Using memset for integer array in C

Question

char str[] = "beautiful earth";
memset(str, '*', 6);
printf("%s", str);

Output:
******ful earth

Like the above use of memset, can we initialize only a few integer array index values to 1 as given below?

int arr[15];
memset(arr, 1, 6);

Answer 1

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.

Answer 2

Short answer, NO.

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

Answer 3

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.

Answer 4

The third argument of memset is byte size. So you should set total byte size of arr[15]

memset(arr, 1, sizeof(arr));

However probably, you should want to set value 1 to whole elements in arr. Then you've better to set in the loop.

for (i = 0; i < sizeof(arr)/sizeof(arr[0]); i++) {
    arr[i] = 1;
}

Because memset() set 1 in each bytes. So it's not your expected.