gdb watchpoint为什么a--，还不会触发

#include<stdio.h>
void printNum(int a)
{
    printf("printNum\n");
    while(a > 0)
    {
        printf("%d\n",a);
        a--;
    }
}
void printNum2(int a,int num)
{
    printf("printNum\n");
    while(a > num && a>0)
    {
        printf("%d\n",a);
        a--;
    }
}
int div(int a,int b)
{
    printf("a=%d,b=%d\n",a,b);
    int temp = a/b;
    return temp;
}
int main(int argc,char *argv[])
{
    printNum2(12,5);
    printNum(10);
    div(10,0);
    return 0;
}

 

 

 

gdb watchpoint 看a。为什么a--，还不会触发。

答：

watchpoint不应该用run吗

一行一行的执行肯定停止啊

这就是个条件断点
watchpoint就是条件断点，当然是run了，条件满足就暂停

n一行一行没有意义啊

 

---------------------------------------------------
关于gdb的watchpoint再说一点

watchpoint不能检测内核区变量，比如网络协议栈，数据库链接，文件io

因为gdb是运行在用户区的，所以不能监控内核区变量

---------------------------------------------------

 例子《gdb的watchpoint在系统调用中被修改似乎不会被触发》

https://blog.csdn.net/imred/article/details/82563599

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实际打印

(gdb) n
printNum
14 while(a > num && a>0)
(gdb) c
Continuing.
12
Hardware watchpoint 2: a

Old value = 12
New value = 11
printNum2 (a=11, num=5) at test.cpp:14
14 while(a > num && a>0)
(gdb) c
Continuing.
11
Hardware watchpoint 2: a

Old value = 11
New value = 10
printNum2 (a=10, num=5) at test.cpp:14
14 while(a > num && a>0)
(gdb) c
Continuing.
10
Hardware watchpoint 2: a

Old value = 10
New value = 9
printNum2 (a=9, num=5) at test.cpp:14
14 while(a > num && a>0)
(gdb)

-----------------------------------------------------

 

问2：watchpoint如果监视一个结构体，有指针，是深监视还是浅监视？

待解。