1

我的方法有问题。我希望它接受一个字符串数组作为它的第一个参数，而不是一个向量字符串。但是，当我尝试使用字符串数组并在主函数中创建一个字符串时，我会遇到各种错误。我不知道我是否应该使用指向字符串数组的指针作为我的参数或只是一个字符串。有什么帮助吗？

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <sstream>
#include<iostream> 
using namespace std;
class UserName
{
    public:
    string newMember(string* exist, string newname) { 
    bool found = false;
    bool match = false;
    stringstream ss;
    string result;
    string othername;
    for(int i = 0; i < exist.size(); i++){
        if(exist[i] == newname){
            found = true;
        break;
        }
    }
    if(found){
        for(int x = 1;   ; x++){
            match = false;
        ss.str("");
        ss << newname << x;
        for(int i = 0; i < exist.size();i++){
            //cout << ss.str() << endl;
            othername = ss.str();
            if(exist[i] == othername){
                match = true;
            break;
            }
        }
        if(!match){
            result = ss.str();
            break;
        }
        }
        return result;
    }
    else return newname;
    }   
};
int main(){
    UserName u;
    string Database [4];
    Database[0] == "Justin";
    Database[1] == "Justin1";
    Database[2] == "Justin2";
    Database[3] == "Justin3";
    cout << u.newMember(Database, "Justin") << endl;
    return 0;
}
4

2 回答 2

6

不幸的是，C++ 中的数组是一种特殊情况，并且在许多方面表现得不像正确的值。几个例子：

void foo(int c[10]); // looks like we're taking an array by value.
// Wrong, the parameter type is 'adjusted' to be int*

int bar[3] = {1,2};
foo(bar); // compile error due to wrong types (int[3] vs. int[10])?
// No, compiles fine but you'll probably get undefined behavior at runtime

// if you want type checking, you can pass arrays by reference (or just use std::array):
void foo2(int (&c)[10]); // paramater type isn't 'adjusted'
foo2(bar); // compiler error, cannot convert int[3] to int (&)[10]

int baz()[10]; // returning an array by value?
// No, return types are prohibited from being an array.

int g[2] = {1,2};
int h[2] = g; // initializing the array? No, initializing an array requires {} syntax
h = g; // copying an array? No, assigning to arrays is prohibited

（取自这里）

如果您想要一个行为类似于正确值的数组，请使用std::array.

#include <array>
#include <string>

void foo(std::array<std::string, 10> arr) { /* ... */ }

int main() {
  std::array<std::string, 10> arr = {"Justin", "Justin1", "Justin2", "Justin3"};
  foo(arr);
}
于 2013-08-18T14:15:01.613 回答
4

使用如下：

std::string Database[] ={ "Justin", "Justin1", "Justin2","Justin3" };

newmember作为

string newMember(std::string exist[], std::size_t n, string newname)

替换exist.size()为n

在main：

cout << u.newMember(Database, 4,"Justin") << endl;

同样根据您编辑的帖子

运算符=与运算符不同==，第一个是赋值运算符（将其右侧的值赋给左侧的变量），另一个==是相等运算符

所以你需要用作：

Database[0] = "Justin";
Database[1] = "Justin1";
Database[2] = "Justin2";
Database[3] = "Justin3";
于 2013-08-18T14:15:53.380 回答