__m64_shladd
Microsoft Specific
Emits the IPF Shift Left and Add (shladd) instruction.
__m64 __m64_shladd(
__m64 a,
const int nBit,
__m64 b
);
Parameters
- [in] a
An __m64 union containing a 64-bit integer.
- [in] nBit
The number of bits shifted left. Valid values are 1, 2, 3, or 4.
- [in] b
An __m64 union containing a 64-bit integer.
Return Value
The result of the left shift of a by nBit bytes, followed by addition of b.
Requirements
| Intrinsic | Architecture |
|---|---|
__m64_shladd |
IPF |
Header file <intrin.h>
Remarks
The result computed and returned is a * 2^nBit + b.
Example
// shladd.cpp
// processor: IPF
#include <stdio.h>
#include <intrin.h>
#pragma intrinsic(__m64_shladd)
void print(__int64 a, int nBit, __int64 b, __int64 c)
{
printf_s("__m64_shladd(%I64d, %d, %I64d) returns %I64d\n",
a, nBit, b, c);
}
int main()
{
__m64 m, n, result;
m.m64_i64 = 15;
n.m64_i64 = 7;
result = __m64_shladd(m, 1, n);
print(m.m64_i64, 1, n.m64_i64, result.m64_i64);
result = __m64_shladd(m, 2, n);
print(m.m64_i64, 2, n.m64_i64, result.m64_i64);
result = __m64_shladd(m, 3, n);
print(m.m64_i64, 3, n.m64_i64, result.m64_i64);
result = __m64_shladd(m, 4, n);
print(m.m64_i64, 4, n.m64_i64, result.m64_i64);
}
Output
__m64_shladd(15, 1, 7) returns 37 __m64_shladd(15, 2, 7) returns 67 __m64_shladd(15, 3, 7) returns 127 __m64_shladd(15, 4, 7) returns 247