当我调用multi()时,它将很好地计算二进制数,但不会计算出adding()。如果我按顺序切换两个,它将计算adding(),而不计算multi()。第二个函数调用未显示其值是否有原因?它将打印printf,但不打印值。任何帮助都会很棒。
#include <stdio.h>
long binary1, binary2, binary3, binary4, multiply = 0;
int binaryproduct(int, int);
int digit, factor = 1;
int multi();
int adding();
int main()
{
binary3 = binary1;
binary4 = binary1;
printf("Enter the first binary number: ");
scanf("%ld", &binary1);
printf("Enter the second binary number: ");
scanf("%ld", &binary2);
multi();
printf("\n");
adding();
}
int adding()
{
int i = 0, remainder = 0, sum[20];
while (binary3 != 0 || binary4 != 0) {
sum[i++] =(binary3 % 10 + binary4 % 10 + remainder) % 2;
remainder =(binary3 % 10 + binary4 % 10 + remainder) / 2;
binary1 = binary3 / 10;
binary2 = binary4 / 10;
}
if (remainder != 0) {
sum[i++] = remainder;
}
--i;
printf("Sum of two binary numbers: ");
while (i >= 0) {
printf("%d", sum[i--]);
}
return 0;
}
int multi()
{
while (binary2 != 0) {
digit = binary2 % 10;
if (digit == 1) {
binary1 = binary1 * factor;
multiply = binaryproduct(binary1, multiply);
} else {
binary1 = binary1 * factor;
}
binary2 = binary2 / 10;
factor = 10;
}
printf("Product of two binary numbers: %ld", multiply);
return 0;
}
int binaryproduct(int binary3, int binary4)
{
int i = 0, remainder = 0, sum[20];
int binaryprod = 0;
while (binary1 != 0 || binary2 != 0) {
sum[i++] =(binary1 % 10 + binary2 % 10 + remainder) % 2;
remainder =(binary1 % 10 + binary2 % 10 + remainder) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (remainder != 0) {
sum[i++] = remainder;
}
--i;
while (i >= 0) {
binaryprod = binaryprod * 10 + sum[i--];
}
return binaryprod;
}
最佳答案
您的全局变量尚未初始化。在adding()
函数中,您可以使用:
二进制1 =二进制3/10
二进制2 =二进制4/10
在binaryproduct函数中,有两个不用的本地int命名为binary3,binary4。
您应该使用唯一的名称,初始化变量,在不需要全局变量的地方使用局部变量。