[Discrete Mathematics] Practice 1 Implementing a binary adder in C language

topics and requirements

Each student uses C language to complete the design of an 8-bit adder independently , and the calculation is only realized by three logical operations: NOT AND OR ( !, &&, ‖ ). (Boolean data type in C language: bool, reference: https://www.javatpoint.com/c-boolean)

Requirements: Input: two addends (8-bit binary number)

Output: sum of addends (8-bit or 9-bit binary number)

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References

Adder principle illustration:

half adder

full adder

Example of adding two three-bit binary numbers (consisting of a half adder and two full adders)

Adder literal explanation:

1. The one in the picture represents the first digit of the first number , the second digit of the first number.... represents the i + 1th digit;

2. synonymous meaning;

3. Represents the i + 1th bit of the summed binary number ;

4. Represents the carry situation of the upper bit after the sum is added ;

The addition of two finite-bit binary numbers can be regarded as composed of several full adders and half adders. The half adder is responsible for calculating the sum and carry of the first bit, that is, the sum (when i = 0 ). The following bits are all in charge of the full adder. The full adder has three entry parameters , , , and two result parameters and will be passed to the next adder.

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Experimental principle

应用半加器和全加器的有效组合来达到将两个有限位的二进制数相加，并且得出一个有限位的结果。同时采用了与“&& ”、非“ ! ”的逻辑实现，由于不能使用异或” ^ ”，故没有设计成超前加法器来提升其效率。

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加法器设计与实现

该加法器的逻辑电路图：

两个八位的二进制数相加的加法器

设计与实现：

• First数组：用于存第一个数的二进制形式

• Second数组：用于存第二个数的二进制形式

• Sum数组：用于存两数相加的二进制形式

• CarryBit数组：用于存两数相加进位情况，0为未进位，1为已进位

• RevCarry数组：用于存放carryBit数组的非

• 运算式子：

• 若i = 0

• Sum [ 0 ] = ( first[ 0 ] + second [ 0 ] ) * revCarry[ 0 ]

• 若i > 0

• Sum [ i ] = ( (first [ i ] + second [ i ]) * revCarry[ i ] ) + carryBit[ i - 1]

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运行界面，结果截图

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源代码：adder.cpp

//每个学生利用 C 语言独自完成设计一个 8 位加法器，计算只用非与或(!，
//&&，‖)三种逻辑运算实现。
//要求：输入：两个加数（8 位的二进制数）
//输出：加数的和（8 位或者 9 位的二进制数）
#include <stdio.h>
#include <iostream>
using namespace std;

const int N1 = 7;
const int N2 = 8;

int first[N1], second[N1], carryBit[N1], revCarry[N1], sum[N2];
int n,m;

//判断输入的数字是否合法 
void isLegal(int a[])
{
    for(int i = N1; i >= 0; i--)
    {
        if((a[i] != 0 && a[i] != 1))
        {
            printf("输入字符%d错误！请输入二进制数",a[i]);
            break;
        }
        else  continue;
    }
}
//进位情况取反
void reverseCarry(int b[])
{
    for(int i = 0; i <= N1; i++)
    {
        if(b[i] == 0) revCarry[i] = 1;
        else revCarry[i] = 0;    
    }    
} 
int main(){
    cout << "请输入第一个二进制数" << endl;
    scanf("%d",&n);
    for(int i = 0; i <= N1; i++){
        first[i] = n % 10;
        n = n /10;
    }
    isLegal(first);
    cout << "请输入第二个二进制数" << endl;
    scanf("%d",&m);
    for(int i = 0; i <= N1; i++){
        second[i] = m % 10;
        m = m /10;
    }
    isLegal(second);
    for(int i = 0; i <= N1; i++)
    {
        carryBit[i] = first[i] * second[i];
//    可验证carry数组里面的数字分布的情况 
//        printf("第%d位carry是%d\n",i,carryBit[i]);
    }
    reverseCarry(carryBit);
//    可验证是否已经取反 
//    for(int i = 0; i <= N1; i++)
//    {
//        printf("取反后的Carry第%d位是%d\n",i,revCarry[i]);
//    }
    for(int i = 0; i <= N2; i++)
    {
        if(i == 0)
        { 
        sum[i] = (first[i] + second[i]) * revCarry[i];
        }
        else if(i <= 7) 
        {
        sum[i] = ((first[i] + second[i]) * revCarry[i]) + carryBit[i - 1];
        }
        else sum[i] = carryBit[i - 1];
    }
    cout << "和是" << endl;
    for(int i = N2; i >= 0 ; i--) printf("%d",sum[i]);
    return 0;
}

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结语

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