使用c ++ stl迭代器而不是传统指针的正确方法是什么？

Question

I have the following very basic question. I want to use stl iterators instead of traditional C-type pointers for filling an array in a function. By the C-style way I mean the following example:

void f(double* v, size_t n) {
    for (int i = 0; i < n; i++) 
         v[i] = 10;    /* a more reasonable value in practice! */
}

I would convert this to the C++ style using the iterators as follows:

void f(vector<double>::const_iterator first, vector<double>::const_iterator last) {
    for(vector<double>::iterator it =  first; it != last; it++)
        *it = 10;
}

But I get compilation errors. If I use iterator instead of const_iterator the problem will be solved. However, I was wondering if that is the correct way? Because I thought vector.begin() and vector.end() iterators are constant.

Thanks in advance!

Answer 1

The difference between

const vector<double>::iterator

and

vector<double>::const_iterator

is roughly the same as between double * const v and const double *v :

• the first says that the iterator must remain constant, but what it points to can be changed
• the second says that the iterator itself is changeable, but what it points to is const .

If you rewrite the function as

void f(const vector<double>::iterator first, const vector<double>::iterator last) {
    for(vector<double>::iterator it =  first; it != last; it++)
        *it = 10;
}

it would compile and run correctly.

Answer 2

What you see is due to the fact that const_iterator 's correspond roughly to pointers to const. So you can change the value of the iterator, ie make it point somewhere else, but you cannot modify what it points to.

This is different from const iterators, which would not allow incrementing or decrementing them. Here is an example:

#include <vector>

int main() {
  std::vector<int> v{ 1, 2, 3 };

  std::vector<int>::const_iterator i = v.begin();
  *i = 10;                                         // ERROR!
  ++i;                                             // OK

  std::vector<int>::iterator const ci = v.begin();
  *ci = 10;                                        // OK
  ++ci;                                            // ERROR!
}

Answer 3

std::fill(my_vector.begin(), my_vector.end(), 10);

Answer 4

Since you're using const_iterator s, you can't modify the vector. Using non-const iterator s is the right thing to do.

In answer to your last question, vector.begin() and vector.end() have both const_ and non- const_ implementations. If your vector is non-const, you'll get a non- const_ iterator. See the documentation for std::vector::begin .

Answer 5

The problem is that your functions takes const_iterator s but your loop needs an iterator , since you want to modify the data. The solution is of course to let your function take iterator s right away, since it is obviously meant to modify the range.

This doesn't have anything to do with what vector.begin() returns. For a const object or reference they will return const_iterator s, otherwise they'll return iterator s. But your function definitely needs iterator s, since it modifies the values in the range passed to it.