C ++中的长手乘法 [英] Long Hand Multplication In C++

问题描述

我编写了这段代码，用于模拟存储在数组BeforeDecimal1&中的4位二进制数的乘法运算。 BeforeDecimal2。乘法的结果存储在数组s []中。问题是这段代码没有提供正确的答案。我试图找出问题但没有价值，所以现在我寻求专家的建议。这是操纵长手乘法的代码：



我正在尝试做这个：

I wrote this code for simulating multiplication of 4 bit binary numbers stored in arrays BeforeDecimal1 & BeforeDecimal2. The result of multiplication is stored in arrays s[]. The problem is this code isn''t providing correct answer. I tried to figure out the problem but worthless, so now I seek the advice of experts.Here is the code for manipulating Long Hand Multiplication:

I am trying to do this:

1000

x

1110
-------------------

   0000

  1000

 1000

1000
-------------
= 1110000

int i=3,x=0,carry=0;
while(true)
{
    if(BeforeDecimal2[i]!=0)
    {
      for(int j=x;j>=0;j--)
      {
        if(s[j]==1 && BeforeDecimal1[j]==1 && carry==0)
        {
          carry=1;
          s[j]=0;
        }
        else if(s[j]==1 && BeforeDecimal1[j]==0 && carry==1)
        {
          carry=1;
          s[j]=0;
        }
        else if(s[j]==0 && BeforeDecimal1[j]==0 && carry==1)
        {
          carry=0;
          s[j]=1;
        }
        else if(s[j]==0 && BeforeDecimal1[j]==0 && carry==0)
        {
          carry=0;
          s[j]=0;
        }

        x++;
      }
    }
    else
    {
      for(int h=7;h>=0;h--)
      {
        if(h==0)
        {
          BeforeDecimal1[0]=0; // that is inserting zeros from the right
        }
        else
        {
          BeforeDecimal1[h]=BeforeDecimal1[h-1];
          BeforeDecimal1[h-1]=0;
        }
      }
    }
    if(i==0)
      break;

    i--;
  }

问候



索引和否则如果在同一行

Regards

Indexation and else if in the same line

推荐答案

我很抱歉这样说，但是...那很漂亮很不可读。



我不确定你使用的是什么算法，但我怀疑它是否包含了很多嵌套，如果是'。

我认为你需要做的第一件事就是修改那一批的缩进，因为它目前非常令人困惑。

首先尝试改变：

I''m sorry to say this, but...that is pretty much unreadable.

I''m not sure what algorithm you are using, but I doubt if it included quite so many nested if''s.
I think the first thing you need to do is look at revising the indentation of that lot, because it is seriously confusing at the moment.
Try starting by changing:

if(reverse==4 && carry==1 && temp==0)
{
    s[reverse]=1;
}
else
    if(reverse==4 && carry==0 && temp==1)
    {
        s[reverse]=1;
    }
    else
        if(reverse==4 && carry==0 && temp==0)

要

To

if(reverse==4 && carry==1 && temp==0)
{
    s[reverse]=1;
}
else if(reverse==4 && carry==0 && temp==1)
{
    s[reverse]=1;
}
else if(reverse==4 && carry==0 && temp==0)

所以如果没有跑掉页面的右侧就更清楚了。

然后看看评论 - 我完全不知道你为什么要做测试你是，并且不知道你想要实现的算法我不能从那个代码中解决它。



对不起，但这需要是在我可以提供帮助之前让它变得可读，并且解释得更好！

So that it is clearer without running off the right hand side of the page.
Then look at commenting - I have absolutely no idea why you are doing the tests you are, and without any idea of the algorithm you are trying to implement I can''t work it out from that code.

Sorry, but this need to be made readable, and explained a bit better before I can help!

有点长，但是应该有效：

Bit long, but should work:

#include <stdio.h>

void mul(int bd1[4], int bd2[4], int s[8])
{
  int i,j;

  for (i=0;i<8;i++)
    s[i] = 0;

  for (i=0; i<4; i++)
  {
    int c = 0;
    if ( bd2[3-i] )
    {
      for (j=0; j<4; j++)
      {
        if ( s[i+3-j] == 1)
        {
          if (bd1[3-j] == 1)
          {
            if (c==1)
            {
              s[i+3-j] = 1;
              c = 1;
            }
            else
            { //c == 0
              s[i+3-j] = 0;
              c = 1;
            }
          }
          else
          { // bd1[3-j] == 0
            if (c==1)
            {
              s[i+3-j] = 0;
              c = 1;
            }
            else
            { //c == 0
              s[i+3-j] = 1;
              c = 0;
            }
          }
        }
        else
        {// s[i+3-j] == 0
          if (bd1[3-j] == 1)
          {
            if (c==1)
            {
              s[i+3-j] = 0;
              c = 1;
            }
            else
            { //c == 0
              s[i+3-j] = 1;
              c = 0;
            }
          }
          else
          { // bd1[3-j] == 0
            if (c==1)
            {
              s[i+3-j] = 1;
              c = 0;
            }
            else
            { //c == 0
              s[i+3-j] = 0;
              c = 0;
            }
          }
        }
      }
    }
  }
}

int main()
{
    int i;
  int bd1[]={1,0,0,0};
  int bd2[]={1,1,1,0};
  int s[8];
  mul(bd1,bd2,s);


  for(i=0;i<8;i++)
    printf("%d", s[i]);
  printf("\n");
  return 0;
}

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