Sar*_*man 3 c malloc bash gcc strtok
我想解析Linux的$ PATH变量,然后将用':'分隔的目录名保存到字符串数组中.
我知道这是一个简单的任务,但我被困住了,任何帮助都会很好.
到目前为止我的代码是这样的,但有些东西是不对的.
char **array;
char *path_string;
char *path_var = getenv("PATH");
int size_of_path_var = strlen(path_var);
path_string = strtok(path_var, ":");
while (path_string != NULL) {
ss = strlen(path_string)
array[i] = (char *)malloc(ss + 1);
array[i] = path_string; //this is actually all i want to do for every path
i++;
path_string = strtok(NULL, ":");
}
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您的代码有两个主要问题,评论几乎总结如下:
getenv
)让我提出一个不使用strtok 的工作实现,从而允许检测空路径(并将其替换.
为Jonathan暗示的).使用gcc -Wall -Wwrite-strings
以下命令编译时没有任何警告:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
const char **array;
const char *orig_path_var = getenv("PATH");
char *path_var = strdup(orig_path_var ? orig_path_var : ""); // just in case PATH is NULL, very unlikely
const char *the_dot = ".";
int j;
int len=strlen(path_var);
int nb_colons=0;
char pathsep = ':';
int current_colon = 0;
// first count how many paths we have, and "split" almost like strtok would do
for (j=0;j<len;j++)
{
if (path_var[j]==pathsep)
{
nb_colons++;
path_var[j] = '\0';
}
}
// allocate the array of strings
array=malloc((nb_colons+1) * sizeof(*array));
array[0] = path_var; // first path
// rest of paths
for (j=0;j<len;j++)
{
if (path_var[j]=='\0')
{
current_colon++;
array[current_colon] = path_var+j+1;
if (array[current_colon][0]=='\0')
{
// special case: add dot if path is empty
array[current_colon] = the_dot;
}
}
}
for (j=0;j<nb_colons+1;j++)
{
printf("Path %d: <%s>\n",j,array[j]);
}
return(0);
}
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操作细节:
;
)和标记.
.可以显示警告,告诉用户这不安全.