C++ – Why Explicitly Write the ‘auto’ Keyword

autoc++c++11

I am moving towards C++11 from C++98 and have become familiar with the auto keyword. I was wondering why we need to explicitly declare auto if the compiler is able to automatically deduce the type. I know C++ is a strongly typed language and this is a rule but was it not possible to achieve the same outcome without explicitly declaring a variable auto?

Best Answer

Dropping the explicit auto would break the language:

e.g.

int main()
{
    int n;
    {
        auto n = 0; // this shadows the outer n.
    }
}

where you can see that dropping the auto would not shadow the outer n.

Related Solutions

C++ STL – Using the auto Keyword

It's additional information, and isn't an answer.

In C++11 you can write:

for (auto& it : s) {
    cout << it << endl;
}

instead of

for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it << endl;
}

It has the same meaning.

Update: See the @Alnitak's comment also.

C++ Auto Keyword – Meaning and Usage in C++11

auto was a keyword that C++ "inherited" from C that had been there nearly forever, but virtually never used because there were only two possible conditions: either it wasn't allowed, or else it was assumed by default.

The use of auto to mean a deduced type was new with C++11.

At the same time, auto x = initializer deduces the type of x from the type of initializer the same way as template type deduction works for function templates. Consider a function template like this:

template<class T>
int whatever(T t) { 
    // point A
};

At point A, a type has been assigned to T based on the value passed for the parameter to whatever. When you do auto x = initializer;, the same type deduction is used to determine the type for x from the type of initializer that's used to initialize it.

This means that most of the type deduction mechanics a compiler needs to implement auto were already present and used for templates on any compiler that even sort of attempted to implement C++98/03. As such, adding support for auto was apparently fairly easy for essentially all the compiler teams--it was added quite quickly, and there seem to have been few bugs related to it either.

When this answer was originally written (in 2011, before the ink was dry on the C++ 11 standard) auto was already quite portable. Nowadays, it's thoroughly portable among all the mainstream compilers. The only obvious reasons to avoid it would be if you need to write code that's compatible with a C compiler, or you have a specific need to target some niche compiler that you know doesn't support it (e.g., a few people still write code for MS-DOS using compilers from Borland, Watcom, etc., that haven't seen significant upgrades in decades). If you're using a reasonably current version of any of the mainstream compilers, there's no reason to avoid it at all though.

More recent revisions of the standard have added a few new places that auto can be used. Starting with C++14, you can use auto for the type of a parameter to a lambda:

    [](auto s) { return s + 1; }

This does essentially the same thing as the example above--even though it doesn't explicitly use template syntax, this is basically a template that deduces the type of the parameter, and instantiates the template over that type.

That was convenient and useful enough that in C++20, the same capability was added for normal functions, not just lambdas.

But, just as before all of this really comes down to using the same basic type deduction mechanism as we've had for function templates since C++98. auto allows that to be used in more places, and more conveniently, but the underlying heavy lifting remains the same.

Best Answer

Related Solutions

C++ STL – Using the auto Keyword

C++ Auto Keyword – Meaning and Usage in C++11

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