blocks|key|1190296|text|cin+>>+number+//+eat+the+numeric+characters
getline(cin,+str)+//+eat+the+remaining+newline|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1190297|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>cin &gt;&gt; number // eat the numeric characters
getline(cin, str) // eat the remaining newline
</code></pre>

blocks|key|1190359|text|cin+>>+number|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1190360|只从缓冲区中抓取数字，它将“回车”留在缓冲区中，然后立即被getline抓取并解释为一个空字符串(或者只有新行的字符串，我忘了)。|unstyled|1190361|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>cin &gt;&gt; number
</code></pre>

only grabs the numbers from the buffer, it leaves the "enter" in the buffer, which is then immediately grabbed up by the getline and interpreted as an empty string (or string with just the new line, i forget).

blocks|key|1405681|text|试一试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1405682|cout+<<+"Enter+name:";
cin.ignore();
getline(cin,+str);|code-block|syntax|javascript|1405683|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try 

<pre><code>cout &lt;&lt; "Enter name:";
cin.ignore();
getline(cin, str);
</code></pre>

blocks|key|1190428|text|我认为问题在于cin+>>传递了换行符(\n)。getline()假定换行符是空格，并将其传递。有人发布了一个你可以使用的解决方案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1190429|您可以使用虚拟的getline(cin,+dummy);或真实的cin.ignore(100,'\n');|1190430|entityMap^0|7|6|J|4|0|8|K|W|L|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|I|$]]

I think the problem is that <code>cin &gt;&gt;</code> passes on the newline character <code>(\n)</code>. The getline() assumes the newline character is whitespace and passes it on. Someone posted a solution you can use.

You can use a dummy <code>getline(cin, dummy);</code> or the real thing <code>cin.ignore(100,'\n');</code>

blocks|key|1405844|text|getline(cin,+str);读取前面读取的数字后面的换行符，并立即返回此“行”。为了避免这种情况，您可以在读取名称之前跳过std::ws中的空格：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1405845|cout+<<+"Enter+number:";
cin+>>+number;
cout+<<+"Enter+name:";
ws(cin);
getline(cin,+str);
...|code-block|syntax|javascript|1405846|请注意，这也会跳过换行符后面的前导空格，因此即使用户输入了空格，str也不会以空格开头。但在这种情况下，这可能是一个功能，而不是一个错误...|1405847|entityMap^0|0|I|1T|7|0|0|W|3|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

The <code>getline(cin, str);</code> reads the newline that comes after the number read previously, and immediately returns with this "line". To avoid this you can skip whitespace with <code>std::ws</code> before reading the name:

<pre><code>cout &lt;&lt; "Enter number:";
cin &gt;&gt; number;
cout &lt;&lt; "Enter name:";
ws(cin);
getline(cin, str);
...
</code></pre>

Note that this also skips leading whitespace after the newline, so <code>str</code> will not start with spaces, even if the user did input them. But in this case that's probably a feature, not a bug...

blocks|key|1190593|text|看起来你想要基于行的阅读。为此，您可能希望始终如一地使用getline，如果需要解析行中的数字，则先解析每一行，然后再读取行。它使输入读数更加一致。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1190594|这样，您就不必手动扫描每一行的末尾，以确保下一次读取操作从新行开始。|1190595|它还使得为重复输入请求添加错误处理变得更简单。|1190596|例如：|1190597|#include+<string>
#include+<iostream>
#include+<istream>
#include+<ostream>
#include+<sstream>

int+parse_integer(const+std::string&+input)
{
++++std::istringstream+iss(input);
++++int+result;
++++if+(!(iss+>>+result))
++++{
++++++++//+error+-+throw+something?
++++}
++++return+result;
}

int+main()
{
++++int+number;
++++std::string+str;
++++int+accountNumber;

++++std::string+inputline;

++++std::cout+<<+"Enter+number:+";

++++if+(!std::getline(std::cin,+inputline))
++++{
++++++++//+error+-+throw+something?
++++}

++++number+=+parse_integer(inputline);

++++std::cout+<<+"Enter+name:";

++++if+(!std::getline(std::cin,+inputline))
++++{
++++++++//+error+-+throw+something?
++++}

++++str+=+inputline;

++++std::cout+<<+"Enter+account+number:";

++++if+(!std::getline(std::cin,+inputline))
++++{
++++++++//+error+-+throw+something?
++++}

++++accountNumber+=+parse_integer(inputline);

++++return+0;
}|code-block|syntax|javascript|1190598|entityMap^0|S|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|X|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|Y|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

It looks like you want line based reading. For this you probably want to use <code>getline</code> consistently and then parse each line if you need to parse a number from then read line. It makes the input reading more consistent.

This way you don't have to manually scan for the end of each line to guarantee that the next read operation starts on a fresh line.

It also makes adding error handling for repeating input requests simpler.

e.g.

<pre><code>#include &lt;string&gt;
#include &lt;iostream&gt;
#include &lt;istream&gt;
#include &lt;ostream&gt;
#include &lt;sstream&gt;

int parse_integer(const std::string&amp; input)
{
 std::istringstream iss(input);
 int result;
 if (!(iss &gt;&gt; result))
 {
 // error - throw something?
 }
 return result;
}

int main()
{
 int number;
 std::string str;
 int accountNumber;

 std::string inputline;

 std::cout &lt;&lt; "Enter number: ";

 if (!std::getline(std::cin, inputline))
 {
 // error - throw something?
 }

 number = parse_integer(inputline);

 std::cout &lt;&lt; "Enter name:";

 if (!std::getline(std::cin, inputline))
 {
 // error - throw something?
 }

 str = inputline;

 std::cout &lt;&lt; "Enter account number:";

 if (!std::getline(std::cin, inputline))
 {
 // error - throw something?
 }

 accountNumber = parse_integer(inputline);

 return 0;
}
</code></pre>

blocks|key|1405933|text|不要使用getline()：这对内存分配是一件坏事。使用fgets()。参见fgets+reference。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1405934|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/cstdio/fgets/^0|4|9|S|7|12|5|12|F|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Don't use <code>getline()</code>: it's a bad thing for memory allocation. Use <code>fgets()</code>. See <a href="http://www.cplusplus.com/reference/cstdio/fgets/" rel="nofollow noreferrer"><code>fgets</code> reference</a>.

Is there a reason why if in my program I am asking the user for input, and I do:

<pre><code>int number;
string str;
int accountNumber;

cout &lt;&lt; "Enter number:";
cin &gt;&gt; number;
cout &lt;&lt; "Enter name:";
getline(cin, str);
cout &lt;&lt; "Enter account number:";
cin &gt;&gt; accountNumber;
</code></pre>

Why after inputting the first number, it outputs "Enter Name", followed immediately by "Enter Account Number" before I even get to input my "str" for the getline(cin, str) line? Thanks!

Need help with getline()

为什么在我的程序中，我要求用户输入，而我确实这样做了，这有什么原因：int number;string str;int accountNumber;cout << "Enter number:";cin >> number;cout << "Enter name:";getline(cin, str);cout << ...

问需要有关getline()的帮助
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