Pointer to const

[LiYS]

void u(const int* cip)
{
//! *cip = 2; // Illegal -- modifies value
int i = *cip; // OK -- copies value
//! int* ip2 = cip; // Illegal: non-const
}

In the snippet above, the argument "const int* cip" is a pointer to const that is to say the thing this pointer pointed to can't be changed but the pointer can be changed right? But I don't understand why the line "int* ip2 = cip;" can't passed the compilation. Is it because after its execution the pointee can be changed through ip2?

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...always look on the bright side of life... (Whistle)

[Bob Stanneveld]

The reason why the compiler complains is that you want to do an assignment where the original value loses qualifiers. If I remember correctly, you can do the following:

void u(const int* cip)
{
int** ppi = &cip;
(*(*ppi)) = 6;
// cip points now to 6
(*(*ppi)) = 8;
// cip points now to 8!
}

This is one of the many reasons why pointer magic should be avoided as much as possible. Behind every great black man...             ... is the police. - Conspiracy brother Blog[^]

[Nish Nishant]

Try :- int* ip2 = const_cast<int*>(cip);

[MaxVampire82]

ip2 isn't a pointer to const.

[LiYS]

Doing "int* ip2 = cip" illegal only because the pointee of pointer cip could be changed through pointer ip2? In that case the constness of pointer to const can't be keep?

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...always look on the bright side of life... (Whistle)

[Bob Stanneveld]

YongSheng Li wrote: Doing "int* ip2 = cip" illegal only because the pointee of pointer cip could be changed through pointer ip2? In that case the constness of pointer to const can't be keep? Thats exactly what I'm saying. Doing that, the original type loses qualifiers, which is illegal.. Behind every great black man...             ... is the police. - Conspiracy brother Blog[^]

[LiYS]

That's exactly what I'm expecting!Thank you!:-D

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...always look on the bright side of life... (Whistle)