const cast and std launder
Modifying a const constructed object after a const_cast is UB (I believe due to constant propagation). Is it still UB even when combined with std::launder (which AFAIK prevents some optimizations such as const propagation)?
#include <new>
#include <iostream>
struct C
{
int i;
};
int main(const int argc, const char * const * const argv)
{
const C c{1};
auto x = std::launder(const_cast<C*>(&c));
++x->i;
std::cout << x->i << std::endl;
std::cout << c.i << std::endl;
return 0;
}
c++ c++17
edited Nov 15 '18 at 10:42
Jarod42
118k12103188
asked Nov 15 '18 at 10:26
mkmostafamkmostafa
1,2431028
add a comment |
Modifying a const constructed object after a const_cast is UB (I believe due to constant propagation). Is it still UB even when combined with std::launder (which AFAIK prevents some optimizations such as const propagation)?
#include <new>
#include <iostream>
struct C
{
int i;
};
int main(const int argc, const char * const * const argv)
{
const C c{1};
auto x = std::launder(const_cast<C*>(&c));
++x->i;
std::cout << x->i << std::endl;
std::cout << c.i << std::endl;
return 0;
}
c++ c++17
edited Nov 15 '18 at 10:42
Jarod42
118k12103188
asked Nov 15 '18 at 10:26
mkmostafamkmostafa
1,2431028
2
Modifying const objects is even given as an example of undefined behavior in standard [intro.execution]/4: "Certain other operations are described in this International Standard as undefined (for example, the effect of attempting to modify a const object)." Why to try to find loopholes in it?
– Öö Tiib
Nov 15 '18 at 10:46
The purpose ofstd::launderis not to "prevent some optimizations". That's the wrong way to look at it. Its purpose is to communicate a specific thing to a compiler which in turn would allow it to do "the right thing".
– Dan M.
Nov 15 '18 at 11:00
I guess I have put it wrong, perhaps 'to prevent the compiler from making the wrong assumptions' would have been a better way to express it.
– mkmostafa
Nov 15 '18 at 11:06
add a comment |
Modifying a const constructed object after a const_cast is UB (I believe due to constant propagation). Is it still UB even when combined with std::launder (which AFAIK prevents some optimizations such as const propagation)?
#include <new>
#include <iostream>
struct C
{
int i;
};
int main(const int argc, const char * const * const argv)
{
const C c{1};
auto x = std::launder(const_cast<C*>(&c));
++x->i;
std::cout << x->i << std::endl;
std::cout << c.i << std::endl;
return 0;
}
c++ c++17
edited Nov 15 '18 at 10:42
Jarod42
118k12103188
asked Nov 15 '18 at 10:26
mkmostafamkmostafa
1,2431028
Modifying a const constructed object after a const_cast is UB (I believe due to constant propagation). Is it still UB even when combined with std::launder (which AFAIK prevents some optimizations such as const propagation)?
#include <new>
#include <iostream>
struct C
{
int i;
};
int main(const int argc, const char * const * const argv)
{
const C c{1};
auto x = std::launder(const_cast<C*>(&c));
++x->i;
std::cout << x->i << std::endl;
std::cout << c.i << std::endl;
return 0;
}
c++ c++17
c++ c++17
edited Nov 15 '18 at 10:42
Jarod42
118k12103188
asked Nov 15 '18 at 10:26
mkmostafamkmostafa
1,2431028
edited Nov 15 '18 at 10:42
Jarod42
118k12103188
asked Nov 15 '18 at 10:26
mkmostafamkmostafa
1,2431028
edited Nov 15 '18 at 10:42
Jarod42
118k12103188
edited Nov 15 '18 at 10:42
Jarod42
118k12103188
edited Nov 15 '18 at 10:42
Jarod42
118k12103188
118k12103188
asked Nov 15 '18 at 10:26
mkmostafamkmostafa
1,2431028
asked Nov 15 '18 at 10:26
mkmostafamkmostafa
1,2431028
asked Nov 15 '18 at 10:26
mkmostafamkmostafa
1,2431028
1,2431028
2
Modifying const objects is even given as an example of undefined behavior in standard [intro.execution]/4: "Certain other operations are described in this International Standard as undefined (for example, the effect of attempting to modify a const object)." Why to try to find loopholes in it?
– Öö Tiib
Nov 15 '18 at 10:46
The purpose ofstd::launderis not to "prevent some optimizations". That's the wrong way to look at it. Its purpose is to communicate a specific thing to a compiler which in turn would allow it to do "the right thing".
– Dan M.
Nov 15 '18 at 11:00
I guess I have put it wrong, perhaps 'to prevent the compiler from making the wrong assumptions' would have been a better way to express it.
– mkmostafa
Nov 15 '18 at 11:06
add a comment |
2
Modifying const objects is even given as an example of undefined behavior in standard [intro.execution]/4: "Certain other operations are described in this International Standard as undefined (for example, the effect of attempting to modify a const object)." Why to try to find loopholes in it?
– Öö Tiib
Nov 15 '18 at 10:46
The purpose ofstd::launderis not to "prevent some optimizations". That's the wrong way to look at it. Its purpose is to communicate a specific thing to a compiler which in turn would allow it to do "the right thing".
– Dan M.
Nov 15 '18 at 11:00
I guess I have put it wrong, perhaps 'to prevent the compiler from making the wrong assumptions' would have been a better way to express it.
– mkmostafa
Nov 15 '18 at 11:06
2
2
Modifying const objects is even given as an example of undefined behavior in standard [intro.execution]/4: "Certain other operations are described in this International Standard as undefined (for example, the effect of attempting to modify a const object)." Why to try to find loopholes in it?
– Öö Tiib
Nov 15 '18 at 10:46
Modifying const objects is even given as an example of undefined behavior in standard [intro.execution]/4: "Certain other operations are described in this International Standard as undefined (for example, the effect of attempting to modify a const object)." Why to try to find loopholes in it?
– Öö Tiib
Nov 15 '18 at 10:46
The purpose of
std::launder is not to "prevent some optimizations". That's the wrong way to look at it. Its purpose is to communicate a specific thing to a compiler which in turn would allow it to do "the right thing".– Dan M.
Nov 15 '18 at 11:00
The purpose of
std::launder is not to "prevent some optimizations". That's the wrong way to look at it. Its purpose is to communicate a specific thing to a compiler which in turn would allow it to do "the right thing".– Dan M.
Nov 15 '18 at 11:00
I guess I have put it wrong, perhaps 'to prevent the compiler from making the wrong assumptions' would have been a better way to express it.
– mkmostafa
Nov 15 '18 at 11:06
I guess I have put it wrong, perhaps 'to prevent the compiler from making the wrong assumptions' would have been a better way to express it.
– mkmostafa
Nov 15 '18 at 11:06
add a comment |
2 Answers
2
active
oldest
votes
const object - an object whose type is const-qualified, or a non-mutable subobject of a const object. Such object cannot be modified: attempt to do so directly is a compile-time error, and attempt to do so indirectly (e.g., by modifying the const object through a reference or pointer to non-const type) results in undefined behavior.
answered Nov 15 '18 at 10:37
P.WP.W
15.6k31453
add a comment |
Yes. Attempting to modify a const object is UB, period.
answered Nov 15 '18 at 10:30
T.C.T.C.
107k14220328
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
const object - an object whose type is const-qualified, or a non-mutable subobject of a const object. Such object cannot be modified: attempt to do so directly is a compile-time error, and attempt to do so indirectly (e.g., by modifying the const object through a reference or pointer to non-const type) results in undefined behavior.
answered Nov 15 '18 at 10:37
P.WP.W
15.6k31453
add a comment |
const object - an object whose type is const-qualified, or a non-mutable subobject of a const object. Such object cannot be modified: attempt to do so directly is a compile-time error, and attempt to do so indirectly (e.g., by modifying the const object through a reference or pointer to non-const type) results in undefined behavior.
answered Nov 15 '18 at 10:37
P.WP.W
15.6k31453
add a comment |
const object - an object whose type is const-qualified, or a non-mutable subobject of a const object. Such object cannot be modified: attempt to do so directly is a compile-time error, and attempt to do so indirectly (e.g., by modifying the const object through a reference or pointer to non-const type) results in undefined behavior.
answered Nov 15 '18 at 10:37
P.WP.W
15.6k31453
const object - an object whose type is const-qualified, or a non-mutable subobject of a const object. Such object cannot be modified: attempt to do so directly is a compile-time error, and attempt to do so indirectly (e.g., by modifying the const object through a reference or pointer to non-const type) results in undefined behavior.
answered Nov 15 '18 at 10:37
P.WP.W
15.6k31453
answered Nov 15 '18 at 10:37
P.WP.W
15.6k31453
answered Nov 15 '18 at 10:37
P.WP.W
15.6k31453
answered Nov 15 '18 at 10:37
P.WP.W
15.6k31453
15.6k31453
add a comment |
add a comment |
Yes. Attempting to modify a const object is UB, period.
answered Nov 15 '18 at 10:30
T.C.T.C.
107k14220328
add a comment |
Yes. Attempting to modify a const object is UB, period.
answered Nov 15 '18 at 10:30
T.C.T.C.
107k14220328
add a comment |
Yes. Attempting to modify a const object is UB, period.
answered Nov 15 '18 at 10:30
T.C.T.C.
107k14220328
Yes. Attempting to modify a const object is UB, period.
answered Nov 15 '18 at 10:30
T.C.T.C.
107k14220328
answered Nov 15 '18 at 10:30
T.C.T.C.
107k14220328
answered Nov 15 '18 at 10:30
T.C.T.C.
107k14220328
answered Nov 15 '18 at 10:30
T.C.T.C.
107k14220328
107k14220328
add a comment |
add a comment |
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2
Modifying const objects is even given as an example of undefined behavior in standard [intro.execution]/4: "Certain other operations are described in this International Standard as undefined (for example, the effect of attempting to modify a const object)." Why to try to find loopholes in it?
– Öö Tiib
Nov 15 '18 at 10:46
The purpose of
std::launderis not to "prevent some optimizations". That's the wrong way to look at it. Its purpose is to communicate a specific thing to a compiler which in turn would allow it to do "the right thing".– Dan M.
Nov 15 '18 at 11:00
I guess I have put it wrong, perhaps 'to prevent the compiler from making the wrong assumptions' would have been a better way to express it.
– mkmostafa
Nov 15 '18 at 11:06