Allocation as default initialization

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












16















Say I have a struct (or class) with a dynamic array, its length, and a constructor:



struct array 
int l;
int* t;
array(int length);
;

array::array(int length)
l=length;
t=new int[l];



I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:



struct array 
int l;
int* t = new int[l];
array(int length);


array::array(int length)
l=length;



It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.










share|improve this question

















  • 7





    don't use 'l' as a variable name, It's easily confused with the number '1',

    – regomodo
    Jan 28 at 13:31











  • @regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

    – Heinzi
    Jan 28 at 14:37







  • 5





    id rather people read "Clean Code" instead

    – regomodo
    Jan 28 at 14:54






  • 2





    @Heinzi Or maybe stop using single letter variables damnit.

    – Bakuriu
    Jan 28 at 20:09















16















Say I have a struct (or class) with a dynamic array, its length, and a constructor:



struct array 
int l;
int* t;
array(int length);
;

array::array(int length)
l=length;
t=new int[l];



I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:



struct array 
int l;
int* t = new int[l];
array(int length);


array::array(int length)
l=length;



It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.










share|improve this question

















  • 7





    don't use 'l' as a variable name, It's easily confused with the number '1',

    – regomodo
    Jan 28 at 13:31











  • @regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

    – Heinzi
    Jan 28 at 14:37







  • 5





    id rather people read "Clean Code" instead

    – regomodo
    Jan 28 at 14:54






  • 2





    @Heinzi Or maybe stop using single letter variables damnit.

    – Bakuriu
    Jan 28 at 20:09













16












16








16








Say I have a struct (or class) with a dynamic array, its length, and a constructor:



struct array 
int l;
int* t;
array(int length);
;

array::array(int length)
l=length;
t=new int[l];



I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:



struct array 
int l;
int* t = new int[l];
array(int length);


array::array(int length)
l=length;



It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.










share|improve this question














Say I have a struct (or class) with a dynamic array, its length, and a constructor:



struct array 
int l;
int* t;
array(int length);
;

array::array(int length)
l=length;
t=new int[l];



I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:



struct array 
int l;
int* t = new int[l];
array(int length);


array::array(int length)
l=length;



It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.







c++ constructor






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 28 at 10:24









tarulentarulen

1,798413




1,798413







  • 7





    don't use 'l' as a variable name, It's easily confused with the number '1',

    – regomodo
    Jan 28 at 13:31











  • @regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

    – Heinzi
    Jan 28 at 14:37







  • 5





    id rather people read "Clean Code" instead

    – regomodo
    Jan 28 at 14:54






  • 2





    @Heinzi Or maybe stop using single letter variables damnit.

    – Bakuriu
    Jan 28 at 20:09












  • 7





    don't use 'l' as a variable name, It's easily confused with the number '1',

    – regomodo
    Jan 28 at 13:31











  • @regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

    – Heinzi
    Jan 28 at 14:37







  • 5





    id rather people read "Clean Code" instead

    – regomodo
    Jan 28 at 14:54






  • 2





    @Heinzi Or maybe stop using single letter variables damnit.

    – Bakuriu
    Jan 28 at 20:09







7




7





don't use 'l' as a variable name, It's easily confused with the number '1',

– regomodo
Jan 28 at 13:31





don't use 'l' as a variable name, It's easily confused with the number '1',

– regomodo
Jan 28 at 13:31













@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

– Heinzi
Jan 28 at 14:37






@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

– Heinzi
Jan 28 at 14:37





5




5





id rather people read "Clean Code" instead

– regomodo
Jan 28 at 14:54





id rather people read "Clean Code" instead

– regomodo
Jan 28 at 14:54




2




2





@Heinzi Or maybe stop using single letter variables damnit.

– Bakuriu
Jan 28 at 20:09





@Heinzi Or maybe stop using single letter variables damnit.

– Bakuriu
Jan 28 at 20:09












2 Answers
2






active

oldest

votes


















27














This code is not correct.



int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



array::array(int length) : llength 


instead would work because l is declared before t.



However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.






share|improve this answer






























    9














    The 2nd code snippet might have undefined behavior.



    The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



    Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



    BTW: With member initializer list the 1st code snippet chould be rewritten as



    array::array(int length) : l(length), t(new int[l]) 






    share|improve this answer




















    • 3





      Why do you say "might"? It does have UB, unconditionally.

      – Ruslan
      Jan 28 at 16:29






    • 1





      @Ruslan Because for static or thread-local objects l will get zero initialized at first.

      – songyuanyao
      Jan 29 at 1:13












    • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

      – Lightness Races in Orbit
      Jan 30 at 16:33










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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    27














    This code is not correct.



    int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



    array::array(int length) : llength 


    instead would work because l is declared before t.



    However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.






    share|improve this answer



























      27














      This code is not correct.



      int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



      array::array(int length) : llength 


      instead would work because l is declared before t.



      However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.






      share|improve this answer

























        27












        27








        27







        This code is not correct.



        int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



        array::array(int length) : llength 


        instead would work because l is declared before t.



        However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.






        share|improve this answer













        This code is not correct.



        int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



        array::array(int length) : llength 


        instead would work because l is declared before t.



        However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 28 at 10:29









        Baum mit AugenBaum mit Augen

        41.2k12118154




        41.2k12118154























            9














            The 2nd code snippet might have undefined behavior.



            The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



            Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



            BTW: With member initializer list the 1st code snippet chould be rewritten as



            array::array(int length) : l(length), t(new int[l]) 






            share|improve this answer




















            • 3





              Why do you say "might"? It does have UB, unconditionally.

              – Ruslan
              Jan 28 at 16:29






            • 1





              @Ruslan Because for static or thread-local objects l will get zero initialized at first.

              – songyuanyao
              Jan 29 at 1:13












            • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

              – Lightness Races in Orbit
              Jan 30 at 16:33















            9














            The 2nd code snippet might have undefined behavior.



            The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



            Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



            BTW: With member initializer list the 1st code snippet chould be rewritten as



            array::array(int length) : l(length), t(new int[l]) 






            share|improve this answer




















            • 3





              Why do you say "might"? It does have UB, unconditionally.

              – Ruslan
              Jan 28 at 16:29






            • 1





              @Ruslan Because for static or thread-local objects l will get zero initialized at first.

              – songyuanyao
              Jan 29 at 1:13












            • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

              – Lightness Races in Orbit
              Jan 30 at 16:33













            9












            9








            9







            The 2nd code snippet might have undefined behavior.



            The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



            Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



            BTW: With member initializer list the 1st code snippet chould be rewritten as



            array::array(int length) : l(length), t(new int[l]) 






            share|improve this answer















            The 2nd code snippet might have undefined behavior.



            The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



            Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



            BTW: With member initializer list the 1st code snippet chould be rewritten as



            array::array(int length) : l(length), t(new int[l]) 







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Jan 28 at 10:33

























            answered Jan 28 at 10:28









            songyuanyaosongyuanyao

            91.6k11175238




            91.6k11175238







            • 3





              Why do you say "might"? It does have UB, unconditionally.

              – Ruslan
              Jan 28 at 16:29






            • 1





              @Ruslan Because for static or thread-local objects l will get zero initialized at first.

              – songyuanyao
              Jan 29 at 1:13












            • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

              – Lightness Races in Orbit
              Jan 30 at 16:33












            • 3





              Why do you say "might"? It does have UB, unconditionally.

              – Ruslan
              Jan 28 at 16:29






            • 1





              @Ruslan Because for static or thread-local objects l will get zero initialized at first.

              – songyuanyao
              Jan 29 at 1:13












            • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

              – Lightness Races in Orbit
              Jan 30 at 16:33







            3




            3





            Why do you say "might"? It does have UB, unconditionally.

            – Ruslan
            Jan 28 at 16:29





            Why do you say "might"? It does have UB, unconditionally.

            – Ruslan
            Jan 28 at 16:29




            1




            1





            @Ruslan Because for static or thread-local objects l will get zero initialized at first.

            – songyuanyao
            Jan 29 at 1:13






            @Ruslan Because for static or thread-local objects l will get zero initialized at first.

            – songyuanyao
            Jan 29 at 1:13














            @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

            – Lightness Races in Orbit
            Jan 30 at 16:33





            @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

            – Lightness Races in Orbit
            Jan 30 at 16:33

















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