Average Error: 0.0 → 0.0
Time: 10.4s
Precision: 64
$0.0 \le a \le 1$
$b + a \cdot \left(c - b\right)$
$b + a \cdot \left(c - b\right)$
b + a \cdot \left(c - b\right)
b + a \cdot \left(c - b\right)
double f(double b, double a, double c) {
double r1195307 = b;
double r1195308 = a;
double r1195309 = c;
double r1195310 = r1195309 - r1195307;
double r1195311 = r1195308 * r1195310;
double r1195312 = r1195307 + r1195311;
return r1195312;
}


double f(double b, double a, double c) {
double r1195313 = b;
double r1195314 = a;
double r1195315 = c;
double r1195316 = r1195315 - r1195313;
double r1195317 = r1195314 * r1195316;
double r1195318 = r1195313 + r1195317;
return r1195318;
}



# Try it out

Results

 In Out
Enter valid numbers for all inputs

# Derivation

1. Initial program 0.0

$b + a \cdot \left(c - b\right)$
2. Final simplification0.0

$\leadsto b + a \cdot \left(c - b\right)$

# Reproduce

herbie shell --seed 1
(FPCore (b a c)
:name "b + a * (c - b) "
:precision binary64
:pre (<= 0.0 a 1)
(+ b (* a (- c b))))