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expter — Sun, 26 Apr 2009 03:05:00 GMT

摘要: �E�住心态，�l�箋前进。�?nbsp; 阅读全文

expter 2009-04-26 11:05 发表评论

expter — Sun, 08 Mar 2009 06:05:00 GMT

1、关于初始化:

deque<string> Deque;// 一个deque容器

deque<string> Deque2(Deque.begin(),Deque.end()); // 一�U�初始化

copy(Deque.begin(),Deque.end(),back_inserter(Deque2) ); // 一�U�初始化

deque<string> Deque3(istream_iterator<string>(cin),istream_iterator<string>()); // 一�U�初始化

现在讨论�W�三�U�初始化方式�Q?br>

typedef istream_iteraor<string> (Func)();

deque<string> Deque3(istream_iteraotr<string> cin, Func);
看上��d��一个函数声明，

typedef istream_iteraor<string> (Func)();
deque<string> f(stream_iteraor<string> ,Func);
现在比较清晰了吧�Q�很像一个函数声明了�Q?br>
所以第三种初始化方式根本就不能做�Q何事�?

//如果真的要达到初始化的目标，可以通过括号来消除二义�?/span>
deque<string> Deque3( (istream_iterator<string> (cin)) , istream_iterator<string>() );

通过加括��h��提示�~�译器我们这里提供的是构造函数的参数而非参数声明�Q�因�?(istream_iterator (cin)) 不能解析为变量声明，�q�个更void f ( (int i)) 不可能是函数声明是一回事。多一�Ҏ(gu��)��号不可能解释为合法的声明语句�?br>

2.关于double 和float 区别

一个例�?

int main()
{

  double  x = 1e8;

  while(x >0)
      x--;

   return 0;
}

现在的编译器一�U�就可能�q�行完成�Q?br>如果吧doubel x = 1e8 ,换�ؓ float x = 1e8;
C++标准�?:
��点�c�d��有三�U? float ,double, long double,关于�_�ֺ�很显�?br>如果换成float �{�案很诡异，因�ؓ取决于float的精度，能否表达0 1e8区间的整�?br>
所以如果换�?float�Q�可能运行一�U�，也可能是��d�@环，因�ؓ如果float不能表达0 1e8区间的整敎ͼ� 因�ؓ其中 n-1 = n (�׃��点�_�ֺ�不够)
�?br>

3. 一些拼写代码错�?br>

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    int x = 1;
    for (int i =0 ; i < 100; i++)
    {
        ;
    }

    //  下面�q�行代码会干些什么？递增???????/
    ++x;
    std::cout << x << std::endl;
    return 0;
}

一个支持标准c++的编译器�{�案输出�?1
原因很诡异，因�ؓ有个有趣的是�Q�注释末��?#8216;??/’会被转换�?\' �Q�写个宏的都知道�q�个��是换行�W�，所�??/换�{换�ؓ\ �Q��?+x��q��当于�_�脓(chu��ng)到注释行的末��?nbsp;�Q�所�?+x.,不会起作用，是注释的一部分�?br>
在看一个例子：
一个符��h��?c++的编译器会报多少错误�?br>

struct X
{
    static bool f(int *p)
    {
        return p &&0[p] and not p[1:>>p[2];
    }
};

先一步一步的分析
1:0[p] 其实和p[0]一�?br>2: and not都是有效关键�?�Q�他们分别是&& �?
3::>竟然也是合法的，他是]的双字符写法�?/p>

所以这个语句会被解析�ؓ return p && p[0] && !p[1]>p[2] �Q�没有语法错误�?br>很诡异�?/p>

expter 2009-03-08 14:05 发表评论

expter — Mon, 22 Dec 2008 14:24:00 GMT

以前做过的笔试题�Q�都是自己曾�l�做�?br>

# include <iostream>
using namespace std;
#define Q(x)  x*x
int main()
{
    int a =10;
    int b = 2;
    int c = 1;

    a /=Q(b+c)/Q(b+c);

    cout << a <<endl;
    return 0;
}

因�ؓ 宏Q(x) 不规范，
所以结�?很迷惑�h

a /=Q(b+c)/Q(b+c);的执行顺序是从右到左

tmp = Q(b+c)/Q(b+c) = 2+1*2+1 / 2+1*2+1 =2+2+0+2+1=7 没有对宏加括��h��l�常�Ҏ(gu��)��犯错�Q?br>
a = a/tmp = 10/7 = 1

所�?nbsp; a = 1;

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

题目2
反�{一个字�W�串�Q�西山居�W�试题）�Q?br>题目�Q�写一个函敎ͼ�把一个以0字符�l�尾的字�W�串str中的'A'字符�U�d��str的末��！

分析�Q?br>1�Q�以旉��换空�?nbsp; 从后向前循环查找字符A
2�Q�以�I�间换时�?nbsp; 一�ơ��@�?

��法1代码�Q?br>

void Deal(char *str )
{
    int len = strlen(str);
    int a= len-1,b=0;
    if (str[a] != '0')
    {
        return;             //如果最后一个字�W�不�?0'
    }
    while (a>=0)
    {
        if (str[a] == 'A')
        {
            int n = a;
            while (n < len-b-1)
            {
                str[n] = str[n+1];  //
                n++;
            }
            str[n] ='A';
            b++;           //b累加�Q�表�C�有b个字�W�A
        }
        a--;
    }
}

��法2:

void Deal(char *str )
{
    int len = strlen(str);
    int a= len-1,b=0,i=0,j=0;
    if (str[a] != '0')
    {
        return;             //如果最后一个字�W�不�?0'
    }
    char * tmp = new char[len+1];
    while(i<len)
    {
        if(str[i] !='A')
            tmp[j++] = str[i];
        else
            b++;          //多少个字�W�A
        i++;
    }

    while (j<len)
    {
        tmp[j++] = 'A';
    }
    tmp[len] = '\0';
    strcpy(str,tmp);


    delete tmp;   //释放�I�间
    tmp = NULL;
}

PS:因�ؓ是直接手写的。。所以难免有错误�Q?br>现在修改了，对不起各位。。�?

expter 2008-12-22 22:24 发表评论

expter — Sat, 06 Dec 2008 02:27:00 GMT

1.指针�q�算

注意 �Q�每个例子不只是一个题目，可能有多个，�?span style="color: #000000;">//....................................来分开�?/span>

# include <iostream>
# include <stdio.h>
using namespace std;

enum string
{
    x1,x2,x3=10,x4,x5,
}X;

int main()
{
    //��试1

    cout << x1 << x5<    unsigned char *p1;
    unsigned long *p2;
    p1 = (unsigned char *)0x801000;
    p2 = (unsigned long *)0x801000;

    printf("%p\n",p1+5);
    printf("%p\n",p2+5);

    char * t [] = { "abccc","daa","caa"};
    char **bb = t;
    cout << t[1] <    cout << sizeof(t)<

//....................................
    int a[5] = {1,2,3,4,5};
    cout << *( (int *)(&a+1)-2) <<endl;

                int  i=0;
    int  j=1;
    long *p = NULL;
    char *c = NULL;
    if(j++ && i++)
        p++,c++;
    if(i++ || j++)
        p++,c++;

    cout << i <<" " << j << " " << *p << " " << (long *)c <<endl;

    //....................................
    int i =1;
    int c = (i++) + (i++) + (i++);

    cout << c <<" "<< i<<endl;

    return 0;
}

# include <iostream>
# include <stdio.h>
using namespace std;

void foo(int k)
{
    char tmp[100];
    //sprintf(tmp,"%d",k);
    printf("%d\n",tmp);
    foo(k-1);
}
int main()
{

    int  i= 0;
    int  x= 0;
    int  y= 0;
    printf("%d %d %d %d\n",x--,++y,i++,i++);
    char tmp[20]="hello\0\t\\\\world";
    cout<<tmp<<endl;
    cout << sizeof(tmp) << " " <<strlen(tmp)<<" "<< tmp<<endl;

    //....................................
    char arr[100]  = "hello world";
    int  *v = (int *)arr;
    v[0] = 0X61626364;
    v[1] = 0X41424344;
    v[2] = 0X31323334;
    v[3] = 0;
    cout<<arr<<endl;

    //....................................
    foo(3);
    return 0;
}

expter 2008-12-06 10:27 发表评论

一道静态变量的�W�试题目

expter — Thu, 23 Oct 2008 13:16:00 GMT

题目大概�q�样子。�?br>

# include <iostream>
using namespace std;

class base
{
public:
    base():m_val2(0)
    {
    }

    int &get1(){return m_val1;}
    int &get2(){return m_val2;}


    static int  getr1(int &a){return a++;}
    static int  getr2(int &a){return a++;}

private:
    static int m_val1;
    int        m_val2;
};
int base::m_val1 = 0;
void print(int val)
{
    static int count = 1;
    cout << count++ <<",val = "<<val<<endl;
}

int main()
{
   base o1;
   base o2;
   print(o1.getr1(o1.get1()));
   print(o1.get1());
   print(o2.get1());

   print(o1.getr1(o1.get1()));
   print(o1.get1());
   print(o2.get1());

   print(o1.getr1(o1.get1()));
   print(o1.get1());
   print(o2.get1());

   print(o1.getr2(o1.get2()));
   print(o1.get2());
   print(o2.get2());
    return 0;
}

主要考虑静态变量的存储�I�间。。�?br>
�{�案很简单。�?

expter 2008-10-23 21:16 发表评论

一道指针笔试题

expter — Mon, 20 Oct 2008 13:14:00 GMT

//已知�l�构�?/span>

struct node
{
    node * link;
    node * next;
};

struct test
{
    int  a1;
    int  a2;
    //
    node *no;
    int  aj;
    //..
};

现在有一函数
int Deal(node * link ,const int a);
假设知道link的地址�Q�且知道link是一个test的一个对�?tmp)的地址�?br>现在求test对象tmp的地址。�?br>
此题的关键就是看no前面有多��个int�Q�然后减��d��U�量卛_��。�?br>因�ؓtest的对齐字节�ؓ4个字节，然后no前面�?j-1)个int 变量。�?br>
所以此题目及node *p = link;
test *t = p - (j-1); //因�ؓ寚w��?个字�?br> 或�?br> int add 得到p的地址�?br> 则t的地址�?add-(j-1)*sizeof(int)

��试�E�序

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

struct node
{
    node * link;
    node * next;
};

struct test
{
    int  a1;
    int  a2;
    //
    node *no;
    int  aj;
    //..
};

int main()
{
    test *t = new test;  //
    node **a = &(t->no);
    printf("t = %p\na = %p\n",t,a);//输出t�Q�a的地址�Q?/span>
    printf("t = %p\n",a-2);//输入我们计算的地址

    return 0;
}

expter 2008-10-20 21:14 发表评论

expter — Mon, 20 Oct 2008 12:30:00 GMT

1# include <iostream>
2using namespace std;
3
4class base
5{
6public:
7    base()
8    {
9        cout <<" base::val "<<endl;
10        val = 12345;
11        print();
12    }
13    base(int x)
14    {
15        val = x;
16        cout << " "<<endl;
17    }
18    virtual void print()
19    {
20        cout << " base:print(), val = "<<val<<endl;
21    }
22    base & operator = (int x)
23    {
24        val = x;
25        cout <<" base::operator, val = "<<x<<endl;
26        return *this;
27    }
28    void relase()
29    {
30        cout << " base::relase" <<endl;
31    }
32    ~base()
33    {
34        cout << " ~base::val = "<< val <<endl;
35    }
36protected:
37    int  val;
38};
39
40class deiver:public base
41{
42public:
43    deiver()
44    {
45        cout << " deiver::val "<<endl;
46    }
47
48    deiver(int x)
49    {
50        val = x;
51        print();
52    }
53    void print()
54    {
55        cout << " deiver::print(), val =  "<< val <<endl;
56    }
57    void relase()
58    {
59        cout << " deiver::relase, val = "<<val<<endl;
60    }
61    deiver & operator = (int x)
62    {
63        val = x;
64        cout <<" deiver::operator, val = "<<x<<endl;
65        return *this;
66    }
67    ~deiver()
68    {
69        cout << " deiver::~deiver "<<endl;
70    }
71};
72void des(base * b)
73{
74    b->print();
75    b->relase();
76}
77deiver  de = 98765;
78int main()
79{
80        deiver  * dee = new deiver;
81    *(base *)dee = 100;
82    des(dee);
83
84    return 0;
85}
86

输出�l�果。。�?br>因�ؓde是全局变量�Q�所以先调用基类构造函敎ͼ�在调用子�cȝ��带int的构造方法。�?br>然后dee同理先基�c�L��造函敎ͼ�然后调用默认构造函数�?br>
最后dee强制转换为基�c�L��针进行复�Ӟ��Ҏ(gu��)��多态，调用基类的operator=�Ҏ(gu��)��
最后调用子�cȝ��print和父�cȝ��relase�Ҏ(gu��)��?br>
因�ؓprint是虚函数

�q�算�l�果�?br> base::val
base:print(), val = 12345
deiver::print(), val = 98765
base::val
base:print(), val = 12345
deiver::val
base::operator, val = 100
deiver::print(), val = 100
base::relase

expter 2008-10-20 20:30 发表评论