Linux stat64 这是什么风格

Question

你好朋友们，我跟踪我的安卓恢复 n 找到了这个

 open("/dev/tty0",
 O_RDWR|O_DSYNC|O_LARGEFILE) = 3
ioctl(3, KDSETMODE, 0x1)                = 0
stat64("/dev/late_display/control",
 {st_mode=S_IFCHR|0660,
 st_rdev=makedev(248, 0), ...}) = 0

write(2, "splash screen display", 21)   = 21
write(2, ": ", 2)                       = 2
write(2, "No such file or directory", 25) = 25
write(2, "n", 1)                       = 1
open("/dev/late_display/control",
 O_RDONLY|O_LARGEFILE) = 4
ioctl(4, SNDCTL_SEQ_SYNC, 0)            = 0

close(4)                                = 0
stat64("/dev/graphics/fb0",
 {st_mode=S_IFCHR|0660,
 st_rdev=makedev(29, 0), ...}) = 0

open("/dev/graphics/fb0",
 O_RDWR|O_LARGEFILE) = 4

我的问题是这是什么程序在 C 中这一行，我写得一样，但它在 { token 之前给出了错误的预期表达式，因此我的构建 CWM 无法找到 FB0 帧缓冲区。谢谢

 stat64("/dev/late_display/control",
 {st_mode=S_IFCHR|0660,
 st_rdev=makedev(248, 0), ...})

Answer 1

这是

使用stat

#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <stdio.h>
int main()
{
    struct stat buf;
    const char *filename = "/dev/late_display/control";
    if(stat(filename, &buf) != 0)
    {
        perror("stat");
        return 1;
    }
    printf("stat succeededn");
    return 0;
}

手册页是这些功能的绝佳参考man 2 stat