print a __m128i variable

ghz 1years ago ⋅ 9070 views

Question

I'm trying to learn to code using intrinsics and below is a code which does addition

compiler used: icc

#include<stdio.h>
#include<emmintrin.h>
int main()
{
        __m128i a = _mm_set_epi32(1,2,3,4);
        __m128i b = _mm_set_epi32(1,2,3,4);
        __m128i c;
        c = _mm_add_epi32(a,b);
        printf("%d\n",c[2]);
        return 0;
}

I get the below error:

test.c(9): error: expression must have pointer-to-object type
        printf("%d\n",c[2]);

How do I print the values in the variable c which is of type __m128i

Answer

Use this function to print them:

#include <stdint.h>
#include <string.h>

void print128_num(__m128i var)
{
    uint16_t val[8];
    memcpy(val, &var, sizeof(val));
    printf("Numerical: %i %i %i %i %i %i %i %i \n", 
           val[0], val[1], val[2], val[3], val[4], val[5], 
           val[6], val[7]);
}

You split 128bits into 16-bits(or 32-bits) before printing them.

This is a way of 64-bit splitting and printing if you have 64-bit support available:

#include <inttypes.h>

void print128_num(__m128i var) 
{
    int64_t v64val[2];
    memcpy(v64val, &var, sizeof(v64val));
    printf("%.16llx %.16llx\n", v64val[1], v64val[0]);
}

Note: casting the &var directly to an int* or uint16_t* would also work MSVC, but this violates strict aliasing and is undefined behaviour. Using memcpy is the standard compliant way to do the same and with minimal optimization the compiler will generate the exact same binary code.

#c