Question

我在main函数中定义了一个字符串数组，我想在另一个函数中更新它，如下所示：

#include <stdio.h>

#define SIZE 15

void read_arrays(char *competitors[SIZE], float points[SIZE], int numOfCompetitors)
{
        for (int cntr = 0; cntr < numOfCompetitors; cntr++)
        {
                printf("Enter the name of competitor %d", cntr+1);
                scanf("%s", &*competitors[cntr]);
                printf("Enter the point of competitor %d", cntr+1);
                scanf("%f", &points[cntr]);
        }
}

int main()
{
        char *competitors[SIZE];
        float points[SIZE];
        int numOfCompetitors = 0;
        while (numOfCompetitors > 15 || numOfCompetitors < 1)
        {
                printf("Enter the number of competitors: ");
                scanf("%d", &numOfCompetitors);
                if (numOfCompetitors > 15) printf("Number of competitors cannot be more than 15!\n");
        }

        read_arrays(&*competitors[SIZE], &points[SIZE], numOfCompetitors);
        printf("%f", points[0]);

}

但是我收到以下错误：

cc     homework2.c   -o homework2
homework2.c: In function ‘main’:
homework2.c:28:14: warning: passing argument 1 of ‘read_arrays’ from incompatible pointer type [-Wincompatible-pointer-types]
  read_arrays(&*competitors[SIZE], &points[SIZE], numOfCompetitors);
              ^
homework2.c:5:6: note: expected ‘char **’ but argument is of type ‘char *’
 void read_arrays(char *competitors[SIZE], float points[SIZE], int numOfCompetitors)

我想在循环中使用scanf分配字符串数组中的值。我怎么能做到这一点？

Answer 1

您可以在将变量传递给函数时使用变量的名称，还需要指定char矩阵的大小（〜字符串数组）。

所以：read_arrays(&*competitors[SIZE], &points[SIZE], numOfCompetitors);

成为：read_arrays(competitors, points, numOfCompetitors);

完整代码：

#include <stdio.h>

#define SIZE 15

void read_arrays(char competitors[SIZE][30], float points[SIZE], int numOfCompetitors)
{
    for (int cntr = 0; cntr < numOfCompetitors; cntr++)
    {
        printf("Enter the name of competitor %d", cntr+1);
        // We read up to 29 characters => no overflow as the size is up to 30
        scanf("%29s", competitors[cntr]);
        printf("Enter the point of competitor %d", cntr+1);
        scanf("%f", &points[cntr]);
    }
}

int main()
{
    char competitors[SIZE][30];
    float points[SIZE];
    int numOfCompetitors = 0;
    while (numOfCompetitors > 15 || numOfCompetitors < 1)
    {
        printf("Enter the number of competitors: ");
        scanf("%d", &numOfCompetitors);
        if (numOfCompetitors > 15) printf("Number of competitors cannot be more than 15!\n");
    }

    read_arrays(competitors, points, numOfCompetitors);

    printf("%s", competitors[0]);
    printf("%s", competitors[1]);
    printf("%f", points[0]);
}

Answer 2

作为Daniel Illescas的替代品，您可以为您输入的每个竞争对手分配空间。请务必稍后释放它们。

#include <stdio.h>
#define SIZE 15

void read_arrays(char *competitors[SIZE], float points[SIZE], int numOfCompetitors)
{
    for (int cntr = 0; cntr < numOfCompetitors; cntr++)
    {
        competitors[cntr] = (char*)calloc(1, 32);
        printf("Enter the name of competitor %d", cntr + 1);
        scanf("%s", competitors[cntr]);
        printf("Enter the point of competitor %d", cntr + 1);
        scanf("%f", &points[cntr]);
    }
}

int main()
{
    char *competitors[SIZE];
    float points[SIZE];
    int numOfCompetitors = 0;
    while (numOfCompetitors > 15 || numOfCompetitors < 1)
    {
        printf("Enter the number of competitors: ");
        scanf("%d", &numOfCompetitors);
        if (numOfCompetitors > 15) printf("Number of competitors cannot be more than 15!\n");
    }

    read_arrays(competitors, points, numOfCompetitors);
    printf("%f", points[0]);

}

更新C中函数内的字符串数组

2 个答案: