如何在没有分段错误的情况下将双指针传递给函数C语言 [英] How to pass a double pointer to a function without segmentation fault C language

问题描述

我正在尝试将双指针作为函数的参数传递，但我看不出为什么发生分段错误...

I'm trying to pass a double pointer as an argument to a function and I can't see why the segmentation fault happen...

这是函数:

void create_path_list(char *path_, char ***path) {
// Convert the path (string) into a list of directories
   char *token = NULL;
   int i = 0;

   *path = (char **) realloc(*path, (i + 1) * sizeof(char *));
   (*path)[i] = (char *) malloc(2);
   strcpy((*path)[0], "/");
   for(token = strtok(path_,"/"), i = 1; token != NULL; token = strtok(NULL, "/"), ++i)
   { 
     *path = (char **) realloc(*path, (i + 1) * sizeof(char *));
     (*path)[i] = (char *) malloc(sizeof(token) + 1);
     strcpy((*path)[i], token);
   }
}

主要内容:

int main(){
   char **path = NULL;
   create_path_list("/dir1/dir2/dir3/file.txt", &path);
   return 0;
}

推荐答案

sizeof(token)

将给出令牌的大小，它是一个指针.这样将不会为复制整个字符串分配足够的空间

Will give the size of token, which is a pointer. That will not allocate enough space to copy for the entire string

malloc(sizeof(token) + 1);
strcpy((*path)[i], token);

您应该用strlen替换sizeof

You should replace sizeof with a strlen

您正在将字符串文字传递给函数，然后尝试使用strtok()对其进行更改.您将必须传递一个可变字符串.

You are passing a string literal to you function and then try to change it with strtok(). You will have to pass a mutable string.

char str[] = "/dir1/dir2/dir3/file.txt" ;
create_path_list( str , &path);

我也看不出如何知道指针分配的数组有多大.您将必须返回大小或使用NULL终止数组.

Also I don't see how can you know how large is your allocated array if pointers. You will have to either return the size or NULL terminate the array.

将最后一个元素设置为null:

Set the last element to null:

 *path = (char **) realloc(*path, (i + 1) * sizeof(char *));
 (*path)[i] = NULL ;

并在函数外部打印

for( size_t i = 0 ; path[i] ; i++ )
{
    printf("%s" , path[i] ) ;
}

这篇关于如何在没有分段错误的情况下将双指针传递给函数C语言的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！