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1

Recently I came across a funny "feature".
The code below compiles equally on both g++ and Visual Studio 2017.

#include <iostream>
#include <list>

int main()

 std::list<int *> l;
 int a = 1, b = 2;
 l.emplace_back(&a);
 auto p = l.front();
 std::cout << p << 'n'; // prints x
 l.erase(l.begin());
 l.emplace_back(&b);
 std::cout << p << 'n'; // prints x
 std::cin.get();

However, if you change line

auto p = l.front();

to

auto & p = l.front();

Visual Studio still gives the same output (considering the address x may change, of course). However, now g++ gives me the output

x
x+4

Obviously, when passing the pointer by reference, g++ recognizes that the first element of the list now has a different value, which is a different address of the stack (offset + 4 compared to the initial one), while Visual Studio 2017 doesn't. So... who's broken?

c++ reference visual-studio-2017 g++ undefined-behavior

share|improve this question

edited Mar 23 at 14:43

songyuanyao

95.6k11187256

asked Mar 23 at 14:20

potatosaladpotatosalad

134

add a comment | 

1

Recently I came across a funny "feature".
The code below compiles equally on both g++ and Visual Studio 2017.

#include <iostream>
#include <list>

int main()

 std::list<int *> l;
 int a = 1, b = 2;
 l.emplace_back(&a);
 auto p = l.front();
 std::cout << p << 'n'; // prints x
 l.erase(l.begin());
 l.emplace_back(&b);
 std::cout << p << 'n'; // prints x
 std::cin.get();

However, if you change line

auto p = l.front();

to

auto & p = l.front();

Visual Studio still gives the same output (considering the address x may change, of course). However, now g++ gives me the output

x
x+4

Obviously, when passing the pointer by reference, g++ recognizes that the first element of the list now has a different value, which is a different address of the stack (offset + 4 compared to the initial one), while Visual Studio 2017 doesn't. So... who's broken?

c++ reference visual-studio-2017 g++ undefined-behavior

share|improve this question

edited Mar 23 at 14:43

songyuanyao

95.6k11187256

asked Mar 23 at 14:20

potatosaladpotatosalad

134

add a comment | 

Recently I came across a funny "feature".
The code below compiles equally on both g++ and Visual Studio 2017.

#include <iostream>
#include <list>

int main()

 std::list<int *> l;
 int a = 1, b = 2;
 l.emplace_back(&a);
 auto p = l.front();
 std::cout << p << 'n'; // prints x
 l.erase(l.begin());
 l.emplace_back(&b);
 std::cout << p << 'n'; // prints x
 std::cin.get();

However, if you change line

auto p = l.front();

to

auto & p = l.front();

Visual Studio still gives the same output (considering the address x may change, of course). However, now g++ gives me the output

x
x+4

Obviously, when passing the pointer by reference, g++ recognizes that the first element of the list now has a different value, which is a different address of the stack (offset + 4 compared to the initial one), while Visual Studio 2017 doesn't. So... who's broken?

c++ reference visual-studio-2017 g++ undefined-behavior

share|improve this question

edited Mar 23 at 14:43

songyuanyao

95.6k11187256

asked Mar 23 at 14:20

potatosaladpotatosalad

134

#include <iostream>
#include <list>

int main()

 std::list<int *> l;
 int a = 1, b = 2;
 l.emplace_back(&a);
 auto p = l.front();
 std::cout << p << 'n'; // prints x
 l.erase(l.begin());
 l.emplace_back(&b);
 std::cout << p << 'n'; // prints x
 std::cin.get();

auto p = l.front();

auto & p = l.front();

x
x+4

share|improve this question

edited Mar 23 at 14:43

songyuanyao

95.6k11187256

asked Mar 23 at 14:20

potatosaladpotatosalad

134

share|improve this question

edited Mar 23 at 14:43

songyuanyao

95.6k11187256

asked Mar 23 at 14:20

potatosaladpotatosalad

134

edited Mar 23 at 14:43

songyuanyao

95.6k11187256

edited Mar 23 at 14:43

songyuanyao

95.6k11187256

asked Mar 23 at 14:20

potatosaladpotatosalad

134

asked Mar 23 at 14:20

potatosaladpotatosalad

134

2 Answers
2

active

oldest

votes

5

who's broken?

Both are correct, because your code has undefined behavior.

After auto & p = l.front();, you erased the element from the list, then p becomes dangled; any dereference on it leads to UB, means anything is possible.

References and iterators to the erased elements are invalidated.

share|improve this answer

edited Mar 23 at 14:31

answered Mar 23 at 14:26

songyuanyaosongyuanyao

95.6k11187256

add a comment | 

4

After l.erase(l.begin()); reference to first item previously obtained at auto & p = l.front(); becomes invalid and accessing value stored in p leads to Undefined Behavior. So it is your code that is broken.

share|improve this answer

answered Mar 23 at 14:25

VTTVTT

27k42651

add a comment | 

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5

who's broken?

Both are correct, because your code has undefined behavior.

After auto & p = l.front();, you erased the element from the list, then p becomes dangled; any dereference on it leads to UB, means anything is possible.

References and iterators to the erased elements are invalidated.

share|improve this answer

edited Mar 23 at 14:31

answered Mar 23 at 14:26

songyuanyaosongyuanyao

95.6k11187256

add a comment | 

5

who's broken?

Both are correct, because your code has undefined behavior.

After auto & p = l.front();, you erased the element from the list, then p becomes dangled; any dereference on it leads to UB, means anything is possible.

References and iterators to the erased elements are invalidated.

share|improve this answer

edited Mar 23 at 14:31

answered Mar 23 at 14:26

songyuanyaosongyuanyao

95.6k11187256

add a comment | 

who's broken?

Both are correct, because your code has undefined behavior.

After auto & p = l.front();, you erased the element from the list, then p becomes dangled; any dereference on it leads to UB, means anything is possible.

References and iterators to the erased elements are invalidated.

share|improve this answer

edited Mar 23 at 14:31

answered Mar 23 at 14:26

songyuanyaosongyuanyao

95.6k11187256

share|improve this answer

edited Mar 23 at 14:31

answered Mar 23 at 14:26

songyuanyaosongyuanyao

95.6k11187256

answered Mar 23 at 14:26

songyuanyaosongyuanyao

95.6k11187256

answered Mar 23 at 14:26

songyuanyaosongyuanyao

95.6k11187256

4

After l.erase(l.begin()); reference to first item previously obtained at auto & p = l.front(); becomes invalid and accessing value stored in p leads to Undefined Behavior. So it is your code that is broken.

share|improve this answer

answered Mar 23 at 14:25

VTTVTT

27k42651

add a comment | 

4

After l.erase(l.begin()); reference to first item previously obtained at auto & p = l.front(); becomes invalid and accessing value stored in p leads to Undefined Behavior. So it is your code that is broken.

share|improve this answer

answered Mar 23 at 14:25

VTTVTT

27k42651

add a comment | 

After l.erase(l.begin()); reference to first item previously obtained at auto & p = l.front(); becomes invalid and accessing value stored in p leads to Undefined Behavior. So it is your code that is broken.

share|improve this answer

answered Mar 23 at 14:25

VTTVTT

27k42651

share|improve this answer

answered Mar 23 at 14:25

VTTVTT

27k42651

answered Mar 23 at 14:25

VTTVTT

27k42651

answered Mar 23 at 14:25

VTTVTT

27k42651