integer pointer conversion

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?

Code:

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unsigned char c = 0x05;

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

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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.

Code:

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unsigned char c = 0x05;

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

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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

Code:

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

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

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.

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.

Quote:

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Originally Posted by ninja9578

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

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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.)

Quote:

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Originally Posted by 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?

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

Quote:

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Originally Posted by Amleto

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

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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.

Quote:

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Originally Posted by 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.

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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.

Quote:

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Originally Posted by laserlight

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.)

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Oops, yup, 64bit, not byte :P

There are 64-bit versions of Windows available.

Quote:

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Originally Posted by laserlight

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.

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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?

Quote:

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Originally Posted by Amleto

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

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Yes.

Quote:

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Originally Posted by Amleto

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

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

thanks :)