| title | __ll_lshift | Microsoft Docs | |||||||||
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| ms.date | 11/04/2016 | |||||||||
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| author | corob-msft | |||||||||
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Microsoft Specific
Shifts the supplied 64-bit value to the left by the specified number of bits.
unsigned __int64 __ll_lshift(
unsigned __int64 Mask,
int nBit
);
[in] Mask
The 64-bit integer value to shift left.
[in] nBit
The number of bits to shift.
The mask shifted left by nBit bits.
| Intrinsic | Architecture |
|---|---|
__ll_lshift |
x86, [!INCLUDEvcprx64] |
Header file <intrin.h>
If you compile your program using the 64 bit architecture and nBit is larger than 63, the number of bits to shift is nBit modulo 64. If you compile your program using the 32 bit architecture and nBit is larger than 31, the number of bits to shift is nBit modulo 32.
The ll in the name indicates that this is an operation on long long (__int64).
// ll_lshift.cpp
// compile with: /EHsc
// processor: x86, x64
#include <iostream>
#include <intrin.h>
using namespace std;
#pragma intrinsic(__ll_lshift)
int main()
{
unsigned __int64 Mask = 0x100;
int nBit = 8;
Mask = __ll_lshift(Mask, nBit);
cout << hex << Mask << endl;
}
10000
Note There is no unsigned version of the left shift operation. This is because __ll_lshift already uses an unsigned input parameter. Unlike the right shift, there is no sign dependence for the left shift, because the least significant bit in the result is always set to zero regardless of the sign of the value shifted.
END Microsoft Specific