memset(3)

#include <string.h>
void *memset(void *s, int c, size_t n);

Description

memset writes the same value into the first n bytes of the area that s points to. The C23 standard spells out the conversion: the function “copies the value of c (converted to an unsigned char) into each of the first n characters”. c is an int, but only its low 8 bits count. That is why memset(buf, 0, n) sets everything to zero, and memset(buf, '-', n) fills the area with dashes.

The most common use is zeroing a buffer or a structure right after allocating it, when calloc(3) does not fit (for a stack variable, for instance). Watch out for a frequent misconception: memset(array, 1, n) does not set each int to 1, it sets each byte to 0x01, so each int becomes 0x01010101. memset works in bytes, not in elements.

Return value

memset returns s, the address you gave it. As with memcpy and memmove, the value is mostly there for chaining calls.

Example

#include <stdio.h>#include <string.h>
int main(void){    char line[6];
    memset(line, '-', 5);    line[5] = '\0';    printf("%s\n", line);    return 0;}

$ gcc -Wall -Wextra -g memset.c -o memset$ ./memset-----

Pitfalls

• Expecting to fill an array of int with a value: only the low byte of c is used, repeated in every byte. Zero is the only value that gives the expected result for every type.

• Mixing up the argument order: it is memset(s, c, n), the value before the size. Many people expect the opposite.

• Counting on memset(p, 0, n) to erase a secret you are about to free: the compiler may remove the call if it decides the area is not used afterwards. C23 adds memset_explicit for exactly that need.

• Passing a size in elements instead of bytes: for an array of int, it is n * sizeof(int).

memset

Write gm_memset, which stores the low byte of c in exactly n bytes and returns the pointer it was given, not the end of the area.

Open the exercise