#include <stdio.h>

int main()
{
        float num = 0;
        double sum = 0;
        int status,l;

        do
        {
                printf("请输入合法的数字：");
                do
                {
                        sum = sum + num;
                        status = scanf("%f", &num);
                        l++;
                } while (getchar() != '\n');
        } while(status == 1);

        printf("结果是：%.2lf\n", sum);

        return 0;
}
为什么输入非法时,dowhile那个跳不出来

#include <stdio.h>

int main()
{
    float num = 0;
    double sum = 0;
    int status=0,l=0;
    do
    {
        printf("请输入合法的数字：");
        sum = sum + num;
        status = scanf("%f", &num);
        l++;
        while (getchar() != '\n');
    }
    while(status == 1);
    printf("结果是：%.2lf\n", sum);
    return 0;
}

为什么我那个无法跳出呢

1楼和2楼的都无法跳出，这样改动就可以跳出了。l  不是1.

#include <stdio.h>
int main()
{
        float num = 0;
        double sum = 0;
        int status,l;
        do
        {
                printf("请输入合法的数字：");
                do
                {
                        sum = sum + num;
                        status = scanf("%f", &num);
                        l++;
                } while (getchar() != '\n');
        } while(status == l);
        printf("结果是：%.2lf\n", sum);
        return 0;
}

这里可能是我的锅没有讲明代码的要求，这个是实现在不输入非法数字时，程序一行可读几个数字，中间随便隔个符合或字母就行了。跳出就是输入非法数字时。