CODEWITHSUNDEEP
cmemset
user541686
user541686

Reputation: 210352

Why does memset take an int instead of a char?

Why does memset take an int as the second argument instead of a char, whereas wmemset takes a wchar_t instead of something like long or long long?

Upvotes: 67

Views: 11685

Answers (3)

Jerry Coffin
Jerry Coffin

Reputation: 490008

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".

The value memset writes into the destination is the value of its second argument converted to an unsigned char.

Upvotes: 73

Frédéric Hamidi
Frédéric Hamidi

Reputation: 262919

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.

Upvotes: 6

Gui13
Gui13

Reputation: 13541

See fred's answer, it's for performance reasons.

On my side, I tried this code:

#include <stdio.h>
#include <string.h>

int main (int argc, const char * argv[])
{
    char c = 0x00;

    printf("Before: c = 0x%02x\n", c);
    memset( &c, 0xABCDEF54, 1);
    printf("After:  c = 0x%02x\n", c);

    return 0;
}

And it gives me this on a 64bits Mac:

Before: c = 0x00
After:  c = 0x54

So as you see, only the last byte gets written. I guess this is dependent on the architecture (endianness).

Upvotes: -4

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