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Hello everyone,
Operator * on iterator of type T will result in reference to type T (T&), right (i.e. not type T itself or some other types)?
I am looking for some STL implementation code for this (operator * on iterator of type T will result in reference to type T (T&)), but can not find (maybe I search method is not correct). Could anyone post some internal implementation code please?
thanks in advance,
George
I'm not sure what you mean by your question?
Are you referring to the fact that the iterotor returns a reference to the object it refers to, rather than a copy? That is what you want, no? Otherwise, you couldn't use an iterator to modify the content that it refers to.
--
Mats
It depends on the kind of iterator. According to the C++ standard:Quote:
Operator * on iterator of type T will result in reference to type T (T&), right (i.e. not type T itself or some other types)?
Input iterators: result is convertible to T
Output iterators: result is not used
Forward, bi-directional and random access iterators: result is T&
Is that close enough? That's not a real implementation, but it's really close to what it might look like.Code:template <typename T>
class iterator
{
T* data;
public:
T& operator*() { return *data; }
};
Another implementation could be this:In which case the operator* is part of the language so you can't actually see the code that implements it.Code:typedef T* iterator;
Non-const iterators usually return a reference, so that the value can be modified. A const iterator could in theory return a value, not a reference, but in practice usually returns a reference for efficiency.Quote:
Originally Posted by George2
Look in your compiler's STL header files, it's all there.Quote:
I am looking for some STL implementation code for this (operator * on iterator of type T will result in reference to type T (T&)), but can not find (maybe I search method is not correct). Could anyone post some internal implementation code please?