Why does the compiler complain when I do not cast the result of malloc?

问题

I am inspecting code that does not require explicitly casting result of malloc call but whenever I attempt to do this, the compiler throws an error.

i.e.

char *somevar;
somevar = malloc(sizeof(char) * n); //error
somevar = (char *)malloc(sizeof(char) * n); // ok

回答1:

This happens if you use C++ compiler instead of C compiler. As C++ requires explicit casting. The problem is not just with (un)casting malloc result, but any void pointer to other pointer.

回答2:

Did you remember to include the function prototype? For malloc(3), this is:

#include <stdlib.h>

来源：https://stackoverflow.com/questions/10568301/why-does-the-compiler-complain-when-i-do-not-cast-the-result-of-malloc