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comp.lang.c FAQ list · Question 7.19b

Q: I'm dynamically allocating an array, like this:

	int *iarray = (int *)malloc(nints);
malloc isn't returning NULL, but the code isn't working.

A: malloc is a low-level, typeless allocator. It doesn't know how you're going to use the memory; all it does is to allocate as many bytes of memory as you ask it. Therefore (except when you're allocating arrays of char) you must multiply by the size of the elements in the array you're allocating:

	int *iarray = malloc(nints * sizeof(int));
	int *iarray = malloc(nints * sizeof(*iarray));
(The latter fragment can be more reliable if the type of iarray might change, since there's only one place to change it. Also, the casts have been removed; see question 7.7.) Compare question 4.4, and see also question 7.8.

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