Question

我正在尝试完成一项要求我为二进制算术编写三个函数的赋值。 badd（）是为我提供的，所以我用它来帮助编写bsub（）和bmult（）函数。但是，我无法理解如何执行bdiv（）函数。我知道我需要使用右移和我的bsubb（）函数迭代这些位，但我不知道如何实现它。以下是我到目前为止所写的功能。如果你发现我在编写错误时会告诉我（意思是bsub（）和bmult（））。谢谢。

/** This function adds the two arguments using bitwise operators. Your      
* implementation should not use arithmetic operators except for loop
* control. Integers are 32 bits long.  This function prints a message
* saying "Overflow occurred\n" if a two's complement overflow occurs
* during the addition process. The sum is returned as the value of
* the function.
*/
int badd(int x,int y){

int i;

char sum;
char car_in=0;
char car_out;
char a,b;

unsigned int mask=0x00000001;
int result=0;

for(i=0;i<32;i++){

  a=(x&mask)!=0;
  b=(y&mask)!=0;
  car_out=car_in & (a|b) |a&b;
  sum=a^b^car_in;

  if(sum) {
     result|=mask;
  }

  if(i!=31) {
     car_in=car_out;
  } else {
     if(car_in!=car_out) {
 printf("Overflow occurred\n");
     }
  }

  mask<<=1;
}

 return result;
 }

// subracts two integers by finding the compliemnt
// of "y", adding 1, and using the badd() function
// to add "-y" and "x"
int bsub(int x, int y){

return badd(x, badd(~y, 1));
}


//add x to total for however many y
int bmult(int x,int y){

int total;
int i;
for(i=0; i < = y; i++)
{
 total = badd(total,x)
}
 return total;
}

// comment me
unsigned int bdiv(unsigned int dividend, unsigned int divisor){

// write me
return 0;
}

Answer 1

这里不多说，它只是base-2中的一些基本数学：

unsigned int bmult(unsigned int x, unsigned int y)
{
    int total = 0;
    int i;

    /* if the i-th bit is non-zero, add 'x' to total */
    /* Multiple total by 2 each step */
    for(i = 32 ; i >= 0 ; i--)
    {
        total <<= 1;
        if( (y & (1 << i)) >> i )
        {
            total = badd(total, x);
        }
    }

    return total;
}

unsigned int bdiv(unsigned int dividend, unsigned int divisor)
{
    int i, quotient = 0, remainder = 0;

    if(divisor == 0) { printf("div by zero\n"); return 0; }

    for(i = 31 ; i >= 0 ; i--)
    {
        quotient <<= 1;
        remainder <<= 1;
        remainder |= (dividend & (1 << i)) >> i;

        if(remainder >= divisor)
        {
            remainder = bsub(remainder, divisor);
            quotient |= 1;
        }
    }

    return quotient;
}

这两篇文章足以对这些示例进行编码：Div和Mul。

Answer 2

在下面的代码中，我使用与问题相同的想法实现加法和减法。唯一的实际区别在于，在我的实现中，这两个函数还接受一个进位/借位，并产生一个进位/借出位。

进位位用于通过加法实现减法，该位有助于获得进位和借位的正确值。基本上，我使用状态寄存器中的进位标志实现典型的类似CPU的加法和减法。

然后使用进位/借位来通过减法实现比较。我在没有>=运算符的情况下实现了比较，我也考虑算术运算，因为它的性质不太明显。由于使用restoring division algorithm。

，因此在除法函数中需要比较函数

我也避免使用!运算符，而是使用^1。

除法函数将除数作为2 unsigned ints，它是最重要和最不重要的部分。最后，它用剩余部分取代最重要的部分，用商除掉最不重要的部分。因此，它执行除法和模运算，并以典型的CPU方式（例如x86 DIV指令）进行处理。该函数在成功时返回1，在溢出/除0时返回0。

主要功能做了一个简单的测试。它将除法函数的结果与直接除法的结果进行比较，并在不匹配时以错误消息结束。

我在测试部分中使用unsigned long long来测试divisor = UINT_MAX而不会陷入无限循环。可能需要花费太多时间来测试被除数和除数的整个值范围，这就是为什么我将它们分别设置为0xFFFF和0xFF而不是UINT_MAX。

代码：

#include <stdio.h>
#include <limits.h>

unsigned add(unsigned a, unsigned b, unsigned carryIn, unsigned* carryOut)
{
  unsigned sum = a ^ b ^ carryIn;
  unsigned carryOuts = a & b | (a | b) & carryIn;
  *carryOut = 0;
  if (sum & (carryOuts << 1))
    sum = add(sum, carryOuts << 1, 0, carryOut);
  else
    sum |= carryOuts << 1;
  *carryOut |= (carryOuts & (UINT_MAX / 2 + 1)) >> (sizeof(unsigned) * CHAR_BIT - 1); // +-*/ are OK in constants
  return sum;
}

unsigned sub(unsigned a, unsigned b, unsigned borrowIn, unsigned* borrowOut)
{
  unsigned diff = add(a, ~b, borrowIn ^ 1, borrowOut);
  *borrowOut ^= 1;
  return diff;
}

unsigned less(unsigned a, unsigned b)
{
  unsigned borrowOut;
  sub(a, b, 0, &borrowOut);
  return borrowOut;
}

int udiv(unsigned* dividendh, unsigned* dividendl, unsigned divisor)
{
  int i;
  unsigned tmp;

  if (less(*dividendh, divisor) ^ 1/* *dividendh >= divisor */)
    return 0; // overflow

  for (i = 0; i < sizeof(unsigned) * CHAR_BIT; i++)
  {
    if (less(*dividendh, UINT_MAX / 2 + 1) ^ 1/* *dividendh >= 0x80...00 */)
    {
      *dividendh = (*dividendh << 1) | (*dividendl >> (sizeof(unsigned) * CHAR_BIT - 1));
      *dividendl <<= 1;

      *dividendh = sub(*dividendh, divisor, 0, &tmp);/* *dividendh -= divisor; */
      *dividendl |= 1;
    }
    else
    {
      *dividendh = (*dividendh << 1) | (*dividendl >> (sizeof(unsigned) * CHAR_BIT - 1));
      *dividendl <<= 1;

      if (less(*dividendh, divisor) ^ 1/* *dividendh >= divisor */)
      {
        *dividendh = sub(*dividendh, divisor, 0, &tmp);/* *dividendh -= divisor; */
        *dividendl |= 1;
      }
    }
  }

  return 1;
}

int udiv2(unsigned* dividendh, unsigned* dividendl, unsigned divisor)
{
  unsigned long long dividend =
    ((unsigned long long)*dividendh << (sizeof(unsigned) * CHAR_BIT)) | *dividendl;

  if (*dividendh >= divisor)
    return 0; // overflow

  *dividendl = (unsigned)(dividend / divisor);
  *dividendh = (unsigned)(dividend % divisor);

  return 1;
}


int main(void)
{
  unsigned long long dividend, divisor;

  for (dividend = 0; dividend <= /*UINT_MAX*/0xFFFF; dividend++)
    for (divisor = 0; divisor <= /*UINT_MAX*/0xFF; divisor++)
    {
      unsigned divh = 0, divl = (unsigned)dividend, divr = (unsigned)divisor;
      unsigned divh2 = 0, divl2 = (unsigned)dividend;

      printf("0x%08X/0x%08X=", divl, divr);

      if (udiv(&divh, &divl, divr))
        printf("0x%08X.0x%08X", divl, divh);
      else
        printf("ovf");

      printf(" ");

      if (udiv2(&divh2, &divl2, divr))
        printf("0x%08X.0x%08X", divl2, divh2);
      else
        printf("ovf");

      if ((divl != divl2) || (divh != divh2))
      {
        printf(" err");
        return -1;
      }

      printf("\n");
    }

  return 0;
}

Answer 3

while dividend＆lt;除数在商数小数点后加零并将右股息换算为1
现在检查除数中的位数是否等于被除数中的位数如果不是，则将除数向左移，直到被除数中的位数等于除数中的位数。
现在减去红利，除数并在商数上加1，确保你在正确位置（如小数位）

重复此过程，直到被除数为0或1

不使用算术运算符执行位除法

3 个答案: