2017-12-02 163 views
-1

我定义在主函数的字符串数组,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?

+2

'read_arrays(competions,points,numOfCompetitors);'? – melpomene

+1

您尚未将内存分配给指针数组char *竞争者[SIZE];'的元素。因此'scanf'在scanf(“%s”,&*竞争者[cntr])中取得了一个不确定的指针;' - 应该是'scanf(“%s”,竞争者[cntr]);'无论如何,是一个指针数组。 –

回答

1

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]); 
} 
+1

'*&'是毫无意义的。另外,'scanf''%s'是一个bug(缓冲区溢出)。 – melpomene

+0

代码已更新,thx。 –

0

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]); 

}