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Why having const and non-const accessors?

Why do the STL containers define const and non-const versions of accessors ?

What is the advantage of defining const T& at(unsigned int i) const and T& at(unsigned int) and not only the non-const version ?

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Vincent Avatar asked Nov 20 '12 18:11

Vincent


1 Answers

Because you wouldn't be able to call at on a const vector object.

If you only had the non-const version, the following:

const std::vector<int> x(10);
x.at(0);

would not compile. Having the const version makes this possible, and at the same time prevents you from actually changing what at returns - which is by contract, since the vector is const.

The non-const version can be called on a non-const object and allows you to modify the returned element, which is also valid because the vector isn't const.

const std::vector<int> x(10);
      std::vector<int> y(10);

int z = x.at(0);          //calls const version - is valid
x.at(0) = 10;             //calls const version, returns const reference, invalid

z = y.at(0);              //calls non-const version - is valid
y.at(0) = 10;             //calls non-const version, returns non-const reference
                          //is valid
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Luchian Grigore Avatar answered Oct 18 '22 18:10

Luchian Grigore