blocks|key|3162373|text|只需密切关注事件链即可。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3162374|3162375|Grab+10|unordered-list-item|3162376|Grab+20|3162377|Grab+30|3162378|Grab+EOF|3162379|<代码>F29|3162380|看看倒数第二次迭代。你抓取了30个，然后继续检查EOF。您还没有达到EOF，因为EOF标记还没有被读取(“二进制”地说，它的概念位置恰好在30行之后)。因此，您将继续进行下一次迭代。与上一次迭代相比，X仍然是30。现在，您从流中读取并获得EOF。X保持为30，并引发ios::eofbit。输出到stderr+x(它是30，就像上一次迭代一样)。接下来，您在循环条件中检查EOF，这一次您退出了循环。|3162381|试试这个：|3162382|while+(true)+{
++++int+x;
++++iFile+>>+x;
++++if(+iFile.eof()+)+break;
++++cerr+<<+x+<<+endl;
}|code-block|syntax|javascript|3162383|顺便说一下，您的代码中还有另一个bug。你有没有试过在一个空文件上运行它？你得到的行为也是出于同样的原因。|3162384|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|12|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|E|7|14|8|@]|9|@]|A|$]]|$1|J|3|K|5|E|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|16|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|17|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|18|8|@]|9|@]|A|$]]|$1|R|3|S|5|T|7|19|8|@]|9|@]|A|$U|V]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

Just follow closely the chain of events.

<ul>
<li>Grab 10</li>
<li>Grab 20</li>
<li>Grab 30</li>
<li>Grab EOF</li>
</ul>

Look at the second-to-last iteration. You grabbed 30, then carried on to check for EOF. You haven't reached EOF because the EOF mark hasn't been read yet ("binarically" speaking, its conceptual location is just after the 30 line). Therefore you carry on to the next iteration. x is still 30 from previous iteration. Now you read from the stream and you get EOF. x remains 30 and the ios::eofbit is raised. You output to stderr x (which is 30, just like in the previous iteration). Next you check for EOF in the loop condition, and this time you're out of the loop.

Try this:

<pre><code>while (true) {
 int x;
 iFile &gt;&gt; x;
 if( iFile.eof() ) break;
 cerr &lt;&lt; x &lt;&lt; endl;
}
</code></pre>

By the way, there is another bug in your code. Did you ever try to run it on an empty file? The behaviour you get is for the exact same reason.

blocks|key|370066|text|我喜欢这个示例，就目前而言，它省略了您可以在while块中添加的check：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|370067|ifstream+iFile("input.txt");++++++++//+input.txt+has+integers,+one+per+line
int+x;

while+(iFile+>>+x)+
{
++++cerr+<<+x+<<+endl;
}|code-block|syntax|javascript|370068|不确定它有多安全...|370069|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I like this example, which for now, leaves out the check which you could add inside the while block:

<pre><code>ifstream iFile("input.txt"); // input.txt has integers, one per line
int x;

while (iFile &gt;&gt; x) 
{
 cerr &lt;&lt; x &lt;&lt; endl;
}
</code></pre>

Not sure how safe it is...

blocks|key|365268|text|有另一种方法可以解决这个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|365269|#include+<iterator>
#include+<algorithm>

//+...

++++copy(istream_iterator<int>(iFile),+istream_iterator<int>(),
+++++++++ostream_iterator<int>(cerr,+"\n"));|code-block|syntax|javascript|365270|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|$]]

There's an alternative approach to this:

<pre><code>#include &lt;iterator&gt;
#include &lt;algorithm&gt;

// ...

 copy(istream_iterator&lt;int&gt;(iFile), istream_iterator&lt;int&gt;(),
 ostream_iterator&lt;int&gt;(cerr, "\n"));
</code></pre>

blocks|key|3162420|text|如果不对原始代码进行太多修改，它可能会变成：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3162421|while+(!iFile.eof())
{++
++++int+x;
++++iFile+>>+x;
++++if+(!iFile.eof())+break;
++++cerr+<<+x+<<+endl;
}|code-block|syntax|javascript|3162422|但总的来说，我更喜欢上面的另外两个解决方案。|3162423|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Without to much modifications of the original code, it could become :

<pre><code>while (!iFile.eof())
{ 
 int x;
 iFile &gt;&gt; x;
 if (!iFile.eof()) break;
 cerr &lt;&lt; x &lt;&lt; endl;
}
</code></pre>

but I prefer the two other solutions above in general.

blocks|key|3375231|text|int+x;
ifile+>>+x

while+(!iFile.eof())
{++
++++cerr+<<+x+<<+endl;++++++++
++++iFile+>>+x;++++++
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3375232|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>int x;
ifile &gt;&gt; x

while (!iFile.eof())
{ 
 cerr &lt;&lt; x &lt;&lt; endl; 
 iFile &gt;&gt; x; 
}
</code></pre>

blocks|key|365290|text|EOF模式需要一个主读取来“引导”EOF检查过程。考虑到在第一次读取之前，空文件最初不会设置EOF。在这种情况下，主读取将捕获EOF，并完全正确地跳过循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|365291|这里您需要记住的是，直到第一次尝试读取文件的可用数据之前，您才会获得EOF。读取准确的数据量不会标记EOF。|365292|我应该指出，如果文件是空的，您的给定代码将被打印出来，因为EOF将阻止在进入循环时将一个值设置为x。|365293|365294|unordered-list-item|365295|365296|因此，添加一个prime读取并将循环的读取移动到末尾：|365297|int+x;

iFile+>>+x;+//+prime+read+here
while+(!iFile.eof())+{
++++cerr+<<+x+<<+endl;
++++iFile+>>+x;
}|code-block|syntax|javascript|365298|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|V|8|@]|9|@]|A|$]]|$1|G|3|-4|5|H|7|W|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|Z|8|@]|9|@]|A|$O|P]]|$1|Q|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|R|$]]

The EOF pattern needs a prime read to 'bootstrap' the EOF checking process. Consider the empty file will not initially have its EOF set until the first read. The prime read will catch the EOF in this instance and properly skip the loop completely.

What you need to remember here is that you don't get the EOF until the first attempt to read past the available data of the file. Reading the exact amount of data will not flag the EOF.

I should point out if the file was empty your given code would have printed since the EOF will have prevented a value from being set to x on entry into the loop.

<ul>
<li>0</li>
</ul>

So add a prime read and move the loop's read to the end:

<pre><code>int x;

iFile &gt;&gt; x; // prime read here
while (!iFile.eof()) {
 cerr &lt;&lt; x &lt;&lt; endl;
 iFile &gt;&gt; x;
}
</code></pre>

blocks|key|365300|text|在最后一行的末尾，有一个新行字符，它不被>>操作符读取，也不是文件的结尾。请进行实验并删除新行(文件中的最后一个字符)-您不会得到重复项。为了拥有灵活的代码并避免不必要的效果，只需应用其他用户给出的任何解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|365301|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

At the end of the last line, you have a new line character, which is not read by >> operator and it is not an end of file.
Please, make an experiment and delete the new line (thelast character in file) - you will not get the duplication.
To have a flexible code and avoid unwanted effects just apply any solution given by other users.

The following C++ code uses a ifstream object to read integers from a text file (which has one number per line) until it hits EOF. Why does it read the integer on the last line twice? How to fix this?

Code:

<pre><code>#include &lt;iostream&gt;
#include &lt;fstream&gt;
using namespace std;

int main()
{
 ifstream iFile("input.txt"); // input.txt has integers, one per line

 while (!iFile.eof())
 {
 int x;
 iFile &gt;&gt; x;
 cerr &lt;&lt; x &lt;&lt; endl;
 }

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

input.txt:

<pre><code>10 
20 
30
</code></pre>

Output:

<pre><code>10 
20 
30 
30
</code></pre>

Note: I've skipped all error checking code to keep the code snippet small. The above behaviour is seen on Windows (Visual C++), cygwin (gcc) and Linux (gcc).

Reading from text file until EOF repeats last line

Linux

下面的C++代码使用ifstream对象从文本文件(每行一个数字)中读取整数，直到命中EOF。为什么它读取最后一行的整数两次？如何解决这个问题？代码：#include <iostream>#include <fstream>using namespace std;int main(){ ifstream iFil...

问从文本文件中读取，直到EOF重复最后一行
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