How to utilize the CVTDQ2PD instruction?
Henrik Mannerström
henrik.mannerstrom@gmail.com
Fri Sep 7 12:06:00 GMT 2012
Hello!
If I understand the Intel reference correctly, the CVTDQ2PD should allow
me to convert two 64 bit integers into two 64 bit doubles in one blow.
How should I write this in gcc C/C++?
Program 1 uses casting to illustrate the point, it outputs 0 and 1.
Program 2 was my naive attempt to use vectorization, it outputs 0 and
4.94066e-324, which is _not_ what I intended.
Best regards,
Henrik Mannerström
Program 1:
#include <iostream>
typedef union {
int long long i[2];
double d[2];
} v_t;
int main(void) {
v_t a;
for (int k=0;k!=2;k+=1) {
a.i[k] = k;
}
for (int k=0;k!=2;k+=1) {
a.d[k] = double(a.i[k]);
}
for (int k=0;k!=2;k+=1) {
std::cout << a.d[k] << std::endl;
}
return 0;
}
Program 2:
extern "C" {
#include <immintrin.h>
}
#include <iostream>
typedef union {
__v2df vd;
__v2di vi;
int long long i[2];
double d[2];
} v_t;
int main(void) {
v_t a;
for (int k=0;k!=2;k+=1) {
a.i[k] = k;
}
a.vd = __v2df(a.vi);
for (int k=0;k!=2;k+=1) {
std::cout << a.d[k] << std::endl;
}
return 0;
}
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