代码跳过驱动程序 2 和 3 的名称输入，但为驱动程序 1 打印并为所有驱动程序打印 km 的输出为 0.00

Question

我正在编写代码来打印三个驱动程序的详细信息，但问题是程序正确地接受了驱动程序 1 的输入，但跳过了驱动程序 2 和 3 的名称输入，而且尽管提供了输入，它还是打印了 distance travelled 作为 0.0.

#include<stdio.h>
struct data
{
    char name;
    int dlo;
    char route;
    float km;
}s1,s2,s3;
void details(struct data *d)
{
    printf("Enter your name: ");
    scanf(" %s",&d->name);
    printf("Enter your driving license no: ");
    scanf("%d",&d->dlo);
    printf("Enter route taken: ");
    scanf("%s",&d->route);
    printf("Enter distance travelled: \n");
    scanf("%.2f",&d->km);
}
void output(struct data *d)
{
    printf("Driver name: %s\n", &d->name);
    printf("Driving license no: %d \n", d->dlo);
    printf("Route taken: %s\n", &d->route);
    printf("Distance travelled: %.2f \n", d->km);
}
int main()
{
        printf("\n--------------driver 1---------------\n");
        details(&s1);
        printf("\n");
        output(&s1);
        printf("\n--------------driver 2---------------\n");
        details(&s2);
        printf("\n");
        output(&s2);
        printf("\n--------------driver 3---------------\n");
        details(&s3);
        printf("\n");
        output(&s3);
    return 0;
}

输出是这样的：

--------------driver 1---------------
Enter your name: abhishek
Enter your driving licence no: 123
Enter route taken: abcd
Enter distance travelled: 
123

Driver name: abhishek
Driving licence no: 123
Route taken: abcd
Distance travelled: 0.00

--------------driver 2---------------
Enter your name: Enter your driving licence no: 321
Enter route taken: efgh
Enter distance travelled: 
45 

Driver name: 123
Driving licence no: 321
Route taken: efgh
Distance travelled: 0.00

Answer 1

问题是我在 scanf 中使用了 %.2f，这导致运行-时间错误，因此将 scanf("%.2f",&d->km); 更改为 scanf("%f",&d->km);错误已解决我还必须为 name 和 route 使用数组以避免进一步的运行-时间错误。

非常感谢大家的建议。

代码跳过驱动程序 2 和 3 的名称输入，但为驱动程序 1 打印并为所有驱动程序打印 km 的输出为 0.00

Code skips the input for the name of driver 2 & 3 but prints for driver 1 as well prints the output of km as 0.00 for all drivers

c

pointers

structure