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

This is all a matter of coding style. Several styles exist.

To actually answer the question, people write

malloc(n * sizeof(char))

to keep all their code that uses malloc consistent. Next time they might need int and then they can write the code in the very same way,

malloc(n * sizeof(int))

So the reason why it is done to keep the coding style consistent. Even though sizeof(char) is indeed guaranteed to always be 1 and is therefore superfluous. It is a way to write self-documenting code.

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However, the most common way to use malloc in C is perhaps

type* t = malloc(n * sizeof(*t));

or 100% equivalent:

type* t = malloc(n * sizeof *t);

Since the operator of sizeof is not evaluated for side-effects, this code is safe, even though the variable t is not yet initialized.

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The third possible style is the pedantically correct one, which would use array pointers, since what we allocate is actually arrays:

type (*t)[n] = malloc( sizeof(type[n]) );

This is perhaps the most correct way, as far as type correctness is concerned. The size of an array is allocated, and we point at the allocated array with an array pointer.

However, the problem with this style is that array pointers add extra complexity to the syntax: you would have to de-reference this array as (*t)[i] instead of t[i]. It makes the code harder to read. (And also as a side-note, if n is not an integer constant expression, the code won't compile on old, outdated C compilers.)