Just a quick question I couldn't find a question on:
double* array = (double*) malloc(ARRAY_SIZE * sizeof(double));
or can I simply
double* array = malloc(ARRAY_SIZE * sizeof(double));
is there any difference?
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too honest for this site
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Natjo
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2you don't cast: http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc – ForeverStudent Dec 10 '15 at 21:44
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1Your question asks for C, so do not add C++ tag. C is not C++ and this is one of the differences between the languages. – too honest for this site Dec 10 '15 at 22:12
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array is a pointer and not an Array. Malloc's return type is VOID*, so no, there is no need to CAST malloc. – Michi Dec 10 '15 at 22:38
1 Answers
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In C, you shouldn't cast since it can hide certain subtle errors. In C++ (since your question originally had that tag despite no mention in the body), you have to cast.
The "subtle errors" of which I speak include strange behaviour when your integers and pointers are different sizes and you haven't included the header for malloc - in that case, you may find it assuming it returns an int (losing information in the process) and, while the compiler would normally warn about that, the explicit cast will probably prevent that.
paxdiablo
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Actually in C++ you should not use C programming style and not use `malloc` & friends. Btw. The rule can be generalised in C: do not cast to/from `void *` in C. The only exception are variadic arguments if the callee expects a `void *` like the `printf` family. – too honest for this site Dec 10 '15 at 22:16
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Since C99 the compiler must give a diagnostic, even if the cast is present – M.M Dec 10 '15 at 22:17
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@arduic, that's _one_ variant, I believe the Irish one is another. Of course the Scot one is more prevalent so I probably should have used that. – paxdiablo Dec 11 '15 at 03:13