constexpr double lerp( double a, double b, double t ) noexcept;
constexpr long double lerp( long double a, long double b,
lerp( /* floating-point-type */ a,
/* floating-point-type */ b,
constexpr /* common-floating-point-type */
Computes the
linear interpolationbetween
aand
b, if the parameter
tis inside
[
â0â,
1)
(the
linear extrapolationotherwise), i.e. the result of
\(a+t(bâa)\)a+t(bâa)with accounting for floating-point calculation imprecision.
The library provides overloads for all cv-unqualified floating-point types as the type of the parameters a, b and t.(since C++23)A) Additional overloads are provided for all other combinations of arithmetic types.
[edit] Parameters a, b, t - floating-point or integer values [edit] Return value\(a + t(b â a)\)a + t(b â a)
When std::isfinite(a) && std::isfinite(b) is true, the following properties are guaranteed:
Let CMP(x, y) be 1 if x > y, -1 if x < y, and â0â otherwise. For any t1 and t2, the product of
is non-negative. (That is, std::lerp
is monotonic.)
The additional overloads are not required to be provided exactly as (A). They only need to be sufficient to ensure that for their first argument num1, second argument num2 and third argument num3:
If num1, num2 and num3 have arithmetic types, then std::lerp(num1, num2, num3) has the same effect as std::lerp(static_cast</*common-floating-point-type*/>(num1),
static_cast</*common-floating-point-type*/>(num2),
static_cast</*common-floating-point-type*/>(num3)), where /*common-floating-point-type*/ is the floating-point type with the greatest floating-point conversion rank and greatest floating-point conversion subrank among the types of num1, num2 and num3, arguments of integer type are considered to have the same floating-point conversion rank as double.
If no such floating-point type with the greatest rank and subrank exists, then overload resolution does not result in a usable candidate from the overloads provided.
(since C++23) [edit] Example#include <cassert> #include <cmath> #include <iostream> float naive_lerp(float a, float b, float t) { return a + t * (b - a); } int main() { std::cout << std::boolalpha; const float a = 1e8f, b = 1.0f; const float midpoint = std::lerp(a, b, 0.5f); std::cout << "a = " << a << ", " << "b = " << b << '\n' << "midpoint = " << midpoint << '\n'; std::cout << "std::lerp is exact: " << (a == std::lerp(a, b, 0.0f)) << ' ' << (b == std::lerp(a, b, 1.0f)) << '\n'; std::cout << "naive_lerp is exact: " << (a == naive_lerp(a, b, 0.0f)) << ' ' << (b == naive_lerp(a, b, 1.0f)) << '\n'; std::cout << "std::lerp(a, b, 1.0f) = " << std::lerp(a, b, 1.0f) << '\n' << "naive_lerp(a, b, 1.0f) = " << naive_lerp(a, b, 1.0f) << '\n'; assert(not std::isnan(std::lerp(a, b, INFINITY))); // lerp here can be -inf std::cout << "Extrapolation demo, given std::lerp(5, 10, t):\n"; for (auto t{-2.0}; t <= 2.0; t += 0.5) std::cout << std::lerp(5.0, 10.0, t) << ' '; std::cout << '\n'; }
Possible output:
a = 1e+08, b = 1 midpoint = 5e+07 std::lerp is exact?: true true naive_lerp is exact?: true false std::lerp(a, b, 1.0f) = 1 naive_lerp(a, b, 1.0f) = 0 Extrapolation demo, given std::lerp(5, 10, t): -5 -2.5 0 2.5 5 7.5 10 12.5 15[edit] See also midpoint between two numbers or pointers
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