The manpage says about memset
:
#include <string.h> void *memset(void *s, int c, size_t n)
The
memset()
function fills the firstn
bytes of the memory area pointed to bys
with the constant bytec
.
It is obvious 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 is it possible to initialize b
with -1
(a 4 bytes value)?
Oddly, the reason this works with -1
is exactly the same as the reason that this works with zeros: in two's complement binary representation, -1
has 1
s in all its bits, regardless of the size of the integer, so filling in a region with bytes filled with all 1
s produces a region of -1
signed int
s, long
s, and short
s on two's complement hardware.
On hardware that differs from two's complement the result will be different. The -1
integer constant would be converted to an unsigned char
of all ones, because the standard is specific on how the conversion has to be performed. However, a region of bytes with all their bits set to 1
would be interpreted as integral values in accordance with the rules of the platform. For example, on sign-and-magnitude hardware all elements of your array would contain the smallest negative value of the corresponding type.