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SubjectRe: NULL. Again. (was Re: [PATCH] 2.4.22pre10: {,un}likely_p())
Sean Neakums wrote:
> I had thought that the need for writing NULL where a pointer is
> expected in varags functions was because the machine may have
> different sizes for pointers and int. In the case of the second
> printf call above, if pointers are 64-bit and integers are 32-bit,
> printf will read eight bytes from the stack, and only four will have
> been occupied by the integer 0.

It is incorrect to write NULL as a varargs argument. It won't
necessarily pass a pointer because:

1. a valid definition of NULL is "0".

If you want to pass a pointer to varargs, you might have considered
writing "(void *) 0", but that doesn't work because:

2. a machine may have different sizes for different pointer types.

You must write "(type) 0" or "(type) NULL", using the correct pointer type.

-- Jamie
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