On 25 Jul 2009, at 03:56, Uli Kusterer wrote:
I don't think you can use sizeof() on a pointer passed as a
parameter, though. That only works on variables in the current
scope. That pointer could point at anything, so the compiler has no
idea whether it points at the stack or the heap. Even if it
generated specialized code for the stack case, what code would it
generate for the heap case? sizeof() is independent from malloc(),
NewPtr() and all the other allocator schemes the OS knows, sizeof()
wouldn't know which one to use.
sizeof() does exactly one thing, which is return the amount of storage
allocated *by the compiler* for its argument. Despite appearances, it
is not a function: it's a compiler directive, and compiles to a
constant value. For example:
char charArray[24];
sizeof(charArray) = 24
int intArray[16];
sizeof(intArray) = 64
char *charArrayPointer = charArray;
int *intArrayPointer = intArray;
sizeof(charArrayPointer) = sizeof(intArrayPointer) = 4 (8 on 64-bit
builds)
What's happened here? Taking the sizeof() a pointer doesn't tell you
anything about the size of the object it's pointing to - all it can
tell you is how big *the pointer itself* is.
You can dereference a pointer in a sizeof() expression, but only in a
limited fashion:
sizeof(*charArrayPointer) = 1
sizeof(*intArrayPointer) = 4
Confused? Keep in mind that *charArrayPointer is equivalent to
charArrayPointer[0] - it returns the first element, not the whole
original array.
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