Most efficient/elegant way to clip a number?

Question

Given a real (n), a maximum value this real can be (upper), and a minimum value this real can be (lower), how can we most efficiently clip n, such that it remains between lo

Answer 1

If performance really matters to you, what about an inline solution that avoids assignment when not required:

#define clip(n, lower, upper) if (n < lower) n= lower; else if (n > upper) n= upper

Answer 2

UPDATE: C++17's <algorithm> header added std::clamp(value, low, high).

In older C++ versions, I'd very rarely go beyond...

return n <= lower ? lower : n >= upper ? upper : n;

...or, if you find it more readable keeping the left-to-right ordering of lower, n and upper...

return n <= lower ? lower : n <= upper ? n : upper;

(using <= lower is better than < lower because when n == lower it avoids having to compare with upper)

If you know you might have them, you'd want to check if NaN / Inf etc. are preserved....

I say rarely and not never just because sometimes less branching can be faster, but you'd sure want to profile it and prove it helped and mattered....

Answer 3

C++17 is expected to add a clamp function. Courtesy of cppreference.com:

template<class T>
constexpr const T& clamp( const T& v, const T& lo, const T& hi );

template<class T, class Compare>
constexpr const T& clamp( const T& v, const T& lo, const T& hi, Compare comp );

Answer 4

What about boring, old, readable, and shortest yet:

float clip(float n, float lower, float upper) {
  return std::max(lower, std::min(n, upper));
}

?

This expression could also be 'genericized' like so:

template <typename T>
T clip(const T& n, const T& lower, const T& upper) {
  return std::max(lower, std::min(n, upper));
}

Update

Billy ONeal added:

Note that on windows you might have to define NOMINMAX because they define min and max macros which conflict

Answer 5

You might like the ternary operator:

value = value<lower?lower:value;
value = value>upper?upper:value;