Why does memset take an int instead of a char?
memset
predates (by quite a bit) the addition of function prototypes to C. Without a prototype, you can't pass a char
to a function -- when/if you try, it'll be promoted to int
when you pass it, and what the function receives is an int
.
It's also worth noting that in C, (but not in C++) a character literal like 'a'
does not have type char
-- it has type int
, so what you pass will usually start out as an int
anyway. Essentially the only way for it to start as a char and get promoted is if you pass a char
variable.
In theory, memset
could probably be modified so it receives a char
instead of an int
, but there's unlikely to be any benefit, and a pretty decent possibility of breaking some old code or other. With an unknown but potentially fairly high cost, and almost no chance of any real benefit, I'd say the chances of it being changed to receive a char
fall right on the line between "slim" and "none".
Edit (responding to the comments): The CHAR_BIT
least significant bits of the int
are used as the value to write to the target.
Probably the same reason why the functions in <ctypes.h>
take ints
and not chars
.
On most platforms, a char
is too small to be pushed on the stack by itself, so one usually pushes the type closest to the machine's word size, i.e. int
.
As the link in @Gui13's comment points out, doing that also increases performance.