关于无��(f��)时变量的变量��g��换的记录 && 求最大表�C�数

蓝尘 — Thu, 11 Jun 2009 07:15:00 GMT

源于无聊, 又再看了(ji��n)复习(f��n)�?ji��n)�?不用临时变量�q�行整型变量交换"

两种��行�Ҏ(gu��)��:
(1) 加减�?br>

1 void AsExchang(int &x, int &y)
2 {
3      x = x + y;
4      y = x - y;
5      x = x - y;
6 }
7

(2) 异或

1 void XorExchang(int &x, int &y)
2 {
3      x = x ^ y;
4      y = x ^ y;
5      x = x ^ y;
6 }

两种�Ҏ(gu��)��, 都可以认识是密码学的一部分�? 只是用加减法来加�? �q�稚且缺乏安全性而已.
当然, 异或法在现实中也得变�U�才能实际应�?br>
一开始傻��d��联想到方法一是否存在��界问题... //基础知识瞬间薄弱�?..
�pȝ��底层的实现居然一旉��忘记�?..

联想到如何确认一个类型能表示的最大最��?br>(1)
所以想到的用sizeof�q�算�W? 判断�c�d��的位�? 然后�W�笨地左�U�d��U�L��作进行计��?br>

int getMaxInt()
{
    int iBase= 1;
    const int IntSize = sizeof(int) * 8;

    //int是有�W�号�? 最左边一位是�W�号�?br>    //so
    for(int i = 1; i < IntSize - 1 ; i++ )
    {
        iBase = (iBase << i) + 1;
    }

    return iBase;
}

好笨的方�?..
很没效率

(2)
用强制�{换吧, (int)0x*,
*是什�? 0x7fffffff
恶心(j��)-_-||, 我一开始又4个f�? 一个f�?位吧...

(3)
�q�有用bitset, bitsit有to_ulong�Ҏ(gu��)��, 凑合着�?br>
(4)
climits�?INT_MAX
在cplusplus.com上面的reference是显�C?br>

INT_MAX Maximum value for an object of type int = 32767

被他的图表骗�?br>
VC6版本的头文�g里面是用
#define INT_MAX 2147483647
直接定义�?br>查看cywin里面的头文�g
也差不多..
#define __INT_MAX__ 2147483647
#define INT_MAX __INT_MAX__

(5)
C++标准�Ҏ(gu��)��..
//直接贴过来了(ji��n)
//前面有说明说numeric_limits::max()�{�于INT_MAX
//... 只是��装�?ji��n)模板而已

1 // numeric_limits example
2 #include <iostream>
3 #include <limits>
4 using namespace std;
5
6 int main () {
7   cout << boolalpha;
8   cout << "Minimum value for int: " << numeric_limits<int>::min() << endl;
9   cout << "Maximum value for int: " << numeric_limits<int>::max() << endl;
10   cout << "int is signed: " << numeric_limits<int>::is_signed << endl;
11   cout << "Non-sign bits in int: " << numeric_limits<int>::digits << endl;
12   cout << "int has infinity: " << numeric_limits<int>::has_infinity << endl;
13   return 0;
14 }

没去��N��手们的奇�? ��浪荡荡又消费了(ji��n)不少旉��...
�l�箋(hu��)潜修�?br>

蓝尘 2009-06-11 15:15 发表评论

蓝尘 — Wed, 25 Feb 2009 12:35:00 GMT

#include<iostream>
using std::cout;
using std::endl;
class CDust{
public:
    CDust()
    {
        cout << " CDust constructor " << endl;
    }

    ~CDust()
    {

        cout << " ~CFoo destructor " << endl;
    }
};

CDust A;

int main()
{
    return 0;
}

我想�c�M��的代码在�|�上Google下应该有不少

��试�?ji��n)�?在vc6.0的情况下,是没有输�?"~CFoo destructor", 但这�q�不代表 ~CDust() 没有执行.

在~CDust里面讄��断点,�?x��)发��C��实上�E�序�q�行时进入了(ji��n)析构函数�? // -_! 表达好牵�?/div>

本来惛_��在跟�t�这里开始程序发生了(ji��n)什么调用了(ji��n)什�?发现功底不��,完全不明�?��先打断�?/div>

而再�? return 0 ' 语句前面加上断点,�?x��)看到这个新加的断点比析构里面的断点先到�?...

以现在c++的造诣和vc6.0的了(ji��n)解情冉|��?头痛�?/div>

��Z��?return 0 后程序不是正常结束了(ji��n)才去执行全局对象的析�?

改写下代�?br>

#include<iostream.h>

class CDust{
public:
    CDust()
    {
        cout << " CDust constructor " << endl;
    }

    ~CDust()
    {

        cout << " ~CFoo destructor " << endl;
    }
};

CDust A;

int main()
{

    return 0;

}

�q�样用旧版本的头文�g实现,控制台输��Z��(ji��n)意外的两个调�?--- 析构函数输出昄��?.

是cout在头文�g的实现方式不�?

在dev-c++里面,�q�行�W�一代代�?br>�Ҏ(gu��)��断点的设�|�测�?也是先执行了(ji��n) main()函数里面�?return 0 才进入全局函数的析�?也能发现析构函数里面的输�?gu��)��调用�?控制台有明确的显�C?br>�q�样一�?又不明白�?..
在vc6.0里面��Z��么执行了(ji��n)全局函数的析构却没有所谓的输出?
是因为cmd控制台在'return 0'�E�序权限收回 // 好像扯到�pȝ��的一些�؜��q��旧记忆了(ji��n)...

�|�游�?ji��n)一�?br>扑ֈ�暂时比较清晰的说法是:
   In C++, the constructor of a global object is executed before the main() function(of course after the STARTUP code), while the destructor
is invoked after the main() function. So in my humble opinion, the main() function is a bridge between the constructor and the destructor.Further more, the constructor and the destructor is the actual manager of the whole program, because they can manage all the
resources of the program(for instance, the constructor allocate memory units and the destructor free them.I'am not sure of this, any comments will be appreciated in advance.).
4)In C++, is it true that the resources obtained during the constructor and the destructor (both belong to a global object)are managed by themselves and have nothing with the main() function.Therfore, I think the main() function is not the king in C++ as it is in C.
   //感谢提出此说法的朋友

_startup才是用户�E�序的�v点和�l�点? 的确,调用_exit()函数再断�Ҏ(gu��)��?全局对象的destructor是没有进入的��Z��(x��)   //长见识了(ji��n)

��此先打�?.. 再深入今晚就�q�样完了(ji��n).
��此记录�?以后再接�?br>

不小�?j��)又接触�?..
懒得开新的��集中在�q�里�?br>
在main()里面手动调用全局对象的析�? 最后程序都�?x��)执行两�ơ析构调�?.. �?lt;iostream.h>的cout�q�种情况下明�?在std::cout下还得靠断点讄��才能体现�?vc6的情�?
     �q�里是一�U�解�?br>     {
      Once a destructor is invoked for an object, the object no longer exists; the behavior is undefined if the destructor is invoked for an object whose lifetime has ended. [Example: if the destructor for an automatic object is explicitly invoked, and the block is subsequently left in a manner that would ordinarily invoke implicit destruction of the object, the behavior is undefined
}
     非global static object 也会(x��)出现两次调用, 区别只在与一个在main() 退��Z��?一个在之后...
(的确,手工调用析构函数的情况很��出�?nbsp; -_! )
     如果我在析构里面存在释放内存 �q�一�c�d��? 那第二次再次释放不是�Ҏ(gu��)��出问�?!!
     以后遇到�q�种情况得注意检��代码的��d��...

蓝尘 2009-02-25 20:35 发表评论

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关于无��(f��)时变量的变量��g��换的记录 && 求最大表�C�数