Why having const and non-const accessors?

问题

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 ?

回答1:

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

来源：https://stackoverflow.com/questions/13479548/why-having-const-and-non-const-accessors