c - getenv() 的值在 strtok() 中不起作用

Question

char *p = strtok (argv[1], ",")    #  Works perfectly
char *p = strtok (getenv("somestring"), ","); # does not work

在我的程序中，我采用 argv[1] 的值，该值以 "x,y" 格式传递。当未给出 argv[1] 时，我的程序应该从
获取值getenv("somestring") 也返回 "x,y"之后我使用 strtok 解析它们。

我不明白为什么 argv[1] 和 getenv() 的行为方式相同，因为如果我没有记错的话，它们的数据类型相同

Answer 1

摘自 getenv 中的注释手册:

As typically implemented, getenv() returns a pointer to a string within the environment list. The caller must take care not to modify this string, since that would change the environment of the process.

当 strtok 修改字符串时，您必须复制 getenv 返回的字符串，然后使用该副本调用 strtok:

char *str, *ptr;
char *p = getenv("somestring");
str = malloc(strlen(p) + 1);
strcpy(str, p);
ptr = strtok(str, ",");

// Make sure to deallocate the memory once you are done using it.
free(str);

您也可以使用strdup:

char *str, *ptr;
char *p = getenv("somestring");
str = strdup(p);
ptr = strtok(str, ",");

// Make sure to deallocate the memory once you are done using it.
free(str);