< cpp‎ | numeric
Defined in header <numeric>
template< class T >
constexpr T midpoint(T a, T b) noexcept;
(1) (since C++20)
template< class T >
constexpr T* midpoint(T* a, T* b);
(2) (since C++20)

Computes the midpoint of the integers, floating-points, or pointers a and b.

1) This overload only participates in overload resolution if T is an arithmetic type other than bool.
2) This overload only participates in overload resolution if T is an object type. Use of this overload is ill-formed if T is an incomplete type.


[edit] Parameters

a, b - integers, floating-points, or pointer values

[edit] Return value

1) Half the sum of a and b. No overflow occurs. If a and b have integer type and the sum is odd, the result is rounded towards a. If a and b have floating-point type, at most one inexact operation occurs.
2) If a and b point to, respectively, x[i] and x[j] of the same array object x (for the purpose of pointer arithmetic), returns a pointer to x[i+(j-i)/2] where the division rounds towards zero. If a and b do not point to elements of the same array object, the behavior is undefined.

[edit] Exceptions

Throws no exceptions.

[edit] Example

#include <cstdint>
#include <limits>
#include <numeric>
#include <iostream>
int main()
    std::uint32_t a = std::numeric_limits<std::uint32_t>::max();
    std::uint32_t b = std::numeric_limits<std::uint32_t>::max() - 2;
    std::cout << "a: " << a << '\n'
              << "b: " << b << '\n'
              << "Incorrect (overflow and wrapping): " << (a + b) / 2 << '\n'
              << "Correct: " << std::midpoint(a, b) << '\n';


a: 4294967295
b: 4294967293
Incorrect (overflow and wrapping): 2147483646
Correct: 4294967294