How memset initializes an array of integers by -1?

Question

The manpage says about memset:

#include 
void *memset(void *s, int c, size_t n)

The me

Answer 1

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.

Answer 2

When all bits of a number are 0, its value is also 0. However, if all bits are 1 the value is -1.

If we write int a[2], 4x2 bytes of memory is allocated which contains random/garbage bits-

00110000 00100101 11100011 11110010    11110101 10001001 00111000 00010001

Then, we write memset(a, 0, sizeof(a)). Now, memset() works byte by byte, and one byte representation (unsigned char) of 0 is 00000000. So, it becomes-

00000000 00000000 00000000 00000000    00000000 00000000 00000000 00000000

Therefore, both a[0] and a[1] are initialized with 0.

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Now, lets see memset(a, -1, sizeof(a)): one byte for -1 is 11111111. And, we get-

11111111 11111111 11111111 11111111    11111111 11111111 11111111 11111111

Here, both a[0] and a[1] will have the value -1.

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However, for memset(a, 1, sizeof(a)): 1 in a byte is 00000001-

00000001 00000001 00000001 00000001    00000001 00000001 00000001 00000001

So, the value will be- 16843009.