如何摆脱“Intellisense：没有合适的转换函数从”std :: string“到“std :: string *”存在“错误？ [英] How do I get rid of the "Intellisense: no suitable conversion function from "std::string" to "std::string *" exists" error?

问题描述

这里是我的代码，

#include <iostream>
#include <string>

using namespace std;

int main()
{
const int NUM_NAMES = 20;
string names [NUM_NAMES] = {"Collins, Bill", "Smith, Bart" , "Allen, Jim", "Griffin, Jim", "Stamey, Marty", "Rose, Geri","Taylor, Terri",
    "Johnson, Jill", "Allison, Jeff", "Looney, Joe" , "Wolfe, Bill", "Rutherford, Greg", "Javens, Renee","Harrison, Rose","Setzer, Cathy",
    "Pike, Gordon","Holland, Beth"};
string smallest;
int minindex;
void displayNames (string[NUM_NAMES], int);



cout<<"Here are the unalphabetized names";
****displayNames(names[NUM_NAMES], NUM_NAMES);****

for(int i=0; i < (NUM_NAMES -1); i++)
{
    minindex=i;
    smallest=names[i];
    for(int j = (i+1); j<NUM_NAMES; j++)
    {
        if(names[j]<smallest)
        {
            smallest = names[j];
            minindex = j;
        }
    }
    names[minindex] =names[i];
    names[i] = smallest;

    system("pause");
    return 0;
}
}

void displayNames(string names[], int NUM_NAMES)
{
    for (int i =0; i<NUM_NAMES; i ++)
{
    cout<<names[i]<<endl;
}
}

四个星号括起来的行是错误代码引用时，我尝试构建。星号在我的实际代码中不存在。我不明白如何修复错误

The line bracketed by four asterisks is the line that the error code references when I try to build. The asterisks are NOT there in my actual code. I don't understand how to fix the error

Intellisense: no suitable conversion function from "std::string" to "std::string *" exists

我已经浏览了页面和页面的搜索， /答案似乎指从 string到int 或 string到char 等的转换函数not string to string 。

I've gone through pages and pages of searches, however all the questions/answers appear to refer to a conversion function from string to int or string to char etc. not string to string.

在旁注中，我还必须按字母顺序排列这些，我的教授说计算机应该评估字符串的char值。

On a side note, I do also have to alphabetize these, will what I currently have work? My professor was saying that the computer should evaluate the strings' char values.

推荐答案

这是错误的：

displayNames(names[NUM_NAMES], NUM_NAMES);

这是正确的：

displayNames(names, NUM_NAMES);

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