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[QUOTE="Peter Pichler, post: 1694788"]
Yes, but... malloc() returns an address properly aligned for any type.
In other words, even if you cast void* to any wider type*, thus losing
the lowest bits of the address, it's no real loss because those bits
must have been zeros anyway. You have to cast again b efore assigning
to a pointer, otherwise it would not compile at all.

NB: I understand that we are talking about cases like

double *p = (int*)malloc(sizeof *p);

This would not compile, unless your compiler is seriously broken.
To make it compile, you would have to cast again, e.g.

double *p = (double*)(int*)malloc(sizeof *p);

I cannot see any possible problems with that, other than it is UGLY,
provided that stdlib.h is #included.

The problems that Ben encountered may have been something like

double *p = (double*)malloc(sizeof(int));

Because *p used to be int*, but changed to double*, and the programmer
failed to notice that (s)he had to change the parameter as well. This
may even work by accident, if the implementation allocates memory in
segments, making such bugs even more difficult to spot.

Peter Pichler
[/QUOTE]

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