integer pointer conversion

[phenixa]

hi,
is there any way to convert unsigned char to int* on the fly?
like this

unsigned char c = 0x05;

int *a = &(int) c;

this code give me "Must take address of a memory location" error
any ideas?

[GNiewerth]

unsigned char c = 0x05;

int *a = reinterpret_cast<int*>( &c );

[_Superman_]

I'm sure that will produce unexpected results because when you access the value in the pointer a it may not be 0x05.

If you expect the value to be 5 you should use an intermediate int variable.

unsigned char c = 0x05;

int a = c;
int* b = &a;

[Speedo]

Do you want the pointer to point to "c", or to the address 0x05?

GNiewerth showed you how to do the former, but I don't see it being very useful. Trying to dereference the int pointer will give you trash, because it's reading 3 bytes in addition to "c". If you try to assign to it, you're going to corrupt the stack.

To do the latter, just do

int* a = reinterpret_cast<int*>(0x05);

But that's also something you shouldn't really mess with unless you know what you're doing...

[ninja9578]

I'm not sure what you're trying to accomplish here. You can't ever store a pointer in a char. A char is 8 bytes, while a pointer is (mostly) 64 bytes, one Windows they may be 32 bytes depending on the version.

[Lindley]

The size of a pointer depends only on whether you're on a 64- or 32-bit OS. There are other types which vary between Windows/Linux/OSX, but not pointers.

[laserlight]

A char is 8 bytes, while a pointer is (mostly) 64 bytes, one Windows they may be 32 bytes depending on the version.

I think you mean bits rather than bytes, but of course the actual values are implementation defined, within the required limits. (Consequently, if the sizeof a pointer is 1, I suppose you could store a pointer in a char without loss of information.)

[Amleto]

hi,
is there any way to convert unsigned char to int* on the fly?
like this

unsigned char c = 0x05;

int *a = &(int) c;

this code give me "Must take address of a memory location" error
any ideas?

types dont have addressess so you can't take the address of 'int' etc.

[laserlight]

types dont have addressess so you can't take the address of 'int' etc.

Ah, but that does not attempt to take the address of the int type. That attempts to take the address of a temporary that is the result of a cast from unsigned char to int.

[ninja9578]

The size of a pointer depends only on whether you're on a 64- or 32-bit OS. There are other types which vary between Windows/Linux/OSX, but not pointers.

True. I assumed he was using a common OS (OSX, Windows, Desktop Linux, Solaris...) out of them, the only one that is 32 bit is Windows.

I think you mean bits rather than bytes, but of course the actual values are implementation defined, within the required limits. (Consequently, if the sizeof a pointer is 1, I suppose you could store a pointer in a char without loss of information.)

Oops, yup, 64bit, not byte :P

[Lindley]

There are 64-bit versions of Windows available.

[Amleto]

Ah, but that does not attempt to take the address of the int type. That attempts to take the address of a temporary that is the result of a cast from unsigned char to int.

why does that temporary not have an address then? is it because it is an rvalue (if that is the case)? would assigning it to an lvalue first resolve the problem in this case?

[laserlight]

is it because it is an rvalue (if that is the case)?

Yes.

would assigning it to an lvalue first resolve the problem in this case?

Yes, and that is pretty much _Superman_'s example in post #3.

[Amleto]

thanks