Hydrologic Runoff Volume effective runoff depth Solver

            from math import *

def hydrologic_runoff_volume_solve_b(c, a) -> float:
    return (c / a)

assert abs(hydrologic_runoff_volume_solve_b(9000, 250000) - 0.036) < 1e-6 * max(1.0, abs(0.036))
          

            #include <assert.h>
#include <math.h>

double hydrologic_runoff_volume_solve_b(double c, double a) {
    return (c / a);
}

int main(void) {
    const double expected = 0.036;
    const double actual = hydrologic_runoff_volume_solve_b(9000, 250000);
    assert(fabs(actual - expected) < 1e-6 * fmax(1.0, fabs(expected)));
}
          

            #include <cassert>
#include <cmath>
#include <numbers>

double hydrologic_runoff_volume_solve_b(double c, double a) {
    return (c / a);
}

int main() {
    constexpr double expected = 0.036;
    const double actual = hydrologic_runoff_volume_solve_b(9000, 250000);
    assert(std::fabs(actual - expected) < 1e-6 * std::fmax(1.0, std::fabs(expected)));
}
          

x86-64 NASM · System V ABI · Linux · SSE2 with libm where required

            ; double hydrologic_runoff_volume_solve_b(double c, double a)
; Linux x86-64 NASM · System V ABI · first eight doubles in xmm0–xmm7
global hydrologic_runoff_volume_solve_b
section .text

hydrologic_runoff_volume_solve_b:
    push rbp
    mov rbp, rsp
    sub rsp, 32
    movsd [rbp-8], xmm0
    movsd [rbp-16], xmm1
    movsd xmm0, [rbp-8]
    divsd xmm0, [rbp-16]
    movsd [rbp-24], xmm0
    movsd xmm0, [rbp-24]
    leave
    ret
          

            function result = hydrologic_runoff_volume_solve_b(c, a)
    result = (c / a);
end
          

            ClearAll[mwCalculate];
mwCalculate[c_, a_] := (c / a);