C++ Integer從* char []

Question

中刪除字符串我正在嘗試將多個值讀入到我的C++程序中。

當我輸入1位數字（在我的代碼的底部），我很好。

但是，如果輸入2位數字，如「10」，則消息（我輸入的第二個東西）將被刪除。

這裏是我的代碼：

char * args[6]; 
unsigned time = 5; 
char input[5]; // for string input 
string message= "message"; 
//these strings and *chars are tempary strings for the purpose of reading in data 
string temp; 
char *temp2 = " "; 
char *temp3 = "empty pointer"; 

    args[count] = "-m"; 
    count ++; 

    //Prompt for the message 
    cout <<endl<<"Alright, Please enter your message: "<<flush; 
    getline(cin, message); 
    cout <<endl<<endl; 
    message.append("\""); 
    message = "\""+message; 
    //we can't use the string, so we copy it to temp3. 
    strcpy(temp3, message.c_str()); 
    //Now we input the string into our array of arguments 
    args[count] = temp3; 
    count ++; 


    cout <<"Please enter time "<<flush; 
    getline(cin,temp); 

    //validate input utnil its an actual second. 
    bool done = false; 
    while (done == false){ 
     for(unsigned i = 0; i < temp.length() & i < 5; i++){ 
      input[i] = temp[i]; 
     } 
    done = CheckInteger(input, input); 
     time = atoi(input); 
     if (done == true & time < 1) { 
      cout <<"Unable to use a number less than 1 seconds! "<<endl; 
      cout <<"Please enter the number of seconds? "<<flush; 
      done = false; 
     }else if (done == false){ 
      cout <<"Please enter the number of seconds? "<<flush; 
     }else{ 
     break; 
     } 
     getline(cin,temp); 
    } 
    cout <<endl<<endl; 
    time = atoi(input); 
    //timer argument 
    args[count] = "-t"; 
    count ++; 

    // enter the time need to comvert from int to string. 
    ostringstream convert; 
    convert <<time; 
    temp = convert.str(); 
    //need to convert from string to character 
    strcpy(temp2, temp.c_str()); 

    args[count] = temp2; 
    count ++; 

我怎樣才能解決這個問題？

Answer 1

strcpy(char* destination, const char* source)將source字符串複製到destination指向的數組中。但是，你在呼喚strcpy(temp3, message.c_str());試圖將字符串複製到指針常量字符串文字：char *temp3 = "empty pointer";，從而導致未定義行爲 [1]

變化temp3從指針到將與此字符串只是初始化數組字面：

char temp3[] = "empty pointer"; 

或甚至更好：使用std::string來代替。

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[1] C++ 03標準2.13.4字符串文字（選定部分）：

§1一個普通字符串文字已鍵入「的N陣列const char「和靜態存儲時間

§2試圖修改字符串文字是未定義的。