Question

我正在进行课程作业（非评分）并且不清楚为什么这段代码导致我的程序“挂起”而不是在循环中运行。

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int nbStars = 0;        // User defines the number of stars to display
    int nbLines = 0;        // User defines the number of lines on which to print

    // Obtain the number of Stars to display
    printf("Enter the number of Stars to display (1-3): ");
    scanf("%d", &nbStars);
    getchar();

    //   Limit the values entered to between 1 and 3
    do {
        printf("Enter the number of Stars to display (1-3): ");
        scanf("%d", &nbStars);

        if (nbStars < 1 || nbStars > 3) puts("\tENTRY ERROR:  Please limit responses to between 1 and 3.\n");
    } while (nbStars < 1 || nbStars > 3);
}

Answer 1

输出通常是行缓冲的，如果不打印新行（"\n"），则不会看到任何输出。你的程序没有被挂起，它只是在等待输入。

注意：如果你在循环中使用do，为什么在循环之前要求输入？即使输入良好，您的程序也会进入循环。即使没有do，它也会有效，因为nbStars已初始化为0。

while (nbStars < 1 || nbStars > 3) {
    printf("Enter the number of Stars to display (1-3): \n");
    scanf("%d", &nbStars);

    if (nbStars < 1 || nbStars > 3) puts("\tENTRY ERROR:  Please limit responses to between 1 and 3.\n");
}

Answer 2

必须有其他事情发生，因为你的代码适用于带有GCC的Linux和带有GCC的Windows 7 cygwin。您能否提供有关您正在使用的输入和环境的更多详细信息？

尝试使用此代码查看是否有不同的行为：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int nbStars = 0;        // User defines the number of stars to display
    int nbLines = 0;        // User defines the number of lines on which to print

    // Obtain the number of Stars to display
    do
    {
        printf("Enter the number of Stars to display (1-3): ");
        scanf("%d", &nbStars);

        if (nbStars < 1 || nbStars > 3)
        {
            puts("\tENTRY ERROR:  Please limit responses to between 1 and 3.\n");
        }
    }while (nbStars < 1 || nbStars > 3);

    printf("You entered %d\n", nbStars);
    return( 0 );
}

while循环导致程序挂起

2 个答案: