C - 基于用户决策的 do/while 循环无法正常工作

Question

我有一个问题，经过多次测试，我认为这是因为我不了解输入缓冲区的工作原理。

我有一个 while 循环，它应该继续迭代，直到用户输入“否”停止迭代。

我有两个问题。

1. 无论用户输入“否”或任何不等于"is"的内容，while 都不会停止迭代
2. 如您所见，第二个周期的输出有问题。该程序不会要求用户输入字符串并跳过该步骤，就像用户只需键入 ENTER 一样。

代码:

int foo = 0;

do{

  int i, cycles;
  char array[MAX_LENGTH+1];



  for(cycles=0; cycles < MAX_READ_CYCLES; cycles++){

    i=0;

    printf("\n\nEnter a string: ");

    char ch;
    while ((ch = getchar()) != '\n' && ch != EOF) {
      array[i] = ch;
      i++;
    }

    array[i] = '\0'; //string terminator


    printf("String you entered: %s\n", array);

    printf("\nDo you want to continue? 1: yes / 0: no \n");
    scanf("%d", &foo);

  }

} while( foo == 1);

输出

Enter a string: test
String you entered: test

Do you want to continue? 1: yes / 0: no
0

Enter a string: String you entered: 

Do you want to continue? 1: yes / 0: no
3

Enter a string: String you entered: 

Do you want to continue?

Answer 1

由于内部 for 循环，如果用户输入 "yes"，您的程序不会终止:

#include <stdio.h>
#include <string.h>
#define MAX_LENGTH 100
#define MAX_READ_CYCLES 100

int main() {
    int cycles = 0;
    char foo[4];
    do {
        char array[MAX_LENGTH + 1];

        printf("\n\nEnter a string: ");

        char ch;
        int i = 0;
        while ((ch = getchar()) != '\n' && ch != EOF) {
            array[i] = ch;
            i++;
        }

        array[i] = '\0'; //string terminator


        printf("String you entered: %s\n", array);

        printf("\nDo you want to continue?");
        scanf("%s", foo);

        cycles++;

        while ((ch = getchar()) != '\n' && ch != EOF); // force drop stdin

    } while (strcmp(foo, "yes") == 0 && cycles < MAX_READ_CYCLES);
}

另见 I am not able to flush stdin和 http://c-faq.com/stdio/stdinflush2.html