__shiftleft128
Microsoft Specific
Shifts a 128-bit quantity, represented as two 64-bit quantities LowPart and HighPart, to the left by a number of bits specified by Shift and returns the high 64 bits of the result.
unsigned __int64 __shiftleft128(
unsigned __int64 LowPart,
unsigned __int64 HighPart,
unsigned char Shift
);
Parameters
[in] LowPart
The low 64 bits of the 128-bit quantity to shift.[in] HighPart
The high 64 bits of the 128-bit quantity to shift.[in] Shift
The number of bits to shift.
Return Value
The high 64 bits of the result.
Requirements
Intrinsic |
Architecture |
|---|---|
__shiftleft128 |
IPF, x64 |
Header file <intrin.h>
Remarks
The Shift value is always modulo 64 so that, for example, if you call __shiftleft128(1, 0, 64), the function will shift the low part 0 bits left and return a high part of 0 and not 1 as might otherwise be expected.
Example
// shiftleft128.c
// processor: IPF, x64
#include <stdio.h>
#include <intrin.h>
#pragma intrinsic (__shiftleft128, __shiftright128)
int main()
{
unsigned __int64 i = 0x1I64;
unsigned __int64 j = 0x10I64;
unsigned __int64 ResultLowPart;
unsigned __int64 ResultHighPart;
ResultLowPart = i << 1;
ResultHighPart = __shiftleft128(i, j, 1);
// concatenate the low and high parts padded with 0's
// to display correct hexadecimal 128 bit values
printf_s("0x%02I64x%016I64x << 1 = 0x%02I64x%016I64x\n",
j, i, ResultHighPart, ResultLowPart);
ResultHighPart = j >> 1;
ResultLowPart = __shiftright128(i, j, 1);
printf_s("0x%02I64x%016I64x >> 1 = 0x%02I64x%016I64x\n",
j, i, ResultHighPart, ResultLowPart);
}
0x100000000000000001 << 1 = 0x200000000000000002 0x100000000000000001 >> 1 = 0x080000000000000000