带字符串的C++关系运算符==
带字符串的C++关系运算符==
提问于 2017-04-30 10:16:02
我得到一个错误，我不能使用关系运算符"==“来测试字符串与字符串是否匹配。是否因为(字符串数组)而需要不同的运算符？
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;
}回答 3
Stack Overflow用户
发布于 2017-04-30 10:45:31
请考虑以下注意事项：
- 函数原型中不需要
&作为指向数组中第一个元素的指针将传递给 - 您应该将函数赋给一个变量以保存返回值: ex：
pos = searchArray(name, firstNameArray); - 函数调用中不需要
[]。 - 您必须为搜索结果添加输出。
- 同样不需要在数组参数中引用，即不需要
&。在ARR_SIZE. - There中，你应该使用
=而不是==. - You，应该使用
0以外的值来初始化位置，ARR_SIZE. - There不应该是
searchArray内部的 - 。
将您的代码与下面的工作代码进行比较：
#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;
}Stack Overflow用户
发布于 2017-04-30 12:55:20
更正/建议-有关评论的更多详细信息：
//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;
}Stack Overflow用户
发布于 2017-04-30 10:29:49
在searchArray(name, firstNameArray[7]);中，您只需传递一个字符串，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是不正确的。
我换了几个地方，希望这能有所帮助
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;
}参考文献
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/43702274
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