Question

我收到一个错误，我不能使用关系运算符“==”来测试字符串到字符串匹配。由于数组（字符串），是否需要不同的运算符？

int searchArray(string name, string &firstNameArray); // declares the function

int main()
{
    string firstNameArray[7] = { "Jim", "Tuyet", "Ann", "Roberto", "Crystal", "Valla", "Mathilda" }; //declares and intializes the array

    string name = ""; 
    cout << "What's your name?"; 
    getline(cin, name);

    searchArray(name, firstNameArray[7]); // using the function


    return 0;
}

int searchArray(string name, string &firstNameArray) { //defining the function


    int position = 0; //declaring and intializing the return variable - positions 0 thru 6 for array elements and position 7 for not in array

    for (int i = 0; i < 7; i++) { //looping through the array

        if (firstNameArray[i] == name) //**error code "no operator "==" matches these operands
        {
            position == firstNameArray[i];
        }
        else
        {
            position == 7;
        }

    }

    return position;

}

Answer 1

请考虑以下注释：

函数原型中不需要&，因为指向数组中第一个元素的指针将被传递
您应该将函数分配给变量以保存返回的价值：例如：pos = searchArray(name, firstNameArray);
函数调用中不需要[]。
您必须为搜索结果添加输出。
同样不需要数组参数中的引用，即No & 需要。

position == firstNameArray[i];

，您应该使用=而不是==。
您应该使用0到ARR_SIZE以外的值初始化位置。
else内不应有searchArray声明。

将您的代码与此工作代码进行比较：

#include <iostream>
using namespace std;

int searchArray(string , string [], int);

int main()
{
    const int ARR_SIZE = 7;

    string firstNameArray[ARR_SIZE] = { "Jim", "Tuyet", "Ann", "Roberto",
                                 "Crystal", "Valla", "Mathilda" };
    string name = "";
    cout << "\n What's your name? ";
    getline(cin, name);

    int pos = searchArray(name, firstNameArray, ARR_SIZE);

    if (pos == -1)
        cout << "\n Not Found!";
    else
        cout << "\n Fount at position " << pos;

    cout << "\n\n\n";

    return 0;
}

int searchArray(string name, string fNameArray[],const int SIZE) {

    int position = -1;

    for (int i = 0; i < SIZE; i++)
        if (fNameArray[i] == name)
            position = i;

    return position;
}

Answer 2

更正/建议 - 有关评论的更多详情：

    //const references
    //use of vector
    //use of vector size variable

    int searchArray(const string & name, const vector<string> & firstNameArray) { 
            int position = -1; //declaring and intializing the return variable - positions 0 thru 6 for array elements and position 7 for not in array

            for (int i = 0; i < firstNameArray.size(); i++) { //looping through the array
                //== is for comparison
                if (firstNameArray[i] == name)
                {
                    position = i; //= is for assignment
                    break; // without break always returns not found
                }
            }
            return position;
        }

        int main(int argc,const char * argv[]) {

            vector<string> firstNameArray = { "Jim", "Tuyet", "Ann", "Roberto", "Crystal", "Valla", "Mathilda" }; //declares and intializes the array

            string name ;
            cout << "What's your name?";

            getline(cin, name);

            cout << "position: " << searchArray(name, firstNameArray) << endl; 


            return 0;
        }

Answer 3

在searchArray(name, firstNameArray[7]);中，您只传递一个字符串，而不是字符串数组。在int searchArray(string name, string &firstNameArray)中，通过引用传递数组的语法不正确。

我换了几个地方，希望这会有所帮助

int searchArray(string name, string* firstNameArray); // declares the function

int main()
{
    string firstNameArray[7] = { "Jim", "Tuyet", "Ann", "Roberto", "Crystal", "Valla", "Mathilda" }; //declares and intializes the array

    string name = ""; 
    cout << "What's your name?"; 
    getline(cin, name);

    searchArray(name, firstNameArray); // right
    //searchArray(name, firstNameArray[7]); 
    // this will pass a single string, not a string array, the index is out of boundary, the range is 0-6


    return 0;
}

int searchArray(string name, string* firstNameArray) { //pass a pointer
//or int searchArray(string name, string (&firstNameArray)[7])

    int position = 0; //declaring and intializing the return variable - positions 0 thru 6 for array elements and position 7 for not in array

    for (int i = 0; i < 7; i++) { //looping through the array

        if (firstNameArray[i] == name) 
        {
            position = i; //not position == firstNameArray[i]; incompatible type, int and string.
        }
        else
        {
            position = 7;
        }

    }

    return position;

}

参考

Passing an array by reference

C ++关系运算符==带字符串

3 个答案:

参考