Is there a downside to declaring variables with auto in C++?

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 ha

Answer 1

What no one mentioned here so far, but for itself is worth an answer if you asked me.

Since (even if everyone should be aware that C != C++) code written in C can easily be designed to provide a base for C++ code and therefore be designed without too much effort to be C++ compatible, this could be a requirement for design.

I know about some rules where some well defined constructs from C are invalid for C++ and vice versa. But this would simply result in broken executables and the known UB-clause applies which most times is noticed by strange loopings resulting in crashes or whatever (or even may stay undetected, but that doesn't matter here).

But auto is the first time¹ this changes!

Imagine you used auto as storage-class specifier before and transfer the code. It would not even necessarily (depending on the way it was used) "break"; it actually could silently change the behaviour of the program.

That's something one should keep in mind.

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¹_{At least the first time I'm aware of.}

Answer 2

Other answers are mentioning drawbacks like "you don't really know what the type of a variable is." I'd say that this is largely related to sloppy naming convention in code. If your interfaces are clearly-named, you shouldn't need to care what the exact type is. Sure, auto result = callSomeFunction(a, b); doesn't tell you much. But auto valid = isValid(xmlFile, schema); tells you enough to use valid without having to care what its exact type is. After all, with just if (callSomeFunction(a, b)), you wouldn't know the type either. The same with any other subexpression temporary objects. So I don't consider this a real drawback of auto.

I'd say its primary drawback is that sometimes, the exact return type is not what you want to work with. In effect, sometimes the actual return type differs from the "logical" return type as an implementation/optimisation detail. Expression templates are a prime example. Let's say we have this:

SomeType operator* (const Matrix &lhs, const Vector &rhs);

Logically, we would expect SomeType to be Vector, and we definitely want to treat it as such in our code. However, it is possible that for optimisation purposes, the algebra library we're using implements expression templates, and the actual return type is this:

MultExpression<Matrix, Vector> operator* (const Matrix &lhs, const Vector &rhs);

Now, the problem is that MultExpression<Matrix, Vector> will in all likelihood store a const Matrix& and const Vector& internally; it expects that it will convert to a Vector before the end of its full-expression. If we have this code, all is well:

extern Matrix a, b, c;
extern Vector v;

void compute()
{
  Vector res = a * (b * (c * v));
  // do something with res
}

However, if we had used auto here, we could get in trouble:

void compute()
{
  auto res = a * (b * (c * v));
  // Oops! Now `res` is referring to temporaries (such as (c * v)) which no longer exist
}

Answer 3

auto does not have drawbacks per se, and I advocate to (hand-wavily) use it everywhere in new code. It allows your code to consistently type-check, and consistently avoid silent slicing. (If B derives from A and a function returning A suddenly returns B, then auto behaves as expected to store its return value)

Although, pre-C++11 legacy code may rely on implicit conversions induced by the use of explicitly-typed variables. Changing an explicitly-typed variable to auto might change code behaviour, so you'd better be cautious.

Answer 4

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 5

I'm surprised nobody has mentioned this, but suppose you are calculating the factorial of something:

#include <iostream>
using namespace std;

int main() {
    auto n = 40;
    auto factorial = 1;

    for(int i = 1; i <=n; ++i)
    {
        factorial *= i;
    }

    cout << "Factorial of " << n << " = " << factorial <<endl;   
    cout << "Size of factorial: " << sizeof(factorial) << endl; 
    return 0;
}

This code will output this:

Factorial of 40 = 0
Size of factorial: 4

That was definetly not the expected result. That happened because auto deduced the type of the variable factorial as int because it was assigned to 1.

Answer 6

As I described in this answer auto can sometimes result in funky situations you didn't intend. You have to explictly say auto& to have a reference type while doing just auto can create a pointer type. This can result in confusion by omitting the specifier all together, resulting in a copy of the reference instead of an actual reference.