How does the conversion of int value to unsigned char in memset() work?

Question

void * memset ( void * ptr, int value, size_t num );

memset() sets the first num bytes of ptr to value. I want to use memset() to set all bytes of a void buffer to:

00000000
11111111
10101010
01010101

For "00000000" I can use

memset(buffer, '0', BUF_LENGTH);

But am I right to assume

memset(buffer, '1', BUF_LENGTH);

won't end in "11111111", but in "00000001"? I've read that '-1' would do the trick, but why? And what value would i need to set bitwise 10101010 or 01010101?

Answer 1

memset operates on a byte-by-byte basis.

To zero the memory use:

memset(buffer, 0, BUF_LENGTH);

To set each byte to the value 1 use:

memset(buffer, 0x01, BUF_LENGTH);

For the binary pattern 10101010 use:

memset(buffer, 0xAA, BUF_LENGTH);

For the binary pattern 01010101 use:

memset(buffer, 0x55, BUF_LENGTH);

To set all bits in the buffer to 1 use:

memset(buffer, 0xFF, BUF_LENGTH);

BUF_LENGTH would be the length used in the array definition, eg

#define BUF_LENGTH 256
char buffer[BUF_LENGTH];

or what you used as length in dynamic memory allocation:

#define BUF_LENGTH 256
char *buffer = malloc(BUF_LENGTH);

Do not quote the 0 as in your code example, as this represents the ascii character '0', which has the value 0x30 (see http://www.asciitable.com/ ).

Answer 2

If this is C++, use binary literals, for instance 0b10101010 and 0b01010101 .

If it is C, translate to hexadecimal, 0xAA and 0x55 .

(Using '0' and '1' will most likely result in 00110000 and 00110001, respectively.)