c - 我声明指针的顺序在 C 中真的很重要吗？ getcwd() 问题

Question

在 Solaris 5.8 机器上，我有以下代码:

[非工作代码]

char *buf;
char *dir;
size_t psize;

psize = (size_t) 1024;
dir = getcwd(buf, psize);

在这台 unix 机器上，上面的代码不 工作，我在尝试运行该程序时遇到段错误。它只有在我声明 dir before buf 时才有效:

[工作代码]

char *dir;
char *buf;
...
dir = getcwd(buf, psize);

当使用另一种 Unix 版本(例如 Mac OS X)时，我没有得到任何关于如何编写代码的似乎非常严格的规则。任何人都可以解释上面的例子是怎么回事吗？谢谢!

Answer 1

这里来自getcwd(3):

DESCRIPTION     The getcwd() function copies the absolute pathname of the current working     directory into the memory referenced by buf and returns a pointer to buf.     The size argument is the size, in bytes, of the array referenced by buf.     If buf is NULL, space is allocated as necessary to store the pathname.     This space may later be free(3)'d.

That is - set the buf to NULL and free(3) the dir when done; OR allocate space for the buf yourself (since you are telling the getcwd(3) you have 1K there).

Edit:

So to clean it up a bit, it's either:

char *dir = getcwd( NULL, 0 );

if ( dir == NULL ) { /* handle error */ }
/* use dir */
free( dir );

或

char buf[1024]; /* or allocate it with malloc(3) */

if ( getcwd( buf, 1024 ) == NULL ) { /* handle error */ }

/* use buf, DO NOT free it if it's on the stack or static, 
   only if malloc-ed */