给定二进制数的补码可以通过两种方法计算,如下 -
方法 1 − 将给定的二进制数转换为补码,然后加 1。
方法 2 − 从 Least 开始设置的第一个位后面的尾随零有效位 (LSB),包括保持不变的一位,其余全部应补码。
对于给定的二进制数查找二进制补码的逻辑如下 -
for(i = SIZE - 1; i >= 0; i--){
if(one[i] == '1' && carry == 1){
two[i] = '0';
}
else if(one[i] == '0' && carry == 1){
two[i] = '1';
carry = 0;
} else {
two[i] = one[i];
}
}
two[SIZE] = '0';
printf("Two's complement of binary number %s is %s<p>",num, two);</p>
从给定的二进制数中找到补码的逻辑是 −
for(i = 0; i < SIZE; i++){
if(num[i] == '0'){
one[i] = '1';
}
else if(num[i] == '1'){
one[i] = '0';
}
}
one[SIZE] = '0';
printf("Ones' complement of binary number %s is %s<p>",num, one);</p>
示例
以下是查找给定数字的补码的 C 程序 -
现场演示
#include<stdio.h>
#include<stdlib.h>
#define SIZE 8
int main(){
int i, carry = 1;
char num[SIZE + 1], one[SIZE + 1], two[SIZE + 1];
printf("Enter the binary number<p>");
gets(num);
for(i = 0; i < SIZE; i++){
if(num[i] == '0'){
one[i] = '1';
}
else if(num[i] == '1'){
one[i] = '0';
}
}
one[SIZE] = '0';
printf("Ones' complement of binary number %s is %s</p><p>",num, one);
for(i = SIZE - 1; i >= 0; i--){
if(one[i] == '1' && carry == 1){
two[i] = '0';
}
else if(one[i] == '0' && carry == 1){
two[i] = '1';
carry = 0;
}
else{
two[i] = one[i];
}
}
two[SIZE] = '0';
printf("Two's complement of binary number %s is %s</p><p>",num, two);
return 0;
}</p>
输出
当执行上述程序时,会产生以下结果 -
Enter the binary number
1000010
Ones' complement of binary number 1000010 is 0111101
Two's complement of binary number 1000010 is 0111110