Specifically, what’s dangerous about casting the result of malloc?

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You won’t get a compiler error, but a compiler warning. As the sources you cite say (especially the first one), you can get an unpredictable runtime error when using the cast without including stdlib.h.

So the error on your side is not the cast, but forgetting to include stdlib.h. Compilers may assume that malloc is a function returning int, therefore converting the void* pointer actually returned by malloc to int and then to your pointer type due to the explicit cast. On some platforms, int and pointers may take up different numbers of bytes, so the type conversions may lead to data corruption.

Fortunately, modern compilers give warnings that point to your actual error. See the gcc output you supplied: It warns you that the implicit declaration (int malloc(int)) is incompatible to the built-in malloc. So gcc seems to know malloc even without stdlib.h.

Leaving out the cast to prevent this error is mostly the same reasoning as writing

if (0 == my_var)

instead of

if (my_var == 0)

since the latter could lead to a serious bug if one would confuse = and ==, whereas the first one would lead to a compile error. I personally prefer the latter style since it better reflects my intention and I don’t tend to do this mistake.

The same is true for casting the value returned by malloc: I prefer being explicit in programming and I generally double-check to include the header files for all functions I use.

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