更新字符串数组用C

  • Z时代
  • 2024-01-10
  • 分类：问答

我定义在主函数的字符串数组" title="字符串数组">字符串数组，我想更新它的另一个函数内部如下：更新字符串数组用C

#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]); 
}

，但我得到了以下错误：

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)

我想在循环中用scanf分配字符串数组中的值。我如何设法做到这一点？

回答：

您可以在将它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）。

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

变为：read_arrays(competitors, points, numOfCompetitors);

全码：

#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]); 
}

回答：

作为一种替代丹尼尔ILLESCAS你可以只对每个竞争对手，你输入分配空间。一定要稍后释放它们。

#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]); 
}

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