指向本地范围之外 [英] Pointer to local outside scope
本文介绍了指向本地范围之外的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
一个函数返回堆栈变量,这将导致意外的程序行为,通常是在碰撞形式的地址。
下面的函数返回一个堆栈地址:
A function returns the address of a stack variable, which will cause unintended program behavior, typically in the form of a crash. The following function returns a stack address:
int init(char *device, DriverType driver)
{
int rv = -1;
if (autodetect) {
void *md;
const char *p = NULL;
char buf[PATH_MAX];
*device = 0;
md = discover_media_devices();
if (!md) {
fprintf (stderr, "open: Failed to open \"auto\" device");
if (*device)
fprintf (stderr, " at %s\n", device);
else
fprintf (stderr, "\n");
goto failure;
}
while (1) {
p = get_associated_device(md, p, MEDIA_V4L_RADIO, NULL, NONE);
if (!p)
break;
snprintf(buf, sizeof(buf), "/dev/%s", p);
device = &buf[0];
}
free_media_devices(md);
/* out_of_scope: Variable "buf" goes out of scope */
}
switch (driver) {
case DRIVER_ANY:
case DRIVER_V4L2:
default:
goto try_v4l2;
case DRIVER_V4L1:
goto try_v4l1;
}
try_v4l1:
dev = v4l1_radio_dev_new();
/* use_invalid: Using "device", which points to an out-of-scope variable "buf" */
rv = dev->init (dev, device);
----------------------------
try_v4l2:
dev = v4l2_radio_dev_new();
/* use_invalid: Using "device", which points to an out-of-scope variable "buf" */
rv = dev->init (dev, device);
----------------------------
failure:
return rv;
}
请帮助code来解决这个问题。
Please help to solve this issue in code
推荐答案
您大致有两种选择:
-
分配堆栈上的字符调用之前比的的init 的功能:
char ch[PATH_MAX];
init (ch, ...);
使用分配函数内部的字符的的malloc 的和免费分配的内存以外的的init 的功能。
Allocate the char inside the function using malloc, and free the allocated memory outside the init function.
int init(char *device, DriverType driver)
{
/*...*/
device = malloc(PATH_MAX);
/*...*/
}
char* p;
init (p, ...);
free(p);
第一个选项是更优雅和高效。
The first option is more elegant and efficient.
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