我试图用lstat获取/ proc / pid / exe文件的大小(以字节为单位)。这是我的代码:
int main(int argc, char *argv[])
{
struct stat sb;
char *linkname;
ssize_t r;
if (argc != 2)
{
fprintf(stderr, "Usage: %s <pathname>\n", argv[0]);
exit(EXIT_FAILURE);
}
if (lstat(argv[1], &sb) == -1)
{
perror("lstat");
exit(EXIT_FAILURE);
}
printf("sb.st_size %d\n", sb.st_size);
exit(EXIT_SUCCESS);
}
似乎sb.st_size总是等于0,我不明白为什么。另外,此示例是从readlink(2)手册页中提取的。
编辑:我试图让它在openSUSE上运行。
提前感谢您的帮助。
答案 0 :(得分:6)
/proc中的文件不是普通文件。对于他们中的大多数人来说,stat()
等。返回.st_size == 0
。
特别是,/proc/PID/exe
实际上不是符号链接或硬链接,而是一个特殊的伪文件,其行为主要是,如符号链接。
(如果需要,您可以检测检查.st_dev
字段的procfs文件。例如,与从.st_dev
获得的lstat("/proc/self/exe",..)
进行比较。)
要根据其PID获取特定execubtable的路径,我建议采用依赖于readlink()
的返回值的方法:
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <errno.h>
#include <stdio.h>
/* Creative Commons CC0: Public Domain dedication
* (In jurisdictions without public domain, this example program
* is licensed under the Creative Commons CC0 license.)
*
* To the extent possible under law, Nominal Animal has waived all
* copyright and related or neighboring rights to this example program.
*
* In other words, you are free to use it in any way you wish,
* but if it breaks something, you get to keep all the pieces.
*/
/** exe_of() - Obtain the executable path a process is running
* @pid: Process ID
* @sizeptr: If specified, the allocated size is saved here
* @lenptr: If specified, the path length is saved here
* Returns the dynamically allocated pointer to the path,
* or NULL with errno set if an error occurs.
*/
char *exe_of(const pid_t pid, size_t *const sizeptr, size_t *const lenptr)
{
char *exe_path = NULL;
size_t exe_size = 1024;
ssize_t exe_used;
char path_buf[64];
int path_len;
path_len = snprintf(path_buf, sizeof path_buf, "/proc/%ld/exe", (long)pid);
if (path_len < 1 || path_len >= sizeof path_buf) {
errno = ENOMEM;
return NULL;
}
while (1) {
exe_path = malloc(exe_size);
if (!exe_path) {
errno = ENOMEM;
return NULL;
}
exe_used = readlink(path_buf, exe_path, exe_size - 1);
if (exe_used == (ssize_t)-1)
return NULL;
if (exe_used < (ssize_t)1) {
/* Race condition? */
errno = ENOENT;
return NULL;
}
if (exe_used < (ssize_t)(exe_size - 1))
break;
free(exe_path);
exe_size += 1024;
}
/* Try reallocating the exe_path to minimum size.
* This is optional, and can even fail without
* any bad effects. */
{
char *temp;
temp = realloc(exe_path, exe_used + 1);
if (temp) {
exe_path = temp;
exe_size = exe_used + 1;
}
}
if (sizeptr)
*sizeptr = exe_size;
if (lenptr)
*lenptr = exe_used;
exe_path[exe_used] = '\0';
return exe_path;
}
int main(int argc, char *argv[])
{
int arg;
char *exe;
long pid;
char dummy;
if (argc < 2 || !strcmp(argv[1], "-h") || !strcmp(argv[1], "--help")) {
printf("\n");
printf("Usage: %s [ -h | --help ]\n", argv[0]);
printf(" %s PID [ PID ... ]\n", argv[0]);
printf("\n");
return 0;
}
for (arg = 1; arg < argc; arg++)
if (sscanf(argv[arg], " %ld %c", &pid, &dummy) == 1 && pid > 0L) {
exe = exe_of((pid_t)pid, NULL, NULL);
if (exe) {
printf("Process %ld runs '%s'.\n", pid, exe);
free(exe);
} else
printf("Process %ld: %s.\n", pid, strerror(errno));
} else {
printf("%s: Invalid PID.\n", argv[arg]);
return 1;
}
return 0;
}
上面,exe_of()
函数返回伪符号链接/proc/PID/exe
指向的动态分配副本,也可以选择存储分配的大小和/或路径长度。 (上面的示例程序不需要它们,所以它们是NULL。)
这个想法非常简单:为大多数情况分配一个足够大的初始动态指针,但不是非常大。保留字符串结尾NUL字节的最后一个字节。如果readlink()
返回的大小与给它的缓冲区长度相同 - 它不会添加一个终止字符串结尾的NUL字节 - 那么缓冲区可能太短;丢弃它,分配更大的缓冲区,然后重试。
同样,如果您希望阅读/proc/
下的伪文件的完整内容,则无法首先使用lstat()
/ stat()
来了解您可能需要多大的缓冲区;你需要分配一个缓冲区,尽可能多地读取,并在必要时重新分配一个更大的缓冲区。 (我也可以展示示例代码。)
有问题吗?
答案 1 :(得分:1)
无法对我要回复的帖子发表评论,对不起!
Valgrind将在Nominal Animal的帖子中报告exe_path = malloc(exe_size);
,因为'块肯定会丢失'。所以,如果你正在使用那个功能,那么你的记忆力会很快上升。
即使返回NULL也释放内存将解决此问题。当你使用-Wall
时,还应该添加来自malloc调用的显式转换(char *)以使gcc停止抱怨char* exe_of(const pid_t pid, size_t *const sizeptr, size_t *const lenptr)
{
char *exe_path = NULL;
size_t exe_size = 1024;
ssize_t exe_used;
char path_buf[64];
unsigned int path_len;
path_len = snprintf(path_buf, sizeof path_buf, "/proc/%ld/exe", (long)pid);
if (path_len < 1 || path_len >= sizeof path_buf) {
errno = ENOMEM;
return NULL;
}
while (1) {
exe_path = (char*)malloc(exe_size);
if (!exe_path) {
errno = ENOMEM;
return NULL;
}
exe_used = readlink(path_buf, exe_path, exe_size - 1);
if (exe_used == (ssize_t)-1) {
free(exe_path);
return NULL;
}
if (exe_used < (ssize_t)1) {
/* Race condition? */
errno = ENOENT;
free(exe_path);
return NULL;
}
if (exe_used < (ssize_t)(exe_size - 1))
break;
free(exe_path);
exe_size += 1024;
}
/* Try reallocating the exe_path to minimum size.
* This is optional, and can even fail without
* any bad effects. */
{
char *temp;
temp = (char*)realloc(exe_path, exe_used + 1);
if (temp) {
exe_path = temp;
exe_size = exe_used + 1;
}
}
if (sizeptr)
*sizeptr = exe_size;
if (lenptr)
*lenptr = exe_used;
exe_path[exe_used] = '\0';
return exe_path;
}
答案 2 :(得分:0)
一般来说,你不必担心这个问题,但在使用Nominal Animal评论后,似乎linux通过PAGE_SIZE来限制来自/ proc / PID / exe的文件路径。因此,即使文件系统支持比这更长的路径,也无法获得readlink()来为您提供该路径,因为这是作为硬限制实现的。我找到了另一种方式,所以IF readlink与ENAMETOOLONG失败,你可以读/ proc / PID / maps,虽然这可能会被击中或错过,我发现它击中了4个不同发行版中的每个进程除了系统进程没有实际文件名(ENOENT)
int getExecutableFromMaps(char *buf, size_t bufsize) {
FILE *fp;
char *mylinebuf = NULL;
size_t mylinebufsize = 0;
size_t counter = 0;
size_t start = 0;
size_t column = 0;
int result = -1;
fp = fopen("/proc/self/maps", "r");
if( fp != NULL ) {
if( getline(&mylinebuf, &mylinebufsize, fp) >= 0 ) {
if( mylinebuf != NULL ) {
while( column < 5 && counter < mylinebufsize ) {
while( counter < mylinebufsize && mylinebuf[counter] != ' ') {
counter++;
}
while( counter < mylinebufsize && mylinebuf[counter] == ' ') {
counter++;
}
column++;
}
int start = counter;
while( counter < mylinebufsize && (mylinebuf[counter] != '\n' && mylinebuf[counter] != '\r') ) {
counter++;
}
if( counter <= mylinebufsize && start < counter && (counter-start+1)<=bufsize ) {
memcpy(buf, &mylinebuf[start], counter-start);
buf[counter-start+1] = 0;
result = counter-start+1;
}
}
}
if( mylinebuf != NULL ) {
free(mylinebuf);
}
fclose(fp);
}
return result;
}
注意:这是补充代码,只应在readlink失败后调用,因为这很昂贵。