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.

Answer 5

Since nobody mentioned it...

Although you cannot initialize the integers with value 1 using memset, you can initialize them with value -1 and simply change your logic to work with negative values instead.

For example, to initialize the first 6 numbers of your array with -1, you would do

memset(arr,-1,6*(sizeof int));

Furthermore, if you only need to do this initialization once, you can actually declare the array to start with values 1 from compile time.

int arr[15] = {1,1,1,1,1,1};

Answer 6

No, you can't [portably] use memset for that purpose, unless the desired target value is 0. memset treats the target memory region as an array of bytes, not an array of ints.

A fairly popular hack for filling a memory region with a repetitive pattern is actually based on memcpy. It critically relies on the expectation that memcpy copies data in forward direction

int arr[15];

arr[0] = 1;
memcpy(&arr[1], &arr[0], sizeof arr - sizeof *arr);

This is, of course, a pretty ugly hack, since the behavior of standard memcpy is undefined when the source and destination memory regions overlap. You can write your own version of memcpy though, making sure it copies data in forward direction, and use in the above fashion. But it is not really worth it. Just use a simple cycle to set the elements of your array to the desired value.

Answer 7

Actually it is possible with memset_pattern4 which sets 4 bytes at a time.

memset_pattern4(your_array, your_number, sizeof(your_array));

Answer 8

Memset sets values for data types having 1 byte but integers have 4 bytes or more , so it won't work and you'll get garbage values. It's mostly used when you are working with char and string types.

Answer 9

I tried the following program and it appears that you can initialize your array using memset() with -1 and 0 only

#include<stdio.h>
#include<string.h>

void printArray(int arr[], int len)
{
        int i=0;
    for(i=0; i<len; i++)
    {
        printf("%d ", arr[i]);
    }
    puts("");
}

int main()
{
    int arrLen = 15;
    int totalNoOfElementsToBeInitialized = 6;

    int arr[arrLen];
    printArray(arr, arrLen);
    memset(arr, -1, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
    printArray(arr, arrLen);
    memset(arr, 0, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
    printArray(arr, arrLen);
    memset(arr, 1, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
    printArray(arr, arrLen);
    memset(arr, 2, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
    printArray(arr, arrLen);
    memset(arr, -2, totalNoOfElementsToBeInitialized*sizeof(arr[0]));
    printArray(arr, arrLen);
    return 0;
}

Answer 10

Ideally you can not use memset to set your arrary to all 1.
Because memset works on byte and set every byte to 1.

memset(hash, 1, cnt);

So once read, the value it will show 16843009 = 0x01010101 = 1000000010000000100000001
Not 0x00000001
But if your requiremnt is only for bool or binary value then we can set using C99 standard for C library

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>        //Use C99 standard for C language which supports bool variables

int main()
{
    int i, cnt = 5;
    bool *hash = NULL;
    hash = malloc(cnt);

    memset(hash, 1, cnt);
    printf("Hello, World!\n");

    for(i=0; i<cnt; i++)
        printf("%d ", hash[i]);

    return 0;
}

Output:

Hello, World!
1 1 1 1 1