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<th>Bug ID</th>
<td><a class="bz_bug_link
bz_status_NEW "
title="NEW --- - ostream_iterator::operator* is not constant"
href="http://llvm.org/bugs/show_bug.cgi?id=19425">19425</a>
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<th>Summary</th>
<td>ostream_iterator::operator* is not constant
</td>
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<th>Product</th>
<td>libc++
</td>
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<th>Version</th>
<td>3.4
</td>
</tr>
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<th>Hardware</th>
<td>All
</td>
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<th>OS</th>
<td>All
</td>
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<th>Status</th>
<td>NEW
</td>
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<th>Severity</th>
<td>normal
</td>
</tr>
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<th>Priority</th>
<td>P
</td>
</tr>
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<th>Component</th>
<td>All Bugs
</td>
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<th>Assignee</th>
<td>unassignedclangbugs@nondot.org
</td>
</tr>
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<th>Reporter</th>
<td>root@zta.lk
</td>
</tr>
<tr>
<th>CC</th>
<td>llvmbugs@cs.uiuc.edu, mclow.lists@gmail.com
</td>
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<th>Classification</th>
<td>Unclassified
</td>
</tr></table>
<p>
<div>
<pre>There is a trick used in ostream_iterator, that doesn't quite conform to
standard.
ostream_iterator<T>::operator* returns itself instead of something with
operator=(const T &);
It's almost never an issue since the all the stl algorithms do * and ++ at the
same time. But if we separate * and ++ the way that is shown below, the program
won't compile.
---------
#include <iostream>
#include <iterator>
using namespace std;
void write_5(const ostream_iterator<int> &i)
{
*i = 5;
}
int main()
{
ostream_iterator<int> i(cout);
write_5(i);
++i;
}
---------
I believe ostream_iterato<T>::operator* must be defined as const and return a
temporary object, that holds the reference to the stream and have operator=
that do the job. In any case, the code above conforms to standard and doesn't
compile.</pre>
</div>
</p>
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