How does this float square root approximation work?

Question

I found a rather strange but working square root approximation for floats; I really don\'t get it. Can someone explain me why this code works?

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Answer 1

Adding a wiki test harness to test all float.

The approximation is within 4% for many float, but very poor for sub-normal numbers. YMMV

Worst:1.401298e-45 211749.20%
Average:0.63%
Worst:1.262738e-38 3.52%
Average:0.02%

Note that with argument of +/-0.0, the result is not zero.

printf("% e % e\n", sqrtf(+0.0), sqrt_apx(0.0));  //  0.000000e+00  7.930346e-20
printf("% e % e\n", sqrtf(-0.0), sqrt_apx(-0.0)); // -0.000000e+00 -2.698557e+19

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Test code

#include 
#include 
#include 
#include 
#include 
#include 
#include 

float sqrt_apx(float f) {
  const int result = 0x1fbb4000 + (*(int*) &f >> 1);
  return *(float*) &result;
}

double error_value = 0.0;
double error_worst = 0.0;
double error_sum = 0.0;
unsigned long error_count = 0;

void sqrt_test(float f) {
  if (f == 0) return;
  volatile float y0 = sqrtf(f);
  volatile float y1 = sqrt_apx(f);
  double error = (1.0 * y1 - y0) / y0;
  error = fabs(error);
  if (error > error_worst) {
    error_worst = error;
    error_value = f;
  }
  error_sum += error;
  error_count++;
}

void sqrt_tests(float f0, float f1) {
  error_value = error_worst = error_sum = 0.0;
  error_count = 0;
  for (;;) {
    sqrt_test(f0);
    if (f0 == f1) break;
    f0 = nextafterf(f0, f1);
  }
  printf("Worst:%e %.2f%%\n", error_value, error_worst*100.0);
  printf("Average:%.2f%%\n", error_sum / error_count);
  fflush(stdout);
}

int main() {
  sqrt_tests(FLT_TRUE_MIN, FLT_MIN);
  sqrt_tests(FLT_MIN, FLT_MAX);
  return 0;
}