Question

我今年夏天刚刚开始使用操作系统课程，我们需要使用C语言进行编程。

我只用C ++编码，所以我是C的新手。我们的第一个任务是输入多个字符串的简单提示。我们最多只能读取10个字符串。

我的问题是scanf()没有等待循环中的用户输入，程序只是在我有机会输入任何内容之前显示提示10次。我已经读过，因为scanf()在缓冲区中有换行符，所以我在%c语句scanf()之前添加了一个额外的空格，但是没有解决问题。

#include <stdio.h>

int main (void)
{
    int SIZE = 10;
    char* states[SIZE];
    int cnt = 0;

    printf ("Here\n");
    while (cnt < 10)
    {
        printf ("Enter State and Pop: ");
        scanf (" %c ", &states[cnt]);
        cnt++;
        printf ("%d\n", cnt);
    }
    return 0;
}

Answer 1

char* states[SIZE]; //pointers not allocated with memory
scanf (" %c ", &states[cnt]); //wrong format specifier and argument

这是未定义的行为，因为您使用了错误的格式说明符，并且还发送了错误的参数以便在字符串中进行扫描
以及更多指向字符串的指针，即states未分配内存

为避免这种情况，首先为states数组

中的每个字符串分配内存

包含stdlib.h标题文件

为states

中的每个指针分配内存并返回地址

for(int i=0; i<10; i++)
{
    states[i] = malloc(max_size_of_string);
}

或者你可以用这种方式动态分配内存：

char states[10][max_size_of_strings];

最后以这种方式使用scanf：

scanf (" %(size_of_string-1)[^\n]", states[cnt]);

，即如果你将max_size_of_string指定为10则使用这种方式：

scanf (" %9[^\n]", states[cnt]);

这是为了避免覆盖终止空字符

Answer 2

这是一个工作示例

// for printf() and scanf()
#include <stdio.h>
// for malloc() and free()
#include <stdlib.h>

// Global constants should be define'd
#define MAX_STRING_SIZE 100
#define STACK_SIZE 10

int main()
{
  // pointer to an array of strings (roughly)
  char *states[STACK_SIZE];
  // used as the iterator throughout
  int i;
  // allocate memory for every string in the array "states"
  for (i = 0; i < 10; i++) {
    // Normally sizeof(char) == 1, but that's how it works in general:
    // the number of elements times the size of on element, be it
    // a char (1) an integer (2,4 or 8 these times) or some
    // complicated struct
    states[i] = malloc(MAX_STRING_SIZE * sizeof(char));
    // check if it actually happened
    if (states[i] == NULL) {
      // the error message should explain a bit more than just:
      fputs("malloc() failed\n", stderr);
      // free the already allocated memory
      while (i >= 0) {
        free(states[i--]);
      }
      // exit with a return code signaling an error
      exit(EXIT_FAILURE);
    }
  }
  // reset iterator
  i = 0;
  // I would have used a for-loop, but it's a matter of taste, I guess
  while (i < 10) {
    printf("Enter State and Pop: ");
    // scan 99 characters of the input between start of line and line-end
    // one less than MAX_STRING_SIZE to leave room for the string-end '\0'

    // scanf() returns the number of characters read or EOF in case of an error
    // You might find it useful at one time.
    scanf(" %99[^\n]", states[i]);
    // Print what we just gathered
    printf("State %d is %s\n", i, states[i]);
    // increment iterator, if that is not obvious--but sometimes it isn't
    i++;
  }
  // Just a string to stdout, so puts(). The function printf() is too expensive for that
  puts("All states:");
  for (i = 0; i < 10; i++) {
    printf("State %d is %s\n", i, states[i]);
    // normally not necessary, the OS does it here at the end main but
    // it should always be the habit of a good programmer to clean up at the end
    free(states[i]);
  }
  exit(EXIT_SUCCESS);
}

Answer 3

第一个getchar（）获取输入字符串，第二个getchar获取Enter键u按;）并且它是：

#include <stdio.h>
#define SIZE 10

int main(void)
{
    char* states[SIZE];
    int cnt = 0;

    printf("Here\n");
    while (cnt < 10)
    {

        printf("Enter State and Pop: ");
        states[cnt] = (char*)getchar();
        getchar();
        cnt++;
        printf("%d\n", cnt);
    }
    return 0;
}

scanf不等待while循环中的用户输入

3 个答案: