Subtle Bug with std::min/max Function Templates

Suppose you have a function f that returns a double, and you want to store in a variable the value of this function, if this a positive number, or zero if the return value is negative. This line of C++ code tries to do that:

double x = std::max(0, f(/* something */));

Unfortunately, this apparently innocent code won’t compile!

The error message produced by VS2015’s MSVC compiler is not very clear, as often with C++ code involving templates.

So, what’s the problem with that code?

The problem is that the std::max function template is declared something like this:

template <typename T> 
const T& max(const T& a, const T& b)

If you look at the initial code invoking std::max, the first argument is of type int; the second argument is of type double (i.e. the return type of f).

Now, if you look at the declaration of std::max, you’ll see that both parameters are expected to be of the same type T. So, the C++ compiler complains as it’s unable to deduce the type of T in the code calling std::max: should T be int or double?

This ambiguity triggers a compile-time error.

To fix this error, you can use the double literal 0.0 instead of 0.

And, what if instead of 0 there’s a variable of type int?

Well, in this case you can either static_cast that variable to double:

double x = std::max(static_cast<double>(n), f(/* something */));

or, as an alternative, you can explicitly specify the double template type for std::max:

double x = std::max<double>(n, f(/* something */));

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