CppCon 2018, Nicolai Josuttis: Why are these interpreted as iterators?
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Nicolai Josuttis' "The Nightmare of Initialization in C++" presentation at CppCon 2018 had, at one point, the following piece of code:
std::vector< std::string > v07 = "1", "2" ;
Nicolai said the following (transcription mine):
The problem is, what happens here is, we interpret these two parameters as iterators. So these are iterators, so this is the beginning of the range, and this is the end of the range, and they should refer to the same range of characters; because characters convert implicitly to strings this will compile. If you're lucky, you'll get a coredump. If not, you've got a big problem.
He lost me there. Can somebody explain what is going on here, exactly, step by step?
c++ initialization c++17
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up vote
24
down vote
favorite
Nicolai Josuttis' "The Nightmare of Initialization in C++" presentation at CppCon 2018 had, at one point, the following piece of code:
std::vector< std::string > v07 = "1", "2" ;
Nicolai said the following (transcription mine):
The problem is, what happens here is, we interpret these two parameters as iterators. So these are iterators, so this is the beginning of the range, and this is the end of the range, and they should refer to the same range of characters; because characters convert implicitly to strings this will compile. If you're lucky, you'll get a coredump. If not, you've got a big problem.
He lost me there. Can somebody explain what is going on here, exactly, step by step?
c++ initialization c++17
add a comment |
up vote
24
down vote
favorite
up vote
24
down vote
favorite
Nicolai Josuttis' "The Nightmare of Initialization in C++" presentation at CppCon 2018 had, at one point, the following piece of code:
std::vector< std::string > v07 = "1", "2" ;
Nicolai said the following (transcription mine):
The problem is, what happens here is, we interpret these two parameters as iterators. So these are iterators, so this is the beginning of the range, and this is the end of the range, and they should refer to the same range of characters; because characters convert implicitly to strings this will compile. If you're lucky, you'll get a coredump. If not, you've got a big problem.
He lost me there. Can somebody explain what is going on here, exactly, step by step?
c++ initialization c++17
Nicolai Josuttis' "The Nightmare of Initialization in C++" presentation at CppCon 2018 had, at one point, the following piece of code:
std::vector< std::string > v07 = "1", "2" ;
Nicolai said the following (transcription mine):
The problem is, what happens here is, we interpret these two parameters as iterators. So these are iterators, so this is the beginning of the range, and this is the end of the range, and they should refer to the same range of characters; because characters convert implicitly to strings this will compile. If you're lucky, you'll get a coredump. If not, you've got a big problem.
He lost me there. Can somebody explain what is going on here, exactly, step by step?
c++ initialization c++17
c++ initialization c++17
asked 2 days ago
DevSolar
47.1k1293164
47.1k1293164
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1 Answer
1
active
oldest
votes
up vote
30
down vote
accepted
Below code
std::vector< std::string > v07 = "1", "2" ;
is equivalent to
std::string s = "1","2"; // call string(const char*, const char*)
std::vector<std::string> v07 = s; // initializer list with one item
the issue is with
s="1","2";
This calls string(const char* start, const char* end)
constructor,
but start
and end
must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.
1
What exactly does UB stand for?
– John
2 days ago
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
30
down vote
accepted
Below code
std::vector< std::string > v07 = "1", "2" ;
is equivalent to
std::string s = "1","2"; // call string(const char*, const char*)
std::vector<std::string> v07 = s; // initializer list with one item
the issue is with
s="1","2";
This calls string(const char* start, const char* end)
constructor,
but start
and end
must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.
1
What exactly does UB stand for?
– John
2 days ago
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
2 days ago
add a comment |
up vote
30
down vote
accepted
Below code
std::vector< std::string > v07 = "1", "2" ;
is equivalent to
std::string s = "1","2"; // call string(const char*, const char*)
std::vector<std::string> v07 = s; // initializer list with one item
the issue is with
s="1","2";
This calls string(const char* start, const char* end)
constructor,
but start
and end
must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.
1
What exactly does UB stand for?
– John
2 days ago
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
2 days ago
add a comment |
up vote
30
down vote
accepted
up vote
30
down vote
accepted
Below code
std::vector< std::string > v07 = "1", "2" ;
is equivalent to
std::string s = "1","2"; // call string(const char*, const char*)
std::vector<std::string> v07 = s; // initializer list with one item
the issue is with
s="1","2";
This calls string(const char* start, const char* end)
constructor,
but start
and end
must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.
Below code
std::vector< std::string > v07 = "1", "2" ;
is equivalent to
std::string s = "1","2"; // call string(const char*, const char*)
std::vector<std::string> v07 = s; // initializer list with one item
the issue is with
s="1","2";
This calls string(const char* start, const char* end)
constructor,
but start
and end
must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.
edited 2 days ago
MSalters
132k8115267
132k8115267
answered 2 days ago
rafix07
5,7731513
5,7731513
1
What exactly does UB stand for?
– John
2 days ago
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
2 days ago
add a comment |
1
What exactly does UB stand for?
– John
2 days ago
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
2 days ago
1
1
What exactly does UB stand for?
– John
2 days ago
What exactly does UB stand for?
– John
2 days ago
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
2 days ago
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
2 days ago
add a comment |
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