Usage of memset() in c

Question

I have written a small program to get used to memset() operation:

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> int main() { int arr[10], i; int t = INT_MAX; memset(arr, t, sizeof(arr)); for (i = 0; i < 10; i++) printf("%d\t",arr[i]); printf("%d",t); return 0; }

The result of the above program is:

-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 2147483647

What is the behaviour of the memset() in the above program? Why is it setting the array elements to -1?

Jonas Schäfer · Accepted Answer · 2013-11-23 10:29:38Z

10

memset only takes the lower eight bits of the value and fills the whole buffer with these bits. The lower eight bits of MAX_INT are all ones (0xFF), and thus the array is afterwards filled with all ones. For signed integers, this is -1.

answered Nov 23, 2013 at 10:29

Jonas Schäfer

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3 Comments

icbytes Over a year ago

Assuming int is 32 bits, a 4 byte, you mean, memset only uses bits 0 to 7 to determine the value, which they represent and put this value into target memory location? If yes, this means also, as unsigned char could only handle old fashioned ascii?

Jonas Schäfer Over a year ago

I'm not sure I understand the question. bits 0 to 7 are eight bits, which is one bit more than ascii (if you don't take parity into account). But yes, if you want a pattern larger than 8 bits, you need to use a different function (wmemset comes to mind) or a for loop.

icbytes Over a year ago

All right. You understood well and answered my question. Thx.

greken · Accepted Answer · 2013-11-23 10:45:25Z

Memset sets the first sizeof(arr) bytes of the block of memory pointed by arr to t interpreted as an unsigned char. So what u get when you read out ints from arr depends in the interpretation of those bytes on your platform.

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