如何将多个用户输入添加到数组中？

Question

如何向数组添加多个字符串或用户输入？我正在尝试创建一个通讯录，要求用户添加最多 10 个联系人。我试图将它们存储为数组或 txt 文件，然后我希望能够使用此输入。

这是我的代码。如果我想说的不清楚，运行代码会有所帮助。

#include <cstdlib>
#include <iostream>
#include <fstream>

using namespace std;

int main(int argc, char *argv[])
{
    // declare two variables;
    char name[20];
    int age;
string ans;
do {    
    // get user to input these;
    cout << "What is your name: ";
    cin >> name;
    cout << "What is your age : ";
    cin >> age;
    cout<<"continue ";cin>>ans;
  }while((ans == "y" || ans=="yes"));  
    // create output stream object for new file and call it fout
    // use this to output data to the file "test.txt"
    char filename[] = "test.txt";
    ofstream fout(filename);
    fout << name << "," << age << "\n";    // name, age to file
    fout.close();   // close file

    // output name and age : as originally entered
    cout << "\n--------------------------------------------------------"
         << "\n name and age data as entered";
    cout << "\n    Your name is: " << name;
    cout << "\n    and your age is: " << age;     

    // output name and age : as taken from file           

    // first display the header
    cout << "\n--------------------------------------------------------"
         << "\n name and age data from file"
         << "\n--------------------------------------------------------";   

      ifstream fin(filename);

       char line[50];               
    fin.getline(line, 50);      


    char fname[20];     
    int count = 0;       
    do
    {
        fname[count] = line[count];       
        count++;
    }
    while (line[count] != ',');        
    fname[count] = '\0';              


    count++;
    char fage_ch[10];    
    int fage_count = 0;   
    do
    {
        fage_ch[fage_count] = line[count];   
        fage_count++; count++;               
    }
    while (line[count] != '\0');             
    fage_ch[fage_count] = '\0';


    int fage_int = 0;        
    int total = 0;           
    char temp;               

    for (int i = 0; i < (fage_count); i++)
    {
        temp = fage_ch[i];
        total = 10*total + atoi(&temp);
    }

    fage_int = total;

    // display data
    cout << "\n\n    Your name is: " << fname;
    cout << "\n    and your age is: " << fage_int;
    cout << "\n\n--------------------------------------------------------";    

      cout <<endl;

    return EXIT_SUCCESS;
}

Answer 1

您可能最好使用结构数组而不是两个单独的数组来存储名称和名称。每个条目的年龄。然后，您可以使用 strcpy 循环将输入字符串从名称复制到结构的名称中。如果您对结构不满意，您也可以使用几个二维数组。

这看起来像是一项家庭作业，所以我不会发布代码，而是提供一个基本算法来帮助您入门（并希望简化您所拥有的内容）：

#define MAX_CONTACTS 10
#define MAX_NAME_LENGTH 20

// 2D array to store up to 10 names of max 20 character length
char nameVar[MAX_CONTACTS][MAX_NAME_LENGTH]
int ageVar[MAX_CONTACTS]

do until end of user input
    read name into nameVar[index]
    read age into ageVar[index]
    index += 1
end loop

while contactCounter < index
    ouput nameVar[contactCounter]
    output age[contactCounter]
    // you could also write to file in this loop if thats what you're trying to do
    // using the fprintf function to write to an opened file
    contactCounter += 1
end loop

另外，我不确定您要做什么通过 atoi 调用，但看起来没有必要。 atoi 的工作原理是，它查看传递的第一个字符并转换所有数字，直到遇到数组中的非数字字符。因此，如果您有 char 数组 c="123h" atoi 将返回 123。如果您传递 atoi "1h2" 它将返回 1。

您还可以使用 fprintf 将 char 数组和 int 打印到文件中。

因此，如果您有 int i 和 char s[10] = "hello" 和文件流，您可以打印到流，例如：

fprintf(stream, "my text to display: %s %i", s,i)

我希望这有帮助。