Is it necessary to multiply by sizeof( char ) when manipulating memory?

Question

When using malloc and doing similar memory manipulation can I rely on sizeof( char ) being always 1?

For example I need to allocate memory for N elements of type char. Is multiplying by sizeof( char ) necessary:

char* buffer = malloc( N * sizeof( char ) );

or can I rely on sizeof( char ) always being 1 and just skip the multiplication

char* buffer = malloc( N );

I understand completely that sizeof is evaluated during compilation and then the compiler might even compile out the multiplication and so the performance penalty will be minimal and most likely zero.

I'm asking mainly about code clarity and portability. Is this multiplication ever necessary for char type?

Answer 1

By definition, sizeof(char) is always equal to 1. One byte is the size of character in C, whatever the numbers of bits in a byte there is (8 on common desktop CPU).

The typical example where one byte is not 8 bits is the PDP-10 and other old, mini-computer-like architectures with 9/36 bits bytes. But bytes which are not 2^N are becoming extremely uncommon I believe

Also, I think this is better style:

char* buf1;
double* buf2;

buf1 = malloc(sizeof(*buf1) * N);
buf2 = malloc(sizeof(*buf2) * N);

because it works whatever the pointer type is.