Is there a downside to declaring variables with au

Question

It seems that auto was a fairly significant feature to be added in C++11 that seems to follow a lot of the newer languages. As with a language like Python, I have not seen any explicit variable declaration (I am not sure if it is possible using Python standards).

Is there a drawback to using auto to declare variables instead of explicitly declaring them?

Answer 1

You've only asked about drawbacks, so I'm highlighting some of those. When used well, auto has several advantages as well. The drawbacks result from ease of abuse, and from increased potential for code to behave in unintended ways.

The main drawback is that, by using auto, you don't necessarily know the type of object being created. There are also occasions where the programmer might expect the compiler to deduce one type, but the compiler adamantly deduces another.

Given a declaration like

auto result = CallSomeFunction(x,y,z);

you don't necessarily have knowledge of what type result is. It might be an int. It might be a pointer. It might be something else. All of those support different operations. You can also dramatically change the code by a minor change like

auto result = CallSomeFunction(a,y,z);

because, depending on what overloads exist for CallSomeFunction() the type of result might be completely different - and subsequent code may therefore behave completely differently than intended. You might suddenly trigger error messages in later code(e.g. subsequently trying to dereference an int, trying to change something which is now const). The more sinister change is where your change sails past the compiler, but subsequent code behaves in different and unknown - possibly buggy - ways.

Not having explicit knowledge of the type of some variables therefore makes it harder to rigorously justify a claim that the code works as intended. This means more effort to justify claims of "fit for purpose" in high-criticality (e.g. safety-critical or mission-critical) domains.

The other, more common drawback, is the temptation for a programmer to use auto as a blunt instrument to force code to compile, rather than thinking about what the code is doing, and working to get it right.

Answer 2

This isn't a drawback of auto in a principled way exactly, but in practical terms it seems to be an issue for some. Basically, some people either: a) treat auto as a savior for types and shut their brain off when using it, or b) forget that auto always deduces to value types. This causes people to do things like this:

auto x = my_obj.method_that_returns_reference();

Oops, we just deep copied some object. It's often either a bug or a performance fail. Then, you can swing the other way too:

const auto& stuff = *func_that_returns_unique_ptr();

Now you get a dangling reference. These problems aren't caused by auto at all, so I don't consider them legitimate arguments against it. But it does seem like auto makes these issue more common (from my personal experience), for the reasons I listed at the beginning.

I think given time people will adjust, and understand the division of labor: auto deduces the underlying type, but you still want to think about reference-ness and const-ness. But it's taking a bit of time.

Answer 3

It makes your code a little harder, or tedious, to read. Imagine something like that:

auto output = doSomethingWithData(variables);

Now, to figure out the type of output, you'd have to track down signature of doSomethingWithData function.

Answer 4

Keyword auto simply deduce the type from the return value. Therefore, it is not equivalent with a Python object, e.g.

# Python
a
a = 10       # OK
a = "10"     # OK
a = ClassA() # OK

// C++
auto a;      // Unable to deduce variable a
auto a = 10; // OK
a = "10";    // Value of const char* can't be assigned to int
a = ClassA{} // Value of ClassA can't be assigned to int
a = 10.0;    // OK, implicit casting warning

Since auto is deduced during compilation, it won't have any drawback at runtime whatsoever.

Answer 5

One of the drawbacks is that sometimes you can't declare const_iterator with auto. You will get ordinary (non const) iterator in this example of code taken from this question:

map<string,int> usa;
//...init usa
auto city_it = usa.find("New York");

Answer 6

One reason that I can think of is that you lose the opportunity to coerce the class that is returned. If your function or method returned a long 64 bit, and you only wanted a 32 unsigned int, then you lose the opportunity to control that.