...one of the most highly
regarded and expertly designed C++ library projects in the
world.
— Herb Sutter and Andrei
Alexandrescu, C++
Coding Standards
This section describes some common pitfalls with BOOST_FOREACH
.
Since BOOST_FOREACH
is a macro, it must have exactly two
arguments, with exactly one comma separating them. That's not always convenient,
especially when the type of the loop variable is a template. Consider trying
to iterate over a std::map
:
std::map<int,int> m; // ERROR! Too many arguments to BOOST_FOREACH macro. BOOST_FOREACH(std::pair<int,int> p, m) // ...
One way to fix this is with a typedef.
std::map<int,int> m; typedef std::pair<int,int> pair_t; BOOST_FOREACH(pair_t p, m) // ...
Another way to fix it is to predeclare the loop variable:
std::map<int,int> m; std::pair<int,int> p; BOOST_FOREACH(p, m) // ...
Under the covers, BOOST_FOREACH
uses iterators to traverse
the element sequence. Before the loop is executed, the end iterator is cached
in a local variable. This is called hoisting, and it is
an important optimization. It assumes, however, that the end iterator of the
sequence is stable. It usually is, but if we modify the sequence by adding
or removing elements while we are iterating over it, we may end up hoisting
ourselves on our own petard.
Consider the following code:
std::vector<int> vect(4, 4); BOOST_FOREACH(int i, vect) { vect.push_back(i + 1); }
This code will compile, but it has undefined behavior. That is because it is logically equivalent to the following:
std::vector<int> vect(4, 4); for(std::vector<int>::iterator it1 = vect.begin(), it2 = vect.end(); it1 != it2; ++it1) { int i = *it1; vect.push_back(i + 1); // Oops! This invalidates it1 and it2! }
The call to vect.push_back()
will cause all iterators into vect
to become invalid, including it1
and it2
. The next iteration
through the loop will cause the invalid iterators to be used. That's bad news.
The moral of the story is to think twice before adding and removing elements
from the sequence over which you are iterating. If doing so could cause iterators
to become invalid, don't do it. Use a regular for
loop instead.
Copyright © 2004 Eric Niebler |