How memset initializes an array of integers by -1?
- by haccks
The manpage says about memset:
#include <string.h>
void *memset(void *s, int c, size_t n)
The memset() function fills the first n bytes of the memory area pointed to by s with the constant byte c.
It is clear that memset can't be used to initialize int array as shown below:
int a[10];
memset(a, 1, sizeof(a));
it is because int is represented by 4 bytes (say) and one can not get the desired value for the integers in array a.
But I often see the programmers use memset to set the int array elements to either 0 or -1.
int a[10];
int b[10];
memset(a, 0, sizeof(a));
memset(b, -1, sizeof(b));
As per my understanding, initializing with integer 0 is OK because 0 can be represented in 1 byte (may be I am wrong in this context). But how it is possible to initialize b with -1 (a 4 bytes value)?
