No, you don't. The following,

    int *i = malloc(sizeof(int));

is valid C, and will compile just fine. I think it's considered good form, too, as the cast doesn't really add anything not already present, and might even drift out of date with the surrounding code. (And just gets in the way of legibility.)

IIRC, it is not valid C++. But malloc()/void * is somewhat rarer to find there, as you have better options.

That's an implicit cast from void * to int * .

    void *malloc(size_t size);

In standard parlance, that is a conversion. Only (this) is a cast.

There's no such thing as an "implicit cast". A cast is an explicit conversion (a cast expression), and the assignment expression discussed here has an implicit conversion.