当我运行以下代码时,
int x[4096];
int *y;
int *m;
void main()
{
y = (int*)malloc( 4096 * sizeof(int) );
m = (int*)mmap( 0, 4096 * sizeof(int), PROT_READ | PROT_WRITE,
MAP_PRIVATE | MAP_ANONYMOUS, -1, 0 );
printf( "x = %p, y = %p, m = %p\n\n", x, y, m );
system( "cat /proc/self/maps" );
}
我得到以下输出,
x = 0x601060, y = 0x606010, m = 0x7ffff7ff5000
00400000-0040b000 r-xp 00000000 08:01 655382 /bin/cat
0060a000-0060b000 r--p 0000a000 08:01 655382 /bin/cat
0060b000-0060c000 rw-p 0000b000 08:01 655382 /bin/cat
0060c000-0062d000 rw-p 00000000 00:00 0 [heap]
7ffff77b5000-7ffff7a59000 r--p 00000000 08:01 395618 /usr/lib/locale/locale-archive
7ffff7a59000-7ffff7bd3000 r-xp 00000000 08:01 1062643 /lib/libc-2.12.1.so
7ffff7bd3000-7ffff7dd2000 ---p 0017a000 08:01 1062643 /lib/libc-2.12.1.so
7ffff7dd2000-7ffff7dd6000 r--p 00179000 08:01 1062643 /lib/libc-2.12.1.so
7ffff7dd6000-7ffff7dd7000 rw-p 0017d000 08:01 1062643 /lib/libc-2.12.1.so
7ffff7dd7000-7ffff7ddc000 rw-p 00000000 00:00 0
7ffff7ddc000-7ffff7dfc000 r-xp 00000000 08:01 1062651 /lib/ld-2.12.1.so
7ffff7fd9000-7ffff7fdc000 rw-p 00000000 00:00 0
7ffff7ff9000-7ffff7ffb000 rw-p 00000000 00:00 0
7ffff7ffb000-7ffff7ffc000 r-xp 00000000 00:00 0 [vdso]
7ffff7ffc000-7ffff7ffd000 r--p 00020000 08:01 1062651 /lib/ld-2.12.1.so
7ffff7ffd000-7ffff7ffe000 rw-p 00021000 08:01 1062651 /lib/ld-2.12.1.so
7ffff7ffe000-7ffff7fff000 rw-p 00000000 00:00 0
7ffffffde000-7ffffffff000 rw-p 00000000 00:00 0 [stack]
ffffffffff600000-ffffffffff601000 r-xp 00000000 00:00 0 [vsyscall]
现在,我的问题是,为什么分配为malloc,mmap或全局数据的内存不在 / proc / self / maps 所示的任何映射范围内?
答案 0 :(得分:12)
当您运行system( "cat /proc/self/maps" );
时,您会显示cat进程的映射。您可以在输出的第一行看到它是映射的cat可执行文件:
00400000-0040b000 r-xp 00000000 08:01 655382 /bin/cat
您需要执行snprintf(..., "cat /proc/%d/maps", getpid());
,或打开/ proc / self / pid文件并在自己的过程中自行阅读。