Understanding heisenbug example: different precision in registers vs main memory

Question

I read the wiki page about heisenbug, but don't understand this example. Can anyone explain it in detail?

One common example of a heisenbug is a bug that appears when the program is compiled with an optimizing compiler, but not when the same program is compiled without optimization (as is often done for the purpose of examining it with a debugger). While debugging, values that an optimized program would normally keep in registers are often pushed to main memory. This may affect, for instance, the result of floating-point comparisons, since the value in memory may have smaller range and accuracy than the value in the register.

Answer 1

Here's a concrete example recently posted:

Infinite loop heisenbug: it exits if I add a printout

It's a really nice specimen because we can all reproduce it: http://ideone.com/rjY5kQ

These bugs are so dependent on very precise features of the platform that people also find them very difficult to reproduce.

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In this case when the 'print-out' is omitted the program performs a high precision comparison inside the CPU registers (higher than stored in a double). But to print out the value the compiler decides to move the result to main memory which results in an implicit truncation of the precision. When it uses that truncated value for the comparison it succeeds.

#include <iostream>
#include <cmath>
  
double up = 19.0 + (61.0/125.0);
double down = -32.0 - (2.0/3.0);
double rectangle = (up - down) * 8.0;
 
double f(double x) {
    return (pow(x, 4.0)/500.0) - (pow(x, 2.0)/200.0) - 0.012;
}
 
double g(double x) {
    return -(pow(x, 3.0)/30.0) + (x/20.0) + (1.0/6.0);
}
 
double area_upper(double x, double step) {
    return (((up - f(x)) + (up - f(x + step))) * step) / 2.0;
}
 
double area_lower(double x, double step) {
    return (((g(x) - down) + (g(x + step) - down)) * step) / 2.0;
}
 
double area(double x, double step) {
    return area_upper(x, step) + area_lower(x, step);
}
 
int main() {
    double current = 0, last = 0, step = 1.0;
 
    do {
        last = current;
        step /= 10.0;
        current = 0;
 
        for(double x = 2.0; x < 10.0; x += step) current += area(x, step);
 
        current = rectangle - current;
        current = round(current * 1000.0) / 1000.0;
        //std::cout << current << std::endl; //<-- COMMENT BACK IN TO "FIX" BUG
    } while(current != last);
 
    std::cout << current << std::endl;
    return 0;
}

Edit: Verified bug and fix still exhibit: 03-FEB-22, 20-Feb-17