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
asked Nov 15 at 12:39
DevSolar
47.3k1294164
add a comment |
up vote
32
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
asked Nov 15 at 12:39
DevSolar
47.3k1294164
add a comment |
up vote
32
down vote
favorite
up vote
32
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
asked Nov 15 at 12:39
DevSolar
47.3k1294164
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 Nov 15 at 12:39
DevSolar
47.3k1294164
asked Nov 15 at 12:39
DevSolar
47.3k1294164
asked Nov 15 at 12:39
DevSolar
47.3k1294164
asked Nov 15 at 12:39
DevSolar
47.3k1294164
asked Nov 15 at 12:39
DevSolar
47.3k1294164
47.3k1294164
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
40
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.
edited Nov 15 at 15:41
MSalters
133k8115267
answered Nov 15 at 12:52
rafix07
6,0031613
3
What exactly does UB stand for?
– John
Nov 15 at 20:48
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
Nov 15 at 20:51
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
40
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.
edited Nov 15 at 15:41
MSalters
133k8115267
answered Nov 15 at 12:52
rafix07
6,0031613
3
What exactly does UB stand for?
– John
Nov 15 at 20:48
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
Nov 15 at 20:51
add a comment |
up vote
40
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.
edited Nov 15 at 15:41
MSalters
133k8115267
answered Nov 15 at 12:52
rafix07
6,0031613
3
What exactly does UB stand for?
– John
Nov 15 at 20:48
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
Nov 15 at 20:51
add a comment |
up vote
40
down vote
accepted
up vote
40
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.
edited Nov 15 at 15:41
MSalters
133k8115267
answered Nov 15 at 12:52
rafix07
6,0031613
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 Nov 15 at 15:41
MSalters
133k8115267
answered Nov 15 at 12:52
rafix07
6,0031613
edited Nov 15 at 15:41
MSalters
133k8115267
edited Nov 15 at 15:41
MSalters
133k8115267
edited Nov 15 at 15:41
MSalters
133k8115267
133k8115267
answered Nov 15 at 12:52
rafix07
6,0031613
answered Nov 15 at 12:52
rafix07
6,0031613
answered Nov 15 at 12:52
rafix07
6,0031613
6,0031613
3
What exactly does UB stand for?
– John
Nov 15 at 20:48
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
Nov 15 at 20:51
add a comment |
3
What exactly does UB stand for?
– John
Nov 15 at 20:48
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
Nov 15 at 20:51
3
3
What exactly does UB stand for?
– John
Nov 15 at 20:48
What exactly does UB stand for?
– John
Nov 15 at 20:48
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
Nov 15 at 20:51
see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
– Julien Rousé
Nov 15 at 20:51
add a comment |
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