std::atan2, std::atan2f, std::atan2l

Since C++26

Header: <cmath>

1-3) Computes the arc tangent of y / x using the signs of arguments to determine the correct quadrant.The library provides overloads of std::atan2 for all cv-unqualified floating-point types as the type of the parameters.(since C++23)

# Declarations

float atan2 ( float y, float x );
double atan2 ( double y, double x );
long double atan2 ( long double y, long double x );

(until C++23)

/*floating-point-type*/
atan2 ( /*floating-point-type*/ y,
/*floating-point-type*/ x );

(since C++23) (constexpr since C++26)

float atan2f( float y, float x );

(since C++11) (constexpr since C++26)

long double atan2l( long double y, long double x );

(since C++11) (constexpr since C++26)

SIMD overload (since C++26)

template< class V0, class V1 >
constexpr /*math-common-simd-t*/<V0, V1>
atan2 ( const V0& v_y, const V1& v_x );

(since C++26)

Additional overloads (since C++11)

template< class Integer >
double atan2 ( Integer y, Integer x );

(constexpr since C++26)

# Parameters

# Return value

If a domain error occurs, an implementation-defined value is returned (NaN where supported).

# Notes

std::atan2(y, x) is equivalent to std::arg(std::complex<std::common_type_t<decltype(x), decltype(y)»(x, y)).

POSIX specifies that in case of underflow, the value y / x is returned, and if that is not supported, an implementation-defined value no greater than DBL_MIN, FLT_MIN, and LDBL_MIN is returned.

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 and second argument num2:

If num1 and num2 have arithmetic types, then std::atan2(num1, num2) has the same effect as std::atan2(static_cast</common-floating-point-type/>(num1),static_cast</common-floating-point-type/>(num2)), where /common-floating-point-type/ is the floating-point type with the greatest floating-point conversion rank and greatest floating-point conversion subrank between the types of num1 and num2, 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.

# Example

#include <cmath>
#include <iostream>
 
void print_coordinates(int x, int y)
{
    std::cout << std::showpos
              << "(x:" << x << ", y:" << y << ") cartesian is "
              << "(r:" << std::hypot(x, y)
              << ", phi:" << std::atan2(y, x) << ") polar\n";
}
 
int main()
{
    // normal usage: the signs of the two arguments determine the quadrant
    print_coordinates(+1, +1); // atan2( 1,  1) =  +pi/4, Quad I
    print_coordinates(-1, +1); // atan2( 1, -1) = +3pi/4, Quad II
    print_coordinates(-1, -1); // atan2(-1, -1) = -3pi/4, Quad III
    print_coordinates(+1, -1); // atan2(-1,  1) =  -pi/4, Quad IV
 
    // special values
    std::cout << std::noshowpos
              << "atan2(0, 0) = " << atan2(0, 0) << '\n'
              << "atan2(0,-0) = " << atan2(0, -0.0) << '\n'
              << "atan2(7, 0) = " << atan2(7, 0) << '\n'
              << "atan2(7,-0) = " << atan2(7, -0.0) << '\n';
}

# See also