see these reasons can be applied to any arbitrary pointer to pointer (...) to function.
              – ad3angel1s
              2 hours ago






            • 4




              In C, a function name itself is not a function pointer. The function name is an identifier. When used as an expression, it has function type. Per C 6.3.2.1 4, an expression that has function type is called a function designator. When it is not used as the operand of sizeof or unary &, then a function designator is converted to a pointer (again by 6.3.2.1 4). I understand you may think of this conversion as so automatic and ubiquitous that one can think of a function name as a pointer, but we are creating a permanent record of technical answers explaining the C standard here.
              – Eric Postpischil
              1 hour ago










            • @EricPostpischil does this mean that you can only call only function pointers and not functions? Because () operator is not sizeof or &.
              – Ajay Brahmakshatriya
              1 hour ago






            • 1




              @AjayBrahmakshatriya: In a function call expression, such as SomeExpression(), if the SomeExpression is a function designator (any expression that has function type), then it is automatically converted to a pointer to the function. In effect, a “proper” way to write a function call such as f1(3) is (&f1)(3). When you write f1(3), the compiler automatically converts it to (&f1)(3) for you.
              – Eric Postpischil
              1 hour ago
















            • Then why does void some_func(void (*****f3)(void)) here f3 names a different type then f1 and f2.
              – Kamil Cuk
              2 hours ago










            • I don't see why it should be the same type as f1 or f2. There is an historical and practical reason why the function name is a pointer to the function itself, like explained in that answer, and I don't see these reasons can be applied to any arbitrary pointer to pointer (...) to function.
              – ad3angel1s
              2 hours ago






            • 4




              In C, a function name itself is not a function pointer. The function name is an identifier. When used as an expression, it has function type. Per C 6.3.2.1 4, an expression that has function type is called a function designator. When it is not used as the operand of sizeof or unary &, then a function designator is converted to a pointer (again by 6.3.2.1 4). I understand you may think of this conversion as so automatic and ubiquitous that one can think of a function name as a pointer, but we are creating a permanent record of technical answers explaining the C standard here.
              – Eric Postpischil
              1 hour ago










            • @EricPostpischil does this mean that you can only call only function pointers and not functions? Because () operator is not sizeof or &.
              – Ajay Brahmakshatriya
              1 hour ago






            • 1




              @AjayBrahmakshatriya: In a function call expression, such as SomeExpression(), if the SomeExpression is a function designator (any expression that has function type), then it is automatically converted to a pointer to the function. In effect, a “proper” way to write a function call such as f1(3) is (&f1)(3). When you write f1(3), the compiler automatically converts it to (&f1)(3) for you.
              – Eric Postpischil
              1 hour ago















            Then why does void some_func(void (*****f3)(void)) here f3 names a different type then f1 and f2.
            – Kamil Cuk
            2 hours ago




            Then why does void some_func(void (*****f3)(void)) here f3 names a different type then f1 and f2.
            – Kamil Cuk
            2 hours ago












            I don't see why it should be the same type as f1 or f2. There is an historical and practical reason why the function name is a pointer to the function itself, like explained in that answer, and I don't see these reasons can be applied to any arbitrary pointer to pointer (...) to function.
            – ad3angel1s
            2 hours ago




            I don't see why it should be the same type as f1 or f2. There is an historical and practical reason why the function name is a pointer to the function itself, like explained in that answer, and I don't see these reasons can be applied to any arbitrary pointer to pointer (...) to function.
            – ad3angel1s
            2 hours ago




            4




            4




            In C, a function name itself is not a function pointer. The function name is an identifier. When used as an expression, it has function type. Per C 6.3.2.1 4, an expression that has function type is called a function designator. When it is not used as the operand of sizeof or unary &, then a function designator is converted to a pointer (again by 6.3.2.1 4). I understand you may think of this conversion as so automatic and ubiquitous that one can think of a function name as a pointer, but we are creating a permanent record of technical answers explaining the C standard here.
            – Eric Postpischil
            1 hour ago




            In C, a function name itself is not a function pointer. The function name is an identifier. When used as an expression, it has function type. Per C 6.3.2.1 4, an expression that has function type is called a function designator. When it is not used as the operand of sizeof or unary &, then a function designator is converted to a pointer (again by 6.3.2.1 4). I understand you may think of this conversion as so automatic and ubiquitous that one can think of a function name as a pointer, but we are creating a permanent record of technical answers explaining the C standard here.
            – Eric Postpischil
            1 hour ago












            @EricPostpischil does this mean that you can only call only function pointers and not functions? Because () operator is not sizeof or &.
            – Ajay Brahmakshatriya
            1 hour ago




            @EricPostpischil does this mean that you can only call only function pointers and not functions? Because () operator is not sizeof or &.
            – Ajay Brahmakshatriya
            1 hour ago




            1




            1




            @AjayBrahmakshatriya: In a function call expression, such as SomeExpression(), if the SomeExpression is a function designator (any expression that has function type), then it is automatically converted to a pointer to the function. In effect, a “proper” way to write a function call such as f1(3) is (&f1)(3). When you write f1(3), the compiler automatically converts it to (&f1)(3) for you.
            – Eric Postpischil
            1 hour ago




            @AjayBrahmakshatriya: In a function call expression, such as SomeExpression(), if the SomeExpression is a function designator (any expression that has function type), then it is automatically converted to a pointer to the function. In effect, a “proper” way to write a function call such as f1(3) is (&f1)(3). When you write f1(3), the compiler automatically converts it to (&f1)(3) for you.
            – Eric Postpischil
            1 hour ago

















             

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