封裝一個C風格的緩衝區

Question

我有下面的代碼來說明C風格的字符串。這段代碼僅用於說明。構造函數正確地初始化實例，但是在閱讀MyString廢話時會回來。任何人都可以建議或解釋什麼是錯的？

#include <iostream> 
using namespace std; 

class MyString 
{ 
private: 
    char* Buffer; 
public: 
    //Constructor 
    MyString(const char* InitialInput) 
    { 
     char* Buffer = new char [4]; // Only three characters are allowed! 
             // It must end with '\0' or it is not a string 
     for(int i=0 ; i<3 ; ++i){ 
      Buffer[i]=InitialInput[i]; 
     } 
     Buffer[3]='\0';     // Now it is a string. 

     cout << "Constructed the string: " << Buffer << endl; 

    } 

    void ShowString() 
    { 
     cout << "This is the string: " << Buffer << endl; 
    } 
}; 

int main() { 
    MyString line1("abc"); // Only three characters are allowed! 
    MyString line2("def"); 

    cout << endl << "MyString objects: " << endl; 
    line1.ShowString(); 
    line2.ShowString(); 


    return 0; 
} 

這是回來的畫面

構建的字符串上：ABC

構建的字符串：DEF

MyString的對象：

這是字符串：ƒä [<°°）@

這是s tring：「ÿ（

Answer 1

問題是您在構造函數的本地作用域中定義了char *Buffer。因此，不使用數據成員，而是使用局部變量。這裏是正確的代碼

class MyString 
{ 
private: 
    char* Buffer; 
public: 
    //Constructor 
    MyString(const char* InitialInput) 
    { 
     //char* Buffer -> dont define here. If defined, this definition 
     //will hide the data member defintion 
     Buffer = new char [4]; // Only three characters are allowed! 
             // It must end with '\0' or it is not a string 
     for(int i=0 ; i<3 ; ++i){ 
      Buffer[i]=InitialInput[i]; 
     } 
     Buffer[3]='\0';     // Now it is a string. 

     cout << "Constructed the string: " << Buffer << endl; 

    } 

    void ShowString() 
    { 
     cout << "This is the string: " << Buffer << endl; 
    } 
}; 

int main() { 
    MyString line1("abc"); // Only three characters are allowed! 
    MyString line2("def"); 

    cout << endl << "MyString objects: " << endl; 
    line1.ShowString(); 
    line2.ShowString(); 


    return 0; 
}