如何用“y”位移“x”？

Question

我在使用移位程序时遇到了一些问题。

挑战是编写一个程序，它可以将无符号整数向左移动数步。两个整数都由用户输入。因此，给定两个整数（x 和y），x 中的位应向左移动y 步，而左侧丢失的位应向右移动。即，在最高有效位之外丢失的位被放置在最低有效位。

为了解决这个挑战，我做了以下尝试：

#include <stdio.h>
#include <utility.h>

unsigned int bitshift(unsigned int a, unsigned int b)
{
    a<<b;
    b>>a;
    return a,b; 
}

int main (void) 
{
    unsigned int x, y;
    
    printf("Enter two integers (smaller than 32) please:\n");
    scanf("%u%u", &x, &y);
    printf("Your integers are %u and %u.\n", x, y);
    printf("In hexadecimal-format, your integers are %08x and %08x.\n", x, y); 
 
    printf("We are now going to perform a bit shift operation\n");
    printf("The result of the bitshift operation is:\n");
    printf("%u and %u\n", bitshift(x,y));
    printf("In hexadecimal: %08x and %08x\n", bitshift(x,y));
    
    while(!KeyHit()); 
    return 0; 
    
}

但是，我在编译时收到一条错误消息，例如“没有足够的参数”，我不明白。

但我最想知道的是bitshift 函数是否可以完成这项工作？

Answer 1

这是对Barmar（现已删除）函数bitshift 解决方案的修改，并在 cmets 中提出了改进建议。

不幸的是，C 没有像 CPU 的指令集中那样有操作符来循环一个值中的位。这就是为什么可以通过将最低有效位向左移动，将最高有效位向右移动并组合结果来完成操作的原因。

要计算右移最高有效位的移位宽度，必须使用sizeof 计算数据类型中的位数。

请注意，大于或等于值中位数的移位宽度是未定义行为 (UB)。这就是为什么移位宽度是以值中的位数为模来计算的。此外，左移 0 会导致 UB 右移。

#include <stdio.h>
// get CHAR_BITS to make code portable for unusual platforms
#include <limits.h>

unsigned int bitshift(unsigned int a, unsigned int b)
{
    // modulo operation to prevent undefined behavior
    b %= sizeof a * CHAR_BIT; // sizeof a * 8 on usual platforms
    // prevent undefined behavior for right shift
    if(b == 0) return a;

    unsigned int upper = a << b;
    // not portable for unusual platforms
    // unsigned int lower = a >> (sizeof a * 8 - b);
    unsigned int lower = a >> (sizeof a * CHAR_BIT - b);
    
    return upper | lower;
}

int main (void) 
{
    unsigned int x, y;
    
    printf("Enter two integers (smaller than 32) please:\n");
    scanf("%u%u", &x, &y);
    printf("Your integers are %u and %u.\n", x, y);
    printf("In hexadecimal-format, your integers are %08x and %08x.\n", x, y); 
 
    printf("We are now going to perform a bit shift operation\n");
    printf("The result of the bitshift operation is:\n");
    printf("%u\n", bitshift(x,y));
    printf("In hexadecimal: %08x\n", bitshift(x,y));
    
    return 0; 
    
}

输入/输出示例：

Enter two integers (smaller than 32) please:
1234567890 12
Your integers are 1234567890 and 12.
In hexadecimal-format, your integers are 499602d2 and 0000000c.
We are now going to perform a bit shift operation
The result of the bitshift operation is:
1613571225
In hexadecimal: 602d2499

Enter two integers (smaller than 32) please:
246 28
Your integers are 246 and 28.
In hexadecimal-format, your integers are 000000f6 and 0000001c.
We are now going to perform a bit shift operation
The result of the bitshift operation is:
1610612751
In hexadecimal: 6000000f

Answer 2

右移足以得到将被移出的位，或者它的值左移。为避免 UB，应检查位数。

unsigned rol(unsigned val, unsigned nbits)
{
    if(nbits && nbits < CHAR_BIT * sizeof(val))
    {
        val = (val >> (CHAR_BIT * sizeof(val) - nbits)) | (val << nbits);
    }
    return val;
}

Answer 3

您实际上想要旋转位。

// Rotate the bits of a by b to the left and return the result

unsigned int bitshift(unsigned int a, unsigned int b)
{
  for (unsigned int i = 0; i < b; i++)
  {
    unsigned int lowbit = 0;
    if (a & 0x80000000)
      lowbit = 1;

    a <<= 1;
    a |= lowbit;
  }

  return a;
}

这假设 unsigned int 是 32 位。您应该能够将其调整为 64 位。

免责声明：

虽然这是一种非常幼稚的方法。如果你需要速度，你应该避免循环。这可以做到，但比较棘手。