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工作学习 / IT技术讨论 / 国内一家公司招聘C++ Programmer的面试题:以下程序预期得结果15, 15, 16，但实际得结果15, 16, 16，为何？
// start of code
#include <stdio.h>
class X
{
public:
int *px;
X ( int init )
{
px = new int; *px = init;
}
~X () {
delete px;
}
};
void print ( X x )
{
printf ("%d＼n", *x.px);
}
int main ()
{
X x(15); print(x);
X y(16); print(x); print(y);
return 0;
}
// end of code
-xxjjs(东方射日);
2001-8-10
{393}
(#162423@0)
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把对象直接当参数传递，函数里用完就释放了
-000(老九);
2001-8-10
(#162466@0)
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说得再清楚一些X x(15);
print(x);
当到这里完成打印时，delete px 被调用，因为是传对象，函数结束
时要执行一次释放动作。假设指针px经过new之后占有单元0x30009876,那么delete px 后，0x30009876又是第一个可用单元。
X y(16);
px 又指向(占有)0x009876单元， *px=16使这个单元存放值16。
print(x);
*x.px 实际上就是引用0x009876单元里面的值。你说这个值现在是多少呢？
print(y);
实际上在VC++中不工作，得一UNDEFINE的值。
-hannibal(汉尼拔);
2001-8-10
{417}
(#162697@0)
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Sorry
-hannibal(汉尼拔);
2001-8-10
{513}
(#162804@0)
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Since this code has memory leak, I don't think 15,16,16 is guranteed, it depends on the compiler
-flipflop(翻云覆雨);
2001-8-11
(#163311@0)
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you are correct
-hannibal(汉尼拔);
2001-8-11
(#163321@0)
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完全正确！这在UNIX下编译是可以的。修改的方法是将print中参数形式改为const &或const *.或者定义拷贝构造函数，因为题中调用缺省的拷贝构造函数是将成员一一拷贝，而不拷贝指针指向的数值。
-xxjjs(东方射日);
2001-8-12
(#164139@0)
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it is caused by memory destruction
-jfhuang(Zigzag);
2001-8-11
{774}
(#163023@0)
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请问这句话怎么理解?allocates a new memory for its member px at the same address as in object x and stores 16 in it.since allocate a new memory then how come "at the same address" ?
1- why at the same address?
2- how to fix the problem??
thanks!
-bigdesk(NightLight);
2001-8-11
{138}
(#163180@0)
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if you read my article.......when you call int *a =new int(10);,a points to a first available memory unit 0x12345678 which stores 10.
when you call delete a, memry unit 0x12345678 becomes first available memory unit again.
when you call int *b= new (11);, b points to 0x12345678 again.
is that clear?
-hannibal(汉尼拔);
2001-8-11
{283}
(#163295@0)
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so it seems allocating the same address is decided by the run time? why i did not see it in books before when i read the part of dynmaic memory management? i' ll check in again. thanks!
-bigdesk(NightLight);
2001-8-11
(#163383@0)
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It's heap management, a block of the size int was just deleted (first print(x)), then u try to alloc the same size (Y(16)), sure the heap manager will give u the same addr.
-flipflop(翻云覆雨);
2001-8-11
{540}
(#163310@0)
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I agree with you. This is a very bad and wrong program. It will cause memory destruction when print(y) is called.
-xiliu(xiliu72);
2001-8-11
(#163326@0)
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one way to fix itvoid print ( X& x )
{
printf ("%d\n", *x.px);
}
-hannibal(汉尼拔);
2001-8-11
{56}
(#163331@0)
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So always keep this in mind: when there is a pointer member in class, define your own copy constructor and overload assignment operator.
-goodman(goodman);
2001-8-11
(#163347@0)
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good comments. because you will never precisely know when
your object will be copied silenty, secretly
-hannibal(汉尼拔);
2001-8-11
(#163352@0)
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good point
-000(老九);
2001-8-11
(#163354@0)