Question

我在Main中有一个函数（int，string）：

string word("HELLO");
int x = 0;
char choice;
swap(x, word);

我正在尝试传递以下函数，但没有成功：

void swap(int, string) {

int x = 0;
string word = "HELLO";

cout << "Would you like to change a letter? Please enter the letter 
position. " << endl;
cin >> x;

if (x == 1) {
    cout << "What do you want to change it to?" << endl;
    cin >> word[0];

我一直收到这个错误：

错误C2664＆＃39; void std :: swap（std :: exception_ptr＆amp;，std :: exception_ptr＆amp;）throw（）＆＃39;：无法从＆＃39; int＆＃39;转换参数1到＆＃39; std :: exception_ptr＆amp;＆＃39;

是什么给出了？

Answer 1

代码的主要问题是缩进。您的代码不可读，而且很难理解它。美化它。写出漂亮，可读和结构化的代码。您可以在以下链接中阅读有关缩进的更多信息。

https://en.wikipedia.org/wiki/Indentation_style

接下来是函数声明。在定义函数之前，不要声明函数。函数声明应该是main函数的顶部，函数的定义应该在main函数之下。您可以在以下链接中找到有关函数声明的更多信息：

http://en.cppreference.com/w/cpp/language/function

由于您没有使用char array打印string，因此通过循环访问string是没用的。添加<string>库并开始使用string类型。通过在string内传递std::cout变量就足以打印出string。

最后，由于您试图在string函数之外操作main变量，因此需要传递引用参数。

void myFunction(std::string& parameter);

这样，将改变主要内部或任何其他函数内部的原始变量。如果没有引用&，您尝试修改的值将不会更改。

以下链接演示了参考的使用。

http://www.learncpp.com/cpp-tutorial/73-passing-arguments-by-reference/

请阅读下面我为什么要应用某些更改的评论。我对change函数进行了狡猾的更改。您现在有资格为任何人工作任意大小的string类型。

#include <iostream>
#include <string> //When you are working on strings, use the string library.

using namespace std;

//Function declaration is very important. Have the declarations above main.
void change(string&);

int main() {
    string word("HELLO");
    char choice;

    cout << "The word is : " << endl;

    cout << word << endl;

    //No need for the for loop to print out the string as
    // we are working on a string and not a char array.
    //  for (int i = 0; i < word.length(); i++) {
    //      cout << word[i];
    //  }


    change(word);

    cout << "The new word is" << endl << word << endl;

    cout << "Would you like to enter another change ? Enter Y or N ? " << endl;

    cin >> choice;

    if (choice == 'y' || choice == 'Y') {
        change(word);
        cout << word << endl;
    }
    else {
        cout << "Good Bye" << endl;
    }

    system("pause");

    return 0;

}


//When your datatype is to be modified outside the function, use the reference
//parameter type '&'.
//Without the reference type, your modified version of the type will only be modified 
//inside that function.
//The original one will not be altered.

void change(string& word) {
   /*
    *  size_t is simply unsigned int, to work towards manipulation and accessing
    *  of string types, use unsigned int or std::size_t
    */
    size_t x = 0;

    cout << "Would you like to change a letter? Please enter the letter position. " << endl;
    cin >> x;

    //Check to see if the inputted value is within the string length range.
    if(x > 0 && x <= word.length())
        cout << "What do you want to change it to?" << endl;
    else{
        cout << "The entered position is outside the string size range\n";
        return; //Quit from the function if the condition is not met.
    }

    /*
     *   Instead of using if/else if statements,
     *   Just make a normal loop. Much simpler.
     */

     for(size_t i = 0; i < word.length(); i++){
         if((x-1) == i)
             cin >> word[i];
     }
}

如何在函数中传递字符串？

1 个答案: