gdb打印长值，监视使用rand()设置的变量

Question

我使用awatch使用gdb脚本来观察变量的变化

#!/bin/bash

# Compile
gcc -Wall -pedantic -g -o demo demo.c

# Exit on compile error
if [ $? -ne 0 ]; then
    exit
fi

# Overwrite the contents of trace.gdb with a title
echo "# Watch var" > trace.gdb

# Don't stop each time there is a pagination
echo "set pagination off" >> trace.gdb

# Set a breakpoint in order to read the var address
echo "break main" >> trace.gdb

# Run the debugger
echo "run" >> trace.gdb

# Set watchpoint
echo "awatch var" >> trace.gdb

# Don't stop on each watchpoint (just show the trace)
echo "commands" >> trace.gdb
echo "continue" >> trace.gdb
echo "end" >> trace.gdb

# Start monitoring
echo "continue" >> trace.gdb

# Exit the debugger
echo "quit" >> trace.gdb

# Run the generated script
gdb -quiet -command=trace.gdb demo

程序：

/* demo.c */
#include <stdio.h>

int main(void)
{
    int var = 0;

    for (int i = 0; i < 5; i++)
    {
        var++;
    }
    printf("%d\n", var);
    return 0;
}

这个结论似乎是正确的：

Hardware access (read/write) watchpoint 2: var

Value = 0
main () at demo.c:7
7       for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 0
New value = 1
main () at demo.c:7
7       for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 1
New value = 2
main () at demo.c:7
7       for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 2
New value = 3
main () at demo.c:7
7       for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 3
New value = 4
main () at demo.c:7
7       for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 4
New value = 5
main () at demo.c:7
7       for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Value = 5
0x0000555555555176 in main () at demo.c:11
11      printf("%d\n", var);
5

但是如果我使用rand()切换到下面的代码

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(void)
{
    int var = 0;

    srand((unsigned)time(NULL));
    for (int i = 0; i < 10; i++)
    {
        var = rand() % 10;
    }
    printf("%d\n", var);
    return 0;
}

并运行相同的脚本，gdb开始打印错误的值：

Breakpoint 1, main () at demo.c:6
6   {
Hardware access (read/write) watchpoint 2: var

Hardware access (read/write) watchpoint 2: var

Value = 0
main () at demo.c:9
9       srand((unsigned)time(NULL));

Hardware access (read/write) watchpoint 2: var

Old value = 0
New value = 32015002
0x00005555555551f8 in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Value = 32015002
0x00005555555551fb in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Old value = 32015002
New value = 7
main () at demo.c:10
10      for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 7
New value = 84992124
0x00005555555551f8 in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Value = 84992124
0x00005555555551fb in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Old value = 84992124
New value = 6
main () at demo.c:10
10      for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 6
New value = 55442740
0x00005555555551f8 in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Value = 55442740
0x00005555555551fb in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Old value = 55442740
New value = 0
main () at demo.c:10
10      for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 0
New value = 208731384
0x00005555555551f8 in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Value = 208731384
0x00005555555551fb in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Old value = 208731384
New value = 3
main () at demo.c:10
10      for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Old value = 3
New value = 114916873
0x00005555555551f8 in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Value = 114916873
0x00005555555551fb in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Old value = 114916873
New value = 9
main () at demo.c:10
10      for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var

Value = 9
0x0000555555555216 in main () at demo.c:14
14      printf("%d\n", var);
9

我读过这篇文章：gdb prints wrong values when modifying arguments和用-fvar-tracking编译，但它没有帮助。

为什么rand()会有这样的行为？

Answer 1

未经优化的gcc程序集可能会很奇怪：

        jmp     .L2
.L3:
        call    rand
        movl    %eax, %edx
        movslq  %edx, %rax
        imulq   $1717986919, %rax, %rax
        shrq    $32, %rax
        sarl    $2, %eax
        movl    %edx, %ecx
        sarl    $31, %ecx
        subl    %ecx, %eax
        movl    %eax, -4(%rbp)
        movl    -4(%rbp), %ecx
        movl    %ecx, %eax
        sall    $2, %eax
        addl    %ecx, %eax
        addl    %eax, %eax
        subl    %eax, %edx
        movl    %edx, -4(%rbp)
        addl    $1, -8(%rbp)
.L2:
        cmpl    $9, -8(%rbp)
        jle     .L3

看起来你在威胁-4(%rbp)。所以有一个“大数字”放在那里的movl %eax, -4(%rbp)，然后在movl -4(%rbp), %ecx中读取，然后在movl %edx, -4(%rbp)中放入% 10的结果。所以你在计算过程中看到了一些数字。即。一个循环对应于：

New value = 32015002
0x00005555555551f8 in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Value = 32015002
0x00005555555551fb in main () at demo.c:12
12          var = rand() % 10;

Hardware access (read/write) watchpoint 2: var

Old value = 32015002
New value = 7
main () at demo.c:10
10      for (int i = 0; i < 5; i++)

Hardware access (read/write) watchpoint 2: var