Why write `sizeof(char)` if char is 1 by standard?

Question

I was doing some C coding and after reading some C code I\'ve noticed that there are code snippets like

char *foo = (char *)malloc(sizeof(char) * someDynamicAmo         


        

Answer 1

You're correct, by standard, the multiplication is irrelivant. That said, it looks like a habit someone got into to be consistent. If you always use the sizeof(), regardless of type, you never forget.

char *foo = (char *)malloc(sizeof(char) * someDynamicAmount);
int  *bar = (int  *)malloc(sizeof(int)  * someDynamicAmount);