为什么编译器会产生4字节负载而不是1字节负载,而较宽的负载可能会访问未映射的数据？

Question

我有一个填充了可变长度记录的字节缓冲区,其长度由记录的第一个字节决定.用于读取单个记录的缩减版C函数

void mach_parse_compressed(unsigned char* ptr, unsigned long int* val)
{
    if (ptr[0] < 0xC0U) {
        *val = ptr[0] + ptr[1];
        return;
    }

  *val = ((unsigned long int)(ptr[0]) << 24)
      | ((unsigned long int)(ptr[1]) << 16)
      | ((unsigned long int)(ptr[2]) << 8)
      | ptr[3];
}

生成汇编(x86_64上的GCC 5.4 -O2 -fPIC),首先在ptr加载4个字节,将第一个字节与0xC0进行比较,然后处理两个,即四个字节.未定义的字节被正确丢弃,但为什么编译器认为首先加载四个字节是安全的？由于ptr没有例如对齐要求,因此它可能指向存储页面的最后两个字节,这是我们所知道的未映射的存储页面的最后两个字节,从而导致崩溃.

再生都需要-fPIC和-O2或更高.

我在这里错过了什么吗？编译器是否正确执行此操作以及如何解决此问题？

我可以通过mmap/mprotect获得上面显示的Valgrind/AddressSanitiser错误或崩溃:

//#define HEAP
#define MMAP
#ifdef MMAP
#include <unistd.h>
#include <sys/mman.h>
#include <stdio.h>
#elif HEAP
#include <stdlib.h>
#endif

void
mach_parse_compressed(unsigned char* ptr, unsigned long int* val)
{
    if (ptr[0] < 0xC0U) {
        *val = ptr[0] + ptr[1];
        return;
    }

    *val = ((unsigned long int)(ptr[0]) << 24)
        | ((unsigned long int)(ptr[1]) << 16)
        | ((unsigned long int)(ptr[2]) << 8)
        | ptr[3];
}

int main(void)
{
    unsigned long int val;
#ifdef MMAP
    int error;
    long page_size = sysconf(_SC_PAGESIZE);
    unsigned char *buf = mmap(NULL, page_size * 2, PROT_READ | PROT_WRITE,
                              MAP_PRIVATE | MAP_ANONYMOUS, -1, 0);
    unsigned char *ptr = buf + page_size - 2;
    if (buf == MAP_FAILED)
    {
        perror("mmap");
        return 1;
    }
    error = mprotect(buf + page_size, page_size, PROT_NONE);
    if (error != 0)
    {
        perror("mprotect");
        return 2;
    }
    *ptr = 0xBF;
    *(ptr + 1) = 0x10;
    mach_parse_compressed(ptr, &val);
#elif HEAP
    unsigned char *buf = malloc(16384);
    unsigned char *ptr = buf + 16382;
    buf[16382] = 0xBF;
    buf[16383] = 0x10;
#else
    unsigned char buf[2];
    unsigned char *ptr = buf;
    buf[0] = 0xBF;
    buf[1] = 0x10;
#endif
    mach_parse_compressed(ptr, &val);
}

MMAP版本:

Segmentation fault (core dumped)

与Valgrind:

==3540== Process terminating with default action of signal 11 (SIGSEGV)
==3540==  Bad permissions for mapped region at address 0x4029000
==3540==    at 0x400740: mach_parse_compressed (in /home/laurynas/gcc-too-wide-load/gcc-too-wide-load)
==3540==    by 0x40060A: main (in /home/laurynas/gcc-too-wide-load/gcc-too-wide-load)

与ASan:

ASAN:SIGSEGV
=================================================================
==3548==ERROR: AddressSanitizer: SEGV on unknown address 0x7f8f4dc25000 (pc 0x000000400d8a bp 0x0fff884e56c6 sp 0x7ffc4272b620 T0)
    #0 0x400d89 in mach_parse_compressed (/home/laurynas/gcc-too-wide-load/gcc-too-wide-load+0x400d89)
    #1 0x400b92 in main (/home/laurynas/gcc-too-wide-load/gcc-too-wide-load+0x400b92)
    #2 0x7f8f4c72082f in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x2082f)
    #3 0x400c58 in _start (/home/laurynas/gcc-too-wide-load/gcc-too-wide-load+0x400c58)

AddressSanitizer can not provide additional info.
SUMMARY: AddressSanitizer: SEGV ??:0 mach_parse_compressed

与Valgrind的HEAP版本:

==30498== Invalid read of size 4
==30498==    at 0x400603: mach_parse_compressed (mach0data_reduced.c:9)
==30498==    by 0x4004DE: main (mach0data_reduced.c:34)
==30498==  Address 0x520703e is 16,382 bytes inside a block of size 16,384 alloc'd
==30498==    at 0x4C2DB8F: malloc (vg_replace_malloc.c:299)
==30498==    by 0x4004C0: main (mach0data_reduced.c:24)

与ASan的堆栈版本:

==30528==ERROR: AddressSanitizer: stack-buffer-overflow on address
0x7ffd50000440 at pc 0x000000400b63 bp 0x7ffd500003c0 sp
0x7ffd500003b0
READ of size 4 at 0x7ffd50000440 thread T0
    #0 0x400b62 in mach_parse_compressed
CMakeFiles/innobase.dir/mach/mach0data_reduced.c:15
    #1 0x40087e in main CMakeFiles/innobase.dir/mach/mach0data_reduced.c:34
    #2 0x7f3be2ce282f in __libc_start_main
(/lib/x86_64-linux-gnu/libc.so.6+0x2082f)
    #3 0x400948 in _start
(/home/laurynas/obj-percona-5.5-release/storage/innobase/CMakeFiles/innobase.dir/mach/mach0data_test+0x400948)

谢谢

编辑:添加了实际崩溃的MMAP版本,澄清了编译器选项

编辑2:报告为https://gcc.gnu.org/bugzilla/show_bug.cgi?id=77673.要解决此问题asm volatile("": : :"memory");,请在if语句解决问题后插入编译器内存屏障.感谢大家!

Answer 1

恭喜！您发现了真正的编译器错误！

您可以使用http://gcc.godbolt.org来探索不同编译器和选项的汇编输出。

对于 x86 64 位 linux 的 gcc 版本 6.2，使用gcc -fPIC -O2，您的函数确实会编译为不正确的代码：

mach_parse_compressed(unsigned char*, unsigned long*):
    movzbl  (%rdi), %edx
    movl    (%rdi), %eax   ; potentially incorrect load of 4 bytes
    bswap   %eax
    cmpb    $-65, %dl
    jbe     .L5
    movl    %eax, %eax
    movq    %rax, (%rsi)
    ret
.L5:
    movzbl  1(%rdi), %eax
    addl    %eax, %edx
    movslq  %edx, %rdx
    movq    %rdx, (%rsi)
    ret

您正确地诊断了问题，并且该mmap示例提供了良好的回归测试。gcc过于努力地优化这个函数，结果代码肯定是不正确的：对于大多数 X86 操作环境来说，从未对齐的地址读取 4 个字节是可以的，但读取数组末尾则不行。

如果不跨越 32 位甚至 64 位边界，编译器可能会假设读取超过数组末尾的内容是可以的，但这种假设对于您的示例来说是不正确的。malloc如果分配的块足够大，则可能会发生崩溃。malloc用于mmap非常大的块（默认 IRCC >= 128KB）。

请注意，此错误是在 5.1 版编译器中引入的。

clang另一方面没有这个问题，但代码在一般情况下似乎效率较低：

#    @mach_parse_compressed(unsigned char*, unsigned long*)
mach_parse_compressed(unsigned char*, unsigned long*):         
    movzbl  (%rdi), %ecx
    cmpq    $191, %rcx
    movzbl  1(%rdi), %eax
    ja      .LBB0_2
    addq    %rcx, %rax
    movq    %rax, (%rsi)
    retq
.LBB0_2:
    shlq    $24, %rcx
    shlq    $16, %rax
    orq     %rcx, %rax
    movzbl  2(%rdi), %ecx
    shlq    $8, %rcx
    orq     %rax, %rcx
    movzbl  3(%rdi), %eax
    orq     %rcx, %rax
    movq    %rax, (%rsi)
    retq