Question

以下是我的代码：

#include <iostream>
#include <string>

using namespace std;

int main()
{
    int age1;
    int age2;
    string name1;
    string name2;

cout << "Please enter the name for one people: " << "\n";
getline (cin, name1);
cout << "Please enter the age for this people: " << "\n";
cin >> age1;

cout << "Please enter the name for another people: " << "\n";
getline (cin, name2);
cout << "Please enter the age for this people too: " << "\n";
cin >> age2;

if ( (age1 <= 100 || age2 <= 100) && (age1 < age2) )
{
    cout << name1 << " is younger!" << "\n";
}
else if ( (age1 <= 100 || age2 <= 100) && (age1 > age2) )
{
    cout << name2 << " is younder!" << "\n";
}
else if ( (age1 <= 100 || age2 <= 100) && (age1 = age2) )
{
    cout << name1 << " and " << name2 << " are of the same age!" << "\n";
}
else
{
    cout << "You've got some really old people that are well older than 100!";
}
}

第一个getline和cin工作正常。我可以被提示输入。但是，第二个getline和cin会立即被提示，因此我只能输入cin。（跳过第二个getline！）

如果我使用四个cins，程序将正常工作。

Answer 1

;

后需要getline (cin, name);

希望这会有所帮助

Answer 2

cin >> age1;未读取该数字后面的换行符。换行符保留在输入缓冲区中，然后过早地停止第二个getline。

所以，只要你在同一行输入第一个年龄和第二个名字，你的程序就会already works。

一种解决方案是在数字后跳过空格：

cin >> age1 >> ws;

Live demo

Answer 3

首先： cin＆gt;＆gt;年龄;它需要数量和商店的年龄，但同样它将换行符留在缓冲区本身的时间。所以当有下一个名字的提示时，cin发现在缓冲区中留下了换行符并将其作为输入。这就是它逃脱name2提示的原因。

    cout << "Please enter the name for one people: " << "\n";       
    cin>>name1;
    cout << "Please enter the age for this people: " << "\n";
    cin >> age1;<<--**this left the new line character in input buffer**
    cin.get();<<-- **get that newline charachter out of there first**
    cout << "Please enter the name for another people: " << "\n";
    getline (cin, name2);
    cout << "Please enter the age for this people too: " << "\n";
    cin >> age2;

现在我给name1-＆gt; shishir age1-＆gt; 28 name2-＆gt; ccr age-＆gt; 22打印ccr younder！＆lt; - 拼写错误：D

获取更多信息获取getline和get（）阅读c ++ primer plus 列出4.3,4.4,4.5

快乐编码

Answer 4

我建议使用cin.ignore（100，＆＃39; \ n＆＃39;）。它会忽略您在调用时指定的字符数（上例中为100），最多为指定为断点的字符。例如：

cout << "Please enter the name for one people: " << "\n";
getline (cin, name1);
cout << "Please enter the age for this people: " << "\n";
cin >> age1;
cin.ignore(100, '\n');
cout << "Please enter the name for another people: " << "\n";
getline (cin, name2);
cout << "Please enter the age for this people too: " << "\n";
cin >> age2;
cin.ignore(100, '\n');

＆＃34;函数getline＆＃34;提示被跳过，没有按预期工作

4 个答案: