Re: compiler assert for sizeof(char)==1
andrew_nuss@yahoo.com wrote:
Let's make it even simpler:
char pch[6] = { 0, 1, 2, 3, 4, 5 };
char cptr = pch;
Now pch is an array of chars, each of which is 1 byte in size, each of
those bytes containing 8 or more bits. The fact that each of those
chars contains 32 bits does not make them ints or longs, they are
still chars even if they happen to be the same size and have the same
number of bits as ints or longs.
If char has 32 bits, it is still has a size of one byte. If int has
32 bits on the same implementation, sizeof(int) is 1 on that
implementation.
And after:
cptr += 3;
cptr will point to the point to the fourth char in pch, which will be
a char with the value 3.
Yes, but what if I am doing my own heap management and creating blocks
and doing pointer arithmetic on true byte values. Simple example:
enum { header_size = 16 };
char* p = static_cast<char*>(malloc(48));
p += header_size; // then do other stuff and return p
The point is that I need to be confident that I can move the pointer
to the malloc'ed object exactly 16 bytes. Is this trouble on some
platforms? Is there a proper assert for platforms where that won't
work?
Did you read the other responses to your post? What do you mean by a
byte? Do you mean an octet? In which case on a system with a char that
has anything other than 8 bits the answer is that you cannot do this.
If you add n to p it will now point to the (n+1)th char. If header_size
is counted in octets you are out of luck on any system where a char has
more than 8-bits.
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