Constant value changing

up vote
7
down vote

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Suppose I have a struct definition:

struct thing

 thing* x;
 int z;

 thing() : x(this), z(0) 
 void foo() const
 
 this->x->z++;
 
;

Note that I create a mutable pointer to myself (evil laugh)

And then I can use this later like this:

int main()

 const thing c;
 c.foo();
 assert(c.z == 1);
 c.foo();
 assert(c.z == 2);
 return c.z;

And as you can see it seems that I can change a constant value......is this UB?

asked 5 hours ago

DarthRubik

1,625934

1

Yes, it is UB, as c is const.
â€“Â geza
4 hours ago

add a commentÂ |Â

up vote
7
down vote

favorite

Suppose I have a struct definition:

struct thing

 thing* x;
 int z;

 thing() : x(this), z(0) 
 void foo() const
 
 this->x->z++;
 
;

Note that I create a mutable pointer to myself (evil laugh)

And then I can use this later like this:

int main()

 const thing c;
 c.foo();
 assert(c.z == 1);
 c.foo();
 assert(c.z == 2);
 return c.z;

And as you can see it seems that I can change a constant value......is this UB?

asked 5 hours ago

DarthRubik

1,625934

1

Yes, it is UB, as c is const.
â€“Â geza
4 hours ago

add a commentÂ |Â

up vote
7
down vote

favorite

Suppose I have a struct definition:

struct thing

 thing* x;
 int z;

 thing() : x(this), z(0) 
 void foo() const
 
 this->x->z++;
 
;

Note that I create a mutable pointer to myself (evil laugh)

And then I can use this later like this:

int main()

 const thing c;
 c.foo();
 assert(c.z == 1);
 c.foo();
 assert(c.z == 2);
 return c.z;

And as you can see it seems that I can change a constant value......is this UB?

asked 5 hours ago

DarthRubik

1,625934

Suppose I have a struct definition:

struct thing

 thing* x;
 int z;

 thing() : x(this), z(0) 
 void foo() const
 
 this->x->z++;
 
;

Note that I create a mutable pointer to myself (evil laugh)

And then I can use this later like this:

int main()

 const thing c;
 c.foo();
 assert(c.z == 1);
 c.foo();
 assert(c.z == 2);
 return c.z;

And as you can see it seems that I can change a constant value......is this UB?

c++ const undefined-behavior

asked 5 hours ago

DarthRubik

1,625934

asked 5 hours ago

DarthRubik

1,625934

asked 5 hours ago

DarthRubik

1,625934

asked 5 hours ago

DarthRubik

1,625934

asked 5 hours ago

DarthRubik

1,625934

1

Yes, it is UB, as c is const.
â€“Â geza
4 hours ago

add a commentÂ |Â

1

Yes, it is UB, as c is const.
â€“Â geza
4 hours ago

Yes, it is UB, as c is const.
â€“Â geza
4 hours ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
7
down vote

accepted

[dcl.type.cv]p4:

Except that any class member declared mutable ([dcl.stc]) can be
modified, any attempt to modify ([expr.ass], [expr.post.incr],
[expr.pre.incr]) a const object ([basic.type.qualifier]) during its
lifetime ([basic.life]) results in undefined behavior.

[basic.type.qualifier]p1:

A const object is an object of type const T or a non-mutable subobject of such an object.

c.z is a const object, because it is a non-mutable subobject of c. Your code attempts to modify it during its lifetime. It follows that the code has undefined behavior.

answered 4 hours ago

T.C.

102k13206310

add a commentÂ |Â

up vote
0
down vote

The foo-function itself would be OK, as const member functions like T::foo() const just indicate that this is of type const *T; The fact that a (non-const) member points to the same object is irrelevant then.

The object c in first place, however, is declared as const. So it is UB to alter the contents of c through whatever code, including the (per se correct) member function foo.

edited 4 hours ago

answered 5 hours ago

Stephan Lechner

22.6k21637

2

What about the fact that c is const?
â€“Â NathanOliver
5 hours ago

@NathanOliver: yes; you may not change the value of c->x, but you may change the object referenced by c->x; so c->x->z++ is valid even if c is const.
â€“Â Stephan Lechner
5 hours ago

But since c->x->z++ is c.z++ isn't that UB?
â€“Â NathanOliver
5 hours ago

1

@NathanOliver: No, I'm quite sure that it is not UB. const does not mean that the object is immutable in general; it just declares that it must not be changed through the (const) pointer at hand. So you cannot change z through a (const*) this->z++, but through another path to that object, i.e. through this->x->z++
â€“Â Stephan Lechner
5 hours ago

1

Your example is only valid because value was not declared const in the first place.. but his thing instance is declared const.. and if you attempted to const_cast<thing*>(this)->z++; it'd be undefined behaviour.. So why would he be able to modify it through a shallow copy of this.. *Note:( I did not downvote.
â€“Â Brandon
5 hours ago

Â |Â
show 2 more comments

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
7
down vote

accepted

[dcl.type.cv]p4:

Except that any class member declared mutable ([dcl.stc]) can be
modified, any attempt to modify ([expr.ass], [expr.post.incr],
[expr.pre.incr]) a const object ([basic.type.qualifier]) during its
lifetime ([basic.life]) results in undefined behavior.

[basic.type.qualifier]p1:

A const object is an object of type const T or a non-mutable subobject of such an object.

c.z is a const object, because it is a non-mutable subobject of c. Your code attempts to modify it during its lifetime. It follows that the code has undefined behavior.

answered 4 hours ago

T.C.

102k13206310

add a commentÂ |Â

up vote
7
down vote

accepted

[dcl.type.cv]p4:

Except that any class member declared mutable ([dcl.stc]) can be
modified, any attempt to modify ([expr.ass], [expr.post.incr],
[expr.pre.incr]) a const object ([basic.type.qualifier]) during its
lifetime ([basic.life]) results in undefined behavior.

[basic.type.qualifier]p1:

A const object is an object of type const T or a non-mutable subobject of such an object.

c.z is a const object, because it is a non-mutable subobject of c. Your code attempts to modify it during its lifetime. It follows that the code has undefined behavior.

answered 4 hours ago

T.C.

102k13206310

add a commentÂ |Â

up vote
7
down vote

accepted

[dcl.type.cv]p4:

Except that any class member declared mutable ([dcl.stc]) can be
modified, any attempt to modify ([expr.ass], [expr.post.incr],
[expr.pre.incr]) a const object ([basic.type.qualifier]) during its
lifetime ([basic.life]) results in undefined behavior.

[basic.type.qualifier]p1:

A const object is an object of type const T or a non-mutable subobject of such an object.

c.z is a const object, because it is a non-mutable subobject of c. Your code attempts to modify it during its lifetime. It follows that the code has undefined behavior.

answered 4 hours ago

T.C.

102k13206310

[dcl.type.cv]p4:

Except that any class member declared mutable ([dcl.stc]) can be
modified, any attempt to modify ([expr.ass], [expr.post.incr],
[expr.pre.incr]) a const object ([basic.type.qualifier]) during its
lifetime ([basic.life]) results in undefined behavior.

[basic.type.qualifier]p1:

A const object is an object of type const T or a non-mutable subobject of such an object.

c.z is a const object, because it is a non-mutable subobject of c. Your code attempts to modify it during its lifetime. It follows that the code has undefined behavior.

answered 4 hours ago

T.C.

102k13206310

answered 4 hours ago

T.C.

102k13206310

answered 4 hours ago

T.C.

102k13206310

answered 4 hours ago

T.C.

102k13206310

add a commentÂ |Â

up vote
0
down vote

The object c in first place, however, is declared as const. So it is UB to alter the contents of c through whatever code, including the (per se correct) member function foo.

edited 4 hours ago

answered 5 hours ago

Stephan Lechner

22.6k21637

2

What about the fact that c is const?
â€“Â NathanOliver
5 hours ago

@NathanOliver: yes; you may not change the value of c->x, but you may change the object referenced by c->x; so c->x->z++ is valid even if c is const.
â€“Â Stephan Lechner
5 hours ago

But since c->x->z++ is c.z++ isn't that UB?
â€“Â NathanOliver
5 hours ago

1

@NathanOliver: No, I'm quite sure that it is not UB. const does not mean that the object is immutable in general; it just declares that it must not be changed through the (const) pointer at hand. So you cannot change z through a (const*) this->z++, but through another path to that object, i.e. through this->x->z++
â€“Â Stephan Lechner
5 hours ago

1

Your example is only valid because value was not declared const in the first place.. but his thing instance is declared const.. and if you attempted to const_cast<thing*>(this)->z++; it'd be undefined behaviour.. So why would he be able to modify it through a shallow copy of this.. *Note:( I did not downvote.
â€“Â Brandon
5 hours ago

Â |Â
show 2 more comments

up vote
0
down vote

The object c in first place, however, is declared as const. So it is UB to alter the contents of c through whatever code, including the (per se correct) member function foo.

edited 4 hours ago

answered 5 hours ago

Stephan Lechner

22.6k21637

2

What about the fact that c is const?
â€“Â NathanOliver
5 hours ago

@NathanOliver: yes; you may not change the value of c->x, but you may change the object referenced by c->x; so c->x->z++ is valid even if c is const.
â€“Â Stephan Lechner
5 hours ago

But since c->x->z++ is c.z++ isn't that UB?
â€“Â NathanOliver
5 hours ago

1

@NathanOliver: No, I'm quite sure that it is not UB. const does not mean that the object is immutable in general; it just declares that it must not be changed through the (const) pointer at hand. So you cannot change z through a (const*) this->z++, but through another path to that object, i.e. through this->x->z++
â€“Â Stephan Lechner
5 hours ago

1

Your example is only valid because value was not declared const in the first place.. but his thing instance is declared const.. and if you attempted to const_cast<thing*>(this)->z++; it'd be undefined behaviour.. So why would he be able to modify it through a shallow copy of this.. *Note:( I did not downvote.
â€“Â Brandon
5 hours ago

Â |Â
show 2 more comments

up vote
0
down vote

The object c in first place, however, is declared as const. So it is UB to alter the contents of c through whatever code, including the (per se correct) member function foo.

edited 4 hours ago

answered 5 hours ago

Stephan Lechner

22.6k21637

The object c in first place, however, is declared as const. So it is UB to alter the contents of c through whatever code, including the (per se correct) member function foo.

edited 4 hours ago

answered 5 hours ago

Stephan Lechner

22.6k21637

edited 4 hours ago

answered 5 hours ago

Stephan Lechner

22.6k21637

answered 5 hours ago

Stephan Lechner

22.6k21637

answered 5 hours ago

Stephan Lechner

22.6k21637

2

What about the fact that c is const?
â€“Â NathanOliver
5 hours ago

@NathanOliver: yes; you may not change the value of c->x, but you may change the object referenced by c->x; so c->x->z++ is valid even if c is const.
â€“Â Stephan Lechner
5 hours ago

But since c->x->z++ is c.z++ isn't that UB?
â€“Â NathanOliver
5 hours ago

1

@NathanOliver: No, I'm quite sure that it is not UB. const does not mean that the object is immutable in general; it just declares that it must not be changed through the (const) pointer at hand. So you cannot change z through a (const*) this->z++, but through another path to that object, i.e. through this->x->z++
â€“Â Stephan Lechner
5 hours ago

1

Your example is only valid because value was not declared const in the first place.. but his thing instance is declared const.. and if you attempted to const_cast<thing*>(this)->z++; it'd be undefined behaviour.. So why would he be able to modify it through a shallow copy of this.. *Note:( I did not downvote.
â€“Â Brandon
5 hours ago

Â |Â
show 2 more comments

2

What about the fact that c is const?
â€“Â NathanOliver
5 hours ago

@NathanOliver: yes; you may not change the value of c->x, but you may change the object referenced by c->x; so c->x->z++ is valid even if c is const.
â€“Â Stephan Lechner
5 hours ago

But since c->x->z++ is c.z++ isn't that UB?
â€“Â NathanOliver
5 hours ago

1

@NathanOliver: No, I'm quite sure that it is not UB. const does not mean that the object is immutable in general; it just declares that it must not be changed through the (const) pointer at hand. So you cannot change z through a (const*) this->z++, but through another path to that object, i.e. through this->x->z++
â€“Â Stephan Lechner
5 hours ago

1

Your example is only valid because value was not declared const in the first place.. but his thing instance is declared const.. and if you attempted to const_cast<thing*>(this)->z++; it'd be undefined behaviour.. So why would he be able to modify it through a shallow copy of this.. *Note:( I did not downvote.
â€“Â Brandon
5 hours ago

What about the fact that c is const?
â€“Â NathanOliver
5 hours ago

@NathanOliver: yes; you may not change the value of c->x, but you may change the object referenced by c->x; so c->x->z++ is valid even if c is const.
â€“Â Stephan Lechner
5 hours ago

But since c->x->z++ is c.z++ isn't that UB?
â€“Â NathanOliver
5 hours ago

@NathanOliver: No, I'm quite sure that it is not UB. const does not mean that the object is immutable in general; it just declares that it must not be changed through the (const) pointer at hand. So you cannot change z through a (const*) this->z++, but through another path to that object, i.e. through this->x->z++
â€“Â Stephan Lechner
5 hours ago

Your example is only valid because value was not declared const in the first place.. but his thing instance is declared const.. and if you attempted to const_cast<thing*>(this)->z++; it'd be undefined behaviour.. So why would he be able to modify it through a shallow copy of this.. *Note:( I did not downvote.
â€“Â Brandon
5 hours ago

Â |Â
show 2 more comments

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