将字符指针数组传递给C中的函数？

Question

我有以下代码：

int main(){
    char **array;
    char a[5];
    int n = 5;

    array = malloc(n *sizeof *array);

    /*Some code to assign array values*/

    test(a, array);

    return 0;
}

int test(char s1, char **s2){
    if(strcmp(s1, s2[0]) != 0)
        return 1;

    return 0;
}

我正在尝试将char和char指针数组传递给函数，但上面的代码会导致以下错误和警告：

temp.c: In function ‘main’:
temp.c:6:5: warning: implicit declaration of function ‘malloc’ [-Wimplicit-function-declaration]
temp.c:6:13: warning: incompatible implicit declaration of built-in function ‘malloc’ [enabled by default]
temp.c:10:5: warning: implicit declaration of function ‘test’ [-Wimplicit-function-declaration]
temp.c: At top level:
temp.c:15:5: error: conflicting types for ‘test’
temp.c:15:1: note: an argument type that has a default promotion can’t match an empty parameter name list declaration
temp.c:10:5: note: previous implicit declaration of ‘test’ was here
temp.c: In function ‘test’:
temp.c:16:5: warning: implicit declaration of function ‘strcmp’ [-Wimplicit-function-declaration]

我正在努力理解问题是什么。

Answer 1

匿名用户

你有几个问题。首先是原型错误。a的数据类型在传递给函数时衰减为char指针，因此您需要：

int test (char* s1, char** s2) { ... }

然而，即使您解决了这个问题，当您第一次使用它时，测试声明也不在范围内。您应该提供一个原型：

int test (char* s1, char** s2);

在main之前，或者简单地将整个定义（函数）移动到main之前。

此外，不要忘记#include字符串. h和stdlib. h标头，以便strcmp和malloc的原型也可用。

Answer 2

错误1

temp.c:6:13: warning: incompatible implicit declaration of 
built-in function ‘malloc’ [enabled by default]

你是这个意思吗？

array = malloc(n * sizeof(*array));

错误2

temp.c:15:5: error: conflicting types for ‘test’
temp.c:15:1: note: an argument type that has a default promotion can’t 
             match an empty     parameter name list declaration
temp.c:10:5: note: previous implicit declaration of ‘test’ was here

您正在传递数组a的第一个元素的地址：

 test(a, array);

因此，函数签名应该是：

int test(char* s1, char** s2)

Answer 3

首先，您应该包含必要的头文件。对于strcmp，您需要

temp.c: In function ‘test’:
temp.c:20:5: warning: passing argument 1 of ‘strcmp’ makes pointer from integer without a cast [enabled by default]
/usr/include/string.h:143:12: note: expected ‘const char *’ but argument is of type ‘char’

这表明test（）应该有一个char*作为第一个参数。总之，您的代码应该如下所示：

#include <string.h>      /* for strcmp */
#include <malloc.h>      /* for malloc */

int test(char*,char**);  /* added declaration */    

int main(){
    char **array;
    char a[5];
    int n = 5;

    array = malloc(sizeof(*array));
    array[0] = malloc(n * sizeof(**array));

    /*Some code to assign array values*/

    test(a, array);

    free(*array); /* free the not longer needed memory */
    free(array);

    return 0;
}

int test(char * s1, char **s2){ /* changed to char* */
    if(strcmp(s1, s2[0]) != 0) /* have a look at the comment after the code */
        return 1;

    return 0;
}

请注意，strcmp可以处理以null结尾的字节字符串。如果s1和s2都不包含空字节，则调用测试将导致分段错误：

[1]    14940 segmentation fault (core dumped)  ./a.out

要么确保两者都包含空字节，要么使用strncmp并更改测试的签名：

int test(char * s1, char **s2, unsigned count){
    if(strncmp(s1, s2[0], count) != 0)
        return 1;
    return 0;
}

/* don' forget to change the declaration to 
      int test(char*,char**,unsigned)
   and call it with test(a,array,min(sizeof(a),n))
*/

您的内存分配也是错误的。数组是一个char**。您为*数组分配内存，它本身就是一个char*。您永远不会为此特定指针分配内存，您缺少数组[0]=malloc（n*sizeof（**数组））：

array = malloc(sizeof(*array));
*array = malloc(n * sizeof(**array));