fill character array by blocks - c

Question

question of curiosity
suppose I have:
int main(void)
{
char str[32];
for (i = 0; i < 32; i++)
str[i] = 0;
}
but I want to do it 4x faster
int main(void)
{
char str[32];
for (i = 0; i < 32 / 4; i += 4)
str[i] = (int)0;
}
I expect that the whole array will be filled with zeros.
but array is not filled by zeroes
my questions: why array not filled by zeroes? how to fill array per int blocks? my question is Research for c feature, how to tell the compiler - write blocks of 4 bytes, ie integer registers, it will reduce the number of memory accesses by 4 times, on x64 processors reduce 8 times
thanks for all, follows work well:
int main(int argc, char *argv[])
{
char str[32];
int i;
for (i = 0; i < 32; i++)
str[i] = 12;
for (i = 0; i < 32 / sizeof(int); i++)
((int *) str)[i] = 0;
printf("%d\n", i);
for (i = 0; i < 32; i++)
printf("%d\n", str[i]);
return 0;
}

Answer 1

The correct and fastest way is to initialize the array to zero
char str[32] = { 0 } ;
If you want to set the array to zero afterwards, then use memset and enable compiler optimization and intrinsic functions, and the compiler will figure out the fastest way to zero the array.
memset( str , 0 , sizeof( str ) ) ;

Answer 2

As pointed out by #user2501, initializing to {0} or using memset is the fastest and correct way.
If you are tempted to use something like ((int *)str)[i] = 0, do not, this can result in unaligned access.
As an alternative to memset, in C99 (and assuming that int is 4 bytes) you can use the type punning feature of unions:
#include <stdio.h>
typedef union {
char as_string[32];
int as_int[8];
} foo;
int main(void)
{
foo x;
int i;
for (i = 0; i < 8; i++)
x.as_int[i] = 0;
for (i = 0; i < 32; i++)
printf("%d", x.as_string[i]);
printf("\n");
return 0;
}
Output:
00000000000000000000000000000000

Answer 3

Seems that you are forced to use pointers and casts, this version uses the heap in order to allocate str on an address properly aligned for int:
#include <stdio.h>
#include <stdlib.h> /* malloc, free */
#include <stdint.h> /* intptr_t (pointer arithmetic using modulo division) */
#define MAX 32
int main(void)
{
size_t i, align;
char *ptr, *str;
align = __alignof__(int);
ptr = malloc(MAX + align); /* MAX + max align distance */
str = ptr + align - (intptr_t)ptr % align; /* now str is properly aligned */
for (i = 0; i < MAX / sizeof(int); i++)
((int *)str)[i] = 0;
for (i = 0; i < MAX; i++)
printf("%d\n", str[i]);
free(ptr);
return 0;
}
Note that __alignof__ is a gcc extension, change to __alignof if you are under Visual Studio or _Alignof if you are comfortable with C11.

Answer 4

You have to cast the pointer as the int, not the value:
int main(void)
{
char str[32];
for (i = 0; i < 32/sizeof(int); i++)
((int *) str)[i] = 0xAABBCCDD; //Be careful of Endianness
}
Alternatively:
int i;
union block_fill
{
char arr[24];
int iarr[6];
};
union block_fill block_arr;
for(i=0; i<6; i++)
block_arr.iarr[i] = 0x11223344;
for(i=0; i<24; i++)
printf("%x", block_arr.arr[i]);