_rotl8, _rotl16

아티클
06/05/2017

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The latest version of this topic can be found at _rotl8, _rotl16.

Microsoft Specific**

Rotate the input values to the left to the most significant bit (MSB) by a specified number of bit positions.

Syntax

unsigned char _rotl8(   
   unsigned char value,   
   unsigned char shift   
);  
unsigned short _rotl16(   
   unsigned short value,   
   unsigned char shift   
);

Parameters

[in] value
The value to rotate.

[in] shift
The number of bits to rotate.

Return Value

The rotated value.

Requirements

Intrinsic	Architecture
`_rotl8`	x86, ARM, x64
`_rotl16`	x86, ARM, x64

Header file <intrin.h>

Remarks

Unlike a left-shift operation, when executing a left rotation, the high order bits that fall off the high end are moved into the least significant bit positions.

Example

// rotl.cpp  
#include <stdio.h>  
#include <intrin.h>  
  
#pragma intrinsic(_rotl8, _rotl16)  
  
int main()  
{  
    unsigned char c = 'A', c1, c2;  
  
    for (int i = 0; i < 8; i++)  
    {  
       printf_s("Rotating 0x%x left by %d bits gives 0x%x\n", c,  
               i, _rotl8(c, i));  
    }  
  
    unsigned short s = 0x12;  
    int nBit = 10;  
  
    printf_s("Rotating unsigned short 0x%x left by %d bits gives 0x%x\n",  
            s, nBit, _rotl16(s, nBit));  
}

Rotating 0x41 left by 0 bits gives 0x41  
Rotating 0x41 left by 1 bits gives 0x82  
Rotating 0x41 left by 2 bits gives 0x5  
Rotating 0x41 left by 3 bits gives 0xa  
Rotating 0x41 left by 4 bits gives 0x14  
Rotating 0x41 left by 5 bits gives 0x28  
Rotating 0x41 left by 6 bits gives 0x50  
Rotating 0x41 left by 7 bits gives 0xa0  
Rotating unsigned short 0x12 left by 10 bits gives 0x4800

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