如果结果是浮点数，如何停止划分？
我已经对此代码进行了编程，但是它部分向我展示了结果50/3向我展示16，而我想介绍以停止这种操作（如果浮动）。该怎么办？
#include <stdio.h>
int main(){
int number, divisor, x, y;
printf("Enter the Natural Number to find Divisors\n");
scanf("%d", &number);
printf("Enter the Natural Number to divide for\n");
scanf("%d", &x);
printf("Divisors of the number %d are \n", number);
y=number/x;
for (divisor = 1; divisor<=number;divisor++){
if((number%divisor)==0){
printf("%d\n", divisor);
}
else{
continue; }
}
printf("The result of the division is %d", y);
return 0;
}
i have programmed this code but it show me partially the result as example 50/3 show me 16 while i want to introduce to stop this kind of operations if float. How to do?
#include <stdio.h>
int main(){
int number, divisor, x, y;
printf("Enter the Natural Number to find Divisors\n");
scanf("%d", &number);
printf("Enter the Natural Number to divide for\n");
scanf("%d", &x);
printf("Divisors of the number %d are \n", number);
y=number/x;
for (divisor = 1; divisor<=number;divisor++){
if((number%divisor)==0){
printf("%d\n", divisor);
}
else{
continue; }
}
printf("The result of the division is %d", y);
return 0;
}
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评论（1）
这就是我必须阅读您的代码多远：
您认为由0的部门做什么？
但是让我们继续前进：
如果字符串常数为null，则该测试是永远不会的。
总体而言，您似乎正在尝试测试
X是素数。在这种情况下，您需要在2上启动循环并测试其余部分：This is how far I have to read your code:
What do you think a division by 0 should do?
But lets keep going:
This tests if the string constant is NULL, which can never be.
Overall it looks like you are trying to test if
xis a prime. In that case you need to start the loop at2and test the remainder: