Question

我编写了这段代码，所有要做的就是输入整数，直到输入否定为止：

int result = 0, input_number = 0;
while(input_number >= 0){
    printf("Please enter students IDs(negetive to stop)= ");
    result = scanf("%d", &input_number);

    if(result != 1){
        printf("Invalid input! try again...\n");
        continue;
    }
}

当输入为int-一切正常时，

但是当输入一个char（任何char不是int）时，

循环变得无穷无尽，scanf不再请求任何输入。

不建议我使用替换扫描，因为它是学校的工作，必须使用scanf。

输入char输出后：

Please enter students IDs(negetive to stop)= 4
Please enter students IDs(negetive to stop)= 5
Please enter students IDs(negetive to stop)= r
Please enter students IDs(negetive to stop)= Invalid input! try again...
Please enter students IDs(negetive to stop)= Invalid input! try again...
Please enter students IDs(negetive to stop)= Invalid input! try again...
Please enter students IDs(negetive to stop)= Invalid input! try again...
Please enter students IDs(negetive to stop)= Invalid input! try again...
Please enter students IDs(negetive to stop)= Invalid input! try again...
...

Answer 1

读取字符串，％s是格式说明符，然后在代码中转换为int。 atoi（）（或strtol（），如果你是C ++）将工作。

http://www.cplusplus.com/reference/cstdlib/atoi/（C或C ++）

http://www.cplusplus.com/reference/cstdlib/strtol/（C ++）

无效输入时忽略C scanf

1 个答案: