迭代 std::vector:无符号与有符号索引变量
在 C++ 中迭代向量的正确方法是什么?
What is the correct way of iterating over a vector in C++?
考虑这两个代码片段，这个可以正常工作:
Consider these two code fragments, this one works fine:
for (unsigned i=0; i < polygon.size(); i++) {
sum += polygon[i];
}
还有这个:
for (int i=0; i < polygon.size(); i++) {
sum += polygon[i];
}
生成警告:有符号和无符号整数表达式之间的比较.
我是 C++ 世界的新手，所以 unsigned 变量对我来说看起来有点吓人，我知道 unsigned 变量如果使用不当会很危险，所以 - 这是正确的吗?
I'm new in the world of C++, so the unsigned variable looks a bit frightening to me and I know unsigned variables can be dangerous if not used correctly, so - is this correct?
推荐答案
要向后迭代，请参阅 这个答案.
For iterating backwards see this answer.
向前迭代几乎相同.只需按增量更改迭代器/交换递减.你应该更喜欢迭代器.有些人告诉您使用 std::size_t 作为索引变量类型.但是，这不是便携式的.始终使用容器的 size_type typedef(虽然在向前迭代的情况下你可以只进行一次转换，但在使用 std 时在向后迭代的情况下它实际上可能一直出错::size_t，以防 std::size_t 比 size_type 的 typedef 更宽):
Iterating forwards is almost identical. Just change the iterators / swap decrement by increment. You should prefer iterators. Some people tell you to use std::size_t as the index variable type. However, that is not portable. Always use the size_type typedef of the container (While you could get away with only a conversion in the forward iterating case, it could actually go wrong all the way in the backward iterating case when using std::size_t, in case std::size_t is wider than what is the typedef of size_type):
for(std::vector<T>::iterator it = v.begin(); it != v.end(); ++it) {
/* std::cout << *it; ... */
}
重要的是，对于您不知道定义的迭代器，始终使用前缀增量形式.这将确保您的代码尽可能通用.
Important is, always use the prefix increment form for iterators whose definitions you don't know. That will ensure your code runs as generic as possible.
for(auto const& value: a) {
/* std::cout << value; ... */
使用索引
for(std::vector<int>::size_type i = 0; i != v.size(); i++) {
/* std::cout << v[i]; ... */
}
<小时>
使用数组
使用迭代器
for(element_type* it = a; it != (a + (sizeof a / sizeof *a)); it++) {
/* std::cout << *it; ... */
}
使用范围 C++11
for(auto const& value: a) {
/* std::cout << value; ... */
使用索引
for(std::size_t i = 0; i != (sizeof a / sizeof *a); i++) {
/* std::cout << a[i]; ... */
}
阅读向后迭代的答案，但是 sizeof 方法可以解决什么问题.
Read in the backward iterating answer what problem the sizeof approach can yield to, though.
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