Question

I want to take input of integer and want to check that input is only one integer. So I have implemented some logic. But it's in infinite loop.

逻辑是：如果scanf返回1表示匹配，否则返回非整数，并且该数字应大于5，因此我也添加了该检查。

#include <stdio.h>
int main()
{
    //Variable Declarations
    int a[1000],n,i,j,int_check=0;

    //Input for number of terms and it should be at least 5
    printf("Enter the number of terms for the array input:");
    int_check=scanf("%d",&n);
    printf("\nscanf=%d and n=%d\n",int_check,n);

    //Input Validity checking. If scanf returns 1 that means it's a match with the input. Number of terms should be greater than 5.
    while(int_check!=1 || n<5)
    {
    printf("Enter the valid number of terms for the array inout. It should be an integer and greater than 5:");
    int_check=scanf("%d",&n);   

    }

    return 0;

}

它应该提供一个输入屏幕。帮助我确定此逻辑中的错误。

Answer 1

如果您没有为scanf("%d",...);输入有效的整数，则会删除 not 无效的输入，因此您将在下一个scanf("%d",...);中得到它，因此(int_check!=1 || n<5)将永远不要虚假

您需要自己清除无效输入，警告不要使用fflush(stdin);，因为这仅适用于文件

请注意查看需要刷新的消息，例如，替换

    printf("Enter the valid number of terms for the array inout. It should be an integer and greater than 5:");

作者

    printf("Enter the valid number of terms for the array inout. It should be an integer and greater than 5:\n");

或在 printf 之后使用fflush(stdout);，是您希望保持一致吗

为什么它处于无限循环中？

1 个答案: