Using memset for integer array in C

No, you cannot use memset() like this. The manpage says (emphasis mine):

The memset() function fills the first n bytes of the memory area pointed to by s with the constant byte c.

Since an int is usually 4 bytes, this won't cut it.

If you (incorrectly!!) try to do this:

int arr[15];memset(arr, 1, 6*sizeof(int));    //wrong!

then the first 6 ints in the array will actually be set to 0x01010101 = 16843009.

The only time it's ever really acceptable to write over a "blob" of data with non-byte datatype(s), is memset(thing, 0, sizeof(thing)); to "zero-out" the whole struture/array. This works because NULL, 0x00000000, 0.0, are all completely zeros.

The solution is to use a for loop and set it yourself:

int arr[15];int i;for (i=0; i<6; ++i)    // Set the first 6 elements in the array    arr[i] = 1;        // to the value 1.

Short answer, NO.

Long answer, memset sets bytes and works for characters because they are single bytes, but integers are not.

On Linux, OSX and other UNIX like operating systems where wchar_t is 32 bits and you can use wmemset() instead of memset().

#include<wchar.h>...int arr[15];wmemset( arr, 1, 6 );

Note that wchar_t on MS-Windows is 16 bits so this trick may not work.