C++ Integer从* char []

Question

中删除字符串我正在尝试将多个值读入到我的C++程序中。

当我输入1位数字（在我的代码的底部），我很好。

但是，如果输入2位数字，如“10”，则消息（我输入的第二个东西）将被删除。

这里是我的代码：

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 ++; 

我怎样才能解决这个问题？

Answer 1

strcpy(char* destination, const char* source)将source字符串复制到destination指向的数组中。但是，你在呼唤strcpy(temp3, message.c_str());试图将字符串复制到指针常量字符串文字：char *temp3 = "empty pointer";，从而导致未定义行为 [1]

变化temp3从指针到将与此字符串只是初始化数组字面：

char temp3[] = "empty pointer"; 

或甚至更好：使用std::string来代替。

---

[1] C++ 03标准2.13.4字符串文字（选定部分）：

§1一个普通字符串文字已键入“的N阵列const char“和静态存储时间

§2试图修改字符串文字是未定义的。