Fast way to replace elements in array - C

Question

Let's say we have an array of ints like this:

const int size = 100000; int array[size]; //set some items to 0 and other items to 1 

I'd like to replace all items that have value of 1 with another value, for example 123456. This can be trivially implemented with:

for(int i = 0; i < size ; i++){     if(array[i] != 0)          array[i] = 123456; } 

Out of curiosity, is there a faster way to do this, by some kind of x86 trickery, or is this the best code for the processor?

Answer 1

For your specific case where you initially have 0 and 1, the following might be faster. You'll have to bench mark it. You probably can't do much better with plain C though; you may need to dive into assembly if you want to take advantage of "x86 trickery" that may exist.

for(int i = 0; i < size ; i++){   array[i] *= 123456; } 

EDIT:

Benchmark code:

#include <time.h> #include <stdlib.h> #include <stdio.h>  size_t diff(struct timespec *start, struct timespec *end) {   return (end->tv_sec - start->tv_sec)*1000000000 + end->tv_nsec - start->tv_nsec; }  int main(void) {   const size_t size = 1000000;   int array[size];    for(size_t i=0; i<size; ++i) {     array[i] = rand() & 1;   }    struct timespec start, stop;    clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start);   for(size_t i=0; i<size; ++i) {     array[i] *= 123456;     //if(array[i]) array[i] = 123456;   }   clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &stop);    printf("size: %zu\t nsec: %09zu\n", size, diff(&start, &stop)); } 

---

my results:

Computer: quad core AMD Phenom @2.5GHz, Linux, GCC 4.7, compiled with

$ gcc arr.c -std=gnu99 -lrt -O3 -march=native 

• if version: ~5-10ms
• *= version: ~1.3ms