How does the Compiler Know Whether to Call the const Overload?

Question

Given string foo, when I call:

auto bar = foo.begin();

There are 2 overloads of string::begin. One returns an string::iterator and the other returns a string::const_iterator. How can I know the type of bar? Is this just based on whether foo is const or not?

You can also call `cbegin` if you know for sure you want a `const_iterator`. — Brian Bi, Jun 01 '16 at 20:42

score 3 · Accepted Answer · edited May 23 '17 at 10:28

3

Is this just based on whether foo is const or not?

Yes

auto is meant for deducing the exact type^*

How to select iterator type using auto variable? indirectly agrees with this answer.

*_{Taken from: How do I get a const_iterator using auto?}

edited May 23 '17 at 10:28

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answered Jun 01 '16 at 20:28

gsamaras

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How does the Compiler Know Whether to Call the const Overload?

1 Answers1