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",num, two);
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",num, one);
以下是查找给定数字的补码的 C 程序 -
现场演示
#include#include #define SIZE 8 int main(){ int i, carry = 1; char num[SIZE + 1], one[SIZE + 1], two[SIZE + 1]; printf("Enter the binary number "); 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
",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
",num, two); return 0; }
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
以上是给定一个数，找到它的二进制补码的C程序的详细内容。更多信息请关注PHP中文网其他相关文章！
