\[ \begin{array}{c}[x, y] = \mathsf{sort}([x, y])\\ \end{array} \]
Math FPCore C Fortran Java Python Julia MATLAB Wolfram TeX \[x \cdot \frac{\frac{y}{z} \cdot t}{t}
\]
↓
\[\begin{array}{l}
t_1 := \frac{y}{z} \cdot x\\
t_2 := \frac{1}{\frac{z}{y \cdot x}}\\
\mathbf{if}\;\frac{y}{z} \leq -1 \cdot 10^{+256}:\\
\;\;\;\;t_2\\
\mathbf{elif}\;\frac{y}{z} \leq -1 \cdot 10^{-242}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;\frac{y}{z} \leq 10^{-253}:\\
\;\;\;\;\frac{y \cdot x}{z}\\
\mathbf{elif}\;\frac{y}{z} \leq 5 \cdot 10^{+109}:\\
\;\;\;\;t_1\\
\mathbf{else}:\\
\;\;\;\;t_2\\
\end{array}
\]
(FPCore (x y z t) :precision binary64 (* x (/ (* (/ y z) t) t))) ↓
(FPCore (x y z t)
:precision binary64
(let* ((t_1 (* (/ y z) x)) (t_2 (/ 1.0 (/ z (* y x)))))
(if (<= (/ y z) -1e+256)
t_2
(if (<= (/ y z) -1e-242)
t_1
(if (<= (/ y z) 1e-253)
(/ (* y x) z)
(if (<= (/ y z) 5e+109) t_1 t_2)))))) double code(double x, double y, double z, double t) {
return x * (((y / z) * t) / t);
}
↓
double code(double x, double y, double z, double t) {
double t_1 = (y / z) * x;
double t_2 = 1.0 / (z / (y * x));
double tmp;
if ((y / z) <= -1e+256) {
tmp = t_2;
} else if ((y / z) <= -1e-242) {
tmp = t_1;
} else if ((y / z) <= 1e-253) {
tmp = (y * x) / z;
} else if ((y / z) <= 5e+109) {
tmp = t_1;
} else {
tmp = t_2;
}
return tmp;
}
real(8) function code(x, y, z, t)
real(8), intent (in) :: x
real(8), intent (in) :: y
real(8), intent (in) :: z
real(8), intent (in) :: t
code = x * (((y / z) * t) / t)
end function
↓
real(8) function code(x, y, z, t)
real(8), intent (in) :: x
real(8), intent (in) :: y
real(8), intent (in) :: z
real(8), intent (in) :: t
real(8) :: t_1
real(8) :: t_2
real(8) :: tmp
t_1 = (y / z) * x
t_2 = 1.0d0 / (z / (y * x))
if ((y / z) <= (-1d+256)) then
tmp = t_2
else if ((y / z) <= (-1d-242)) then
tmp = t_1
else if ((y / z) <= 1d-253) then
tmp = (y * x) / z
else if ((y / z) <= 5d+109) then
tmp = t_1
else
tmp = t_2
end if
code = tmp
end function
public static double code(double x, double y, double z, double t) {
return x * (((y / z) * t) / t);
}
↓
public static double code(double x, double y, double z, double t) {
double t_1 = (y / z) * x;
double t_2 = 1.0 / (z / (y * x));
double tmp;
if ((y / z) <= -1e+256) {
tmp = t_2;
} else if ((y / z) <= -1e-242) {
tmp = t_1;
} else if ((y / z) <= 1e-253) {
tmp = (y * x) / z;
} else if ((y / z) <= 5e+109) {
tmp = t_1;
} else {
tmp = t_2;
}
return tmp;
}
def code(x, y, z, t):
return x * (((y / z) * t) / t)
↓
def code(x, y, z, t):
t_1 = (y / z) * x
t_2 = 1.0 / (z / (y * x))
tmp = 0
if (y / z) <= -1e+256:
tmp = t_2
elif (y / z) <= -1e-242:
tmp = t_1
elif (y / z) <= 1e-253:
tmp = (y * x) / z
elif (y / z) <= 5e+109:
tmp = t_1
else:
tmp = t_2
return tmp
function code(x, y, z, t)
return Float64(x * Float64(Float64(Float64(y / z) * t) / t))
end
↓
function code(x, y, z, t)
t_1 = Float64(Float64(y / z) * x)
t_2 = Float64(1.0 / Float64(z / Float64(y * x)))
tmp = 0.0
if (Float64(y / z) <= -1e+256)
tmp = t_2;
elseif (Float64(y / z) <= -1e-242)
tmp = t_1;
elseif (Float64(y / z) <= 1e-253)
tmp = Float64(Float64(y * x) / z);
elseif (Float64(y / z) <= 5e+109)
tmp = t_1;
else
tmp = t_2;
end
return tmp
end
function tmp = code(x, y, z, t)
tmp = x * (((y / z) * t) / t);
end
↓
function tmp_2 = code(x, y, z, t)
t_1 = (y / z) * x;
t_2 = 1.0 / (z / (y * x));
tmp = 0.0;
if ((y / z) <= -1e+256)
tmp = t_2;
elseif ((y / z) <= -1e-242)
tmp = t_1;
elseif ((y / z) <= 1e-253)
tmp = (y * x) / z;
elseif ((y / z) <= 5e+109)
tmp = t_1;
else
tmp = t_2;
end
tmp_2 = tmp;
end
code[x_, y_, z_, t_] := N[(x * N[(N[(N[(y / z), $MachinePrecision] * t), $MachinePrecision] / t), $MachinePrecision]), $MachinePrecision]
↓
code[x_, y_, z_, t_] := Block[{t$95$1 = N[(N[(y / z), $MachinePrecision] * x), $MachinePrecision]}, Block[{t$95$2 = N[(1.0 / N[(z / N[(y * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(y / z), $MachinePrecision], -1e+256], t$95$2, If[LessEqual[N[(y / z), $MachinePrecision], -1e-242], t$95$1, If[LessEqual[N[(y / z), $MachinePrecision], 1e-253], N[(N[(y * x), $MachinePrecision] / z), $MachinePrecision], If[LessEqual[N[(y / z), $MachinePrecision], 5e+109], t$95$1, t$95$2]]]]]]
x \cdot \frac{\frac{y}{z} \cdot t}{t}
↓
\begin{array}{l}
t_1 := \frac{y}{z} \cdot x\\
t_2 := \frac{1}{\frac{z}{y \cdot x}}\\
\mathbf{if}\;\frac{y}{z} \leq -1 \cdot 10^{+256}:\\
\;\;\;\;t_2\\
\mathbf{elif}\;\frac{y}{z} \leq -1 \cdot 10^{-242}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;\frac{y}{z} \leq 10^{-253}:\\
\;\;\;\;\frac{y \cdot x}{z}\\
\mathbf{elif}\;\frac{y}{z} \leq 5 \cdot 10^{+109}:\\
\;\;\;\;t_1\\
\mathbf{else}:\\
\;\;\;\;t_2\\
\end{array}