__m64_pshradd2
Microsoft Specific
Emits the IPF Parallel Shift Right and Add (pshradd2) instruction.
__m64 __m64_pshradd2(
__m64 a,
const int nBit,
__m64 b
);
Parameters
[in] a
An __m64 union containing an array of four 16-bit signed integers to be shifted right and added to.[in] nBit
The number of bits to right shift the integers in the first array. Valid values from 0 to 3.[in] b
An __m64 union containing an array of four 16-bit signed integers to add.
Return Value
An __m64 union containing an array of four 16-bit signed integers, the elements of which are computed according to the expression described here.
Requirements
Intrinsic |
Architecture |
|---|---|
__m64_pshradd2 |
IPF |
Header file <intrin.h>
Remarks
This instruction computes:
c[i] = a[i] >> nBit + b[i]
for each of the four elements i of a and b and returns the array c as an __m64 union.
Example
// m64_pshradd2.c
// processor: IPF
#include <stdio.h>
#include <intrin.h>
#pragma intrinsic(__m64_pshradd2)
int main()
{
__m64 a, b, result;
a.m64_i16[0] = 0;
a.m64_i16[1] = 2;
a.m64_i16[2] = 16;
a.m64_i16[3] = 100;
b.m64_i16[0] = 1;
b.m64_i16[1] = 2;
b.m64_i16[2] = 3;
b.m64_i16[3] = 4;
printf_s("Input a: %d %d %d %d\n",
a.m64_i16[0], a.m64_i16[1], a.m64_i16[2],
a.m64_i16[3]);
printf_s("Input b: %d %d %d %d\n",
b.m64_i16[0], b.m64_i16[1], b.m64_i16[2],
b.m64_i16[3]);
result = __m64_pshradd2(a, 1, b);
printf_s("Results of pshradd2(a, 1, b): %d %d %d %d\n",
result.m64_i16[0], result.m64_i16[1], result.m64_i16[2],
result.m64_i16[3]);
}
Input a: 0 2 16 100
Input b: 1 2 3 4
Results of pshradd2(a, 1, b): 1 3 11 54