Question

int main()
{
    int num[20];

    printf("Enter 4 integers: ");

    int n = scanf("%d %d %d %d",&num[0],&num[1],&num[2],&num[3]);

        if(n != 4)
        {
            printf("\n");
            printf("Input must consist of 4 integers\n");
            exit(0);
        } else if(n == 4)
        {
            if(sol_for_24(num))
            {
                printf("\n");
                printf("Yes! 24 is reachable from { %d, %d, %d, %d }\n", num[0], num[1], num[2], num[3]);

            } else {
                 printf("\n");
                 printf(sol_for_24(num) ? "\n" : "Noooo :( 24 is unreachable from { %d, %d, %d, %d }\n",num[0],num[1],num[2],num[3]);
            }
            }

    return 0;
}

对于我到目前为止所写的上述主要功能代码，我想要进一步实现的是，如果我输入的数字超过或少于4倍，它应该打印出来并退出程序。但是，即使输入4位以上的数字（但只执行前4位前面的数字），它也会运行程序。我可以获得任何帮助或建议我应该如何解决这个问题吗？我真的在努力解决这个问题..谢谢。

Answer 1

您可以使用fgets和strtok代替scanf来实现您想要的内容。

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

int main()
{
    int num[20];

    printf("Enter 4 integers: ");

    char read[100];
    fgets(read, 100, stdin);

    int count = 0;
    char* p = strtok(read, " ");
    while (p != NULL) {
        if (count == 4) {
            printf("Your input is more than 4 integers.\n");
            exit(EXIT_FAILURE);
        }
        num[count++] = atoi(p);
        p = strtok(NULL, " ");
    }

    if (count < 4) {
        printf("Your input is less than 4 integers.\n");
        exit(EXIT_FAILURE);
    }

    // Do other stuffs....

    return 0;
}

<强>更新

请注意，上面的代码假定它只需要一行用户输入，它将计算该一行输入中的数字，并且用户小心不要输入除数字之外的任何其他内容。

虽然我认为这个假设对于一些简单的家庭作业来说是可行的，但它并不像其他人在评论中所说的那样强健。以下代码是strtol更强大的版本。请注意，它也没有使用strtok。（感谢David C. Rankin的评论！）

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

int main()
{
    int num[20];

    printf("Enter 4 integers: ");

    char read[100];
    fgets(read, 100, stdin);

    int count = 0;
    char* endptr = NULL;
    char* nextptr = read;

    do {
        num[count] = strtol(nextptr, &endptr, 10);

        if (errno == ERANGE) {
            printf("Number out of range.\n");
            exit(EXIT_FAILURE);
        }
    } while (nextptr != endptr                  // strtol found an integer
             && (++count < 20)                  // so we've read (++count) number of integers so far, also check for array out-of-bounds error
             && (nextptr = endptr)              // we'll continue to read from the place where the last strtol stopped
            );

    // while loop condition can be false on several cases
    // check if it were the case where there were non-number input
    char tmp[2];
    if (sscanf(nextptr, "%1s", tmp) != EOF) {   // %s ignores whitespaces(i.e. '\n', ' ', '\t')
        printf("Nothing other than numbers such as '%s' should be in the input\n", tmp);
        exit(EXIT_FAILURE);
    }

    // now we know that user input contained only numbers
    // check how many numbers were given
    if (count < 4) {
        printf("You've entered less than 4 numbers.\n");
        exit(EXIT_FAILURE);
    } else if (count > 4) {
        printf("You've entered more than 4 numbers.\n");
        exit(EXIT_FAILURE);
    }

    // We now have definitely 4 numbers! Do other stuffs with it!

    return 0;
}

Answer 2

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

int main(int argc, char *argv[])
{
    int num[20];
    int i = 0;

    if(argc != 5)
    {
        printf("\n");
        printf("Input must consist of 4 integers\n");
        exit(0);
    }

    while(argc != 1)
    {
        num[i] = atoi(argv[i+1]);
        argc--;
        i++;
    }


    if(sol_for_24(num))
    {
        printf("\n");
        printf("Yes! 24 is reachable from { %d, %d, %d, %d }\n", num[0], num[1], num[2], num[3]);
     }
     else
     {
        printf("\n");
        printf(sol_for_24(num) ? "\n" : "Noooo :( 24 is unreachable from { %d, %d, %d, %d }\n",num[0],num[1],num[2],num[3]);
     }

    return 0;
}

上面解决了你的问题。 Seb和John Forkosh在评论中指出了正确的方向。

Answer 3

如果您不能直接从主列表中使用argv []列表。我建议使用vscanf。

成功时，该函数返回成功填充的参数列表的项目数。

所以你可以用这样的东西替换你的scanf：

int my_scanf( const char * format, ... )
{
  int nb_filled = 0;

  va_list args;
  va_start (args, format);
  nb_filled = vscanf (format, args);
  va_end (args);

  return nb_filled;
}

并使用返回值来检查填充的args数。

Answer 4

如果用户输入的数量超过某些数字，则将计数循环增加一个并返回状态以添加约束使用if else，以获得简单的计数循环。

Answer 5

尝试阅读5个数字......

int tmp;
int n = scanf("%d%d%d%d%d", num, num+1, num+2, num+3, &tmp);
if (n != 4) {
    fprintf(stderr, "error\n");
    exit(EXIT_FAILURE);
}
/* ... rest of code ... */

计算C中的整数输入数

5 个答案: