给一个C++函数传递一个类实例作为参数会引起调用这个类实例的析构函数。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<memory.h>

using namespace std;
class te
{
public:
te(char *chs)
{
if (chs == NULL)
{
return ;
}
m_str = (char *)malloc(strlen(chs) + 1);
sprintf(m_str, "%s", chs);
}

~te()
{
printf("\n call ~te()");
free(m_str);
m_str = NULL;
}

void show()
{
printf("\nsting=%s \n",m_str);
}

private:
char *m_str;

};

void func1(te p)
{
printf("\nin func1 start \n");
//te t1(p);
p.show();
printf("\nin func1 end \n");
}

main()
{

te t2("ddddd");
t2.show();
func1(t2);
printf("\n aaaa \n");
t2.show();
printf("\n cccc \n");
}

 

=======================

运行结果：

./a.out

sting=ddddd

in func1 start

sting=ddddd

in func1 end

call ~te()
aaaa

sting=

cccc

*** Error in `./a.out': double free or corruption (fasttop): 0x0000000001bd6010 ***
======= Backtrace: =========