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How memset initializes an array of integers by -1?

Posted by: admin November 30, 2017 Leave a comment


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 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 1s in all its bits, regardless of the size of the integer, so filling in a region with bytes filled with all 1s produces a region of -1 signed ints, longs, and shorts 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.