File 只能从文本文件中读取一个名称

我有一个包含名称（char name[25]）、年龄（int age）和gpa（float gpa）的结构（RECORD）。我正在尝试从此文本文件读取数据：

Jacqueline Kennedy 33 3.5
克劳迪娅·约翰逊25 2.5
帕特·尼克松33 2.7
罗莎琳·卡特26 2.6
南希·里根193.5
芭芭拉·布什33 3.4
希拉里·克林顿25 2.5

我认为这会有所帮助。首先，您需要读取数组中的整个文件，并将其转换为字符串数组，该数组的字符串长度为您所说的25个字符。然后，您只需遍历该数组即可显示它

if(f.is_open())
{
//file opened successfully so we are here
cout << "File Opened successfully!!!. Reading data from file into array" << endl;
//this loop run until end of file (eof) does not occ
    while(!f.eof() && position < array_size)
    {
        f.get(array[position]); //reading one character from file to array
        position++;
    }
    array[position-1] = '\0'; //placing character array terminating character

cout << "Displaying Array..." << endl << endl;
//this loop display all the charaters in array till \0 
    for(int i = 0; array[i] != '\0'; i++)
    {
        cout << array[i];
       /*then you can divide the array in your desired length strings, which     here is: Jacqueline Kennedy       33 3.5
       Claudia Johnson          25 2.5
       Pat Nixon                33 2.7
       Rosalyn Carter           26 2.6
       Nancy Reagan             19 3.5
       Barbara Bush             33 3.4
       Hillary Clinton          25 2.5*/

    }
}
else //file could not be opened
{
    cout << "File could not be opened." << endl;
}

如果（f.是开着的（））
{
//文件已成功打开，所以我们在这里
库特