Passing double types to ceil results in different values for different optimization levels in GCC

Question

Below, the result1 and result2 variable values are reporting different values depending upon whether or not you compile the code with -g or with -O on GCC 4.2.1 and on GCC 3.2.0 (and I have not tried more recent GCC versions):

double double_identity(double in_double)
{
    return in_double;
}

...

double result1 = ceil(log(32.0) / log(2.0));
std::cout << __FILE__ << ":" << __LINE__ << ":" << "result1==" << result1 << std::endl;

double result2 = ceil(double_identity(log(32.0) / log(2.0)));
std::cout << __FILE__ << ":" << __LINE__ << ":" << "result2==" << result2 << std::endl;

result1 and result2 == 5 only when compiling using -g, but if instead I compile with -O I get result1 == 6 and result2 == 5.

This seems like a difference in how optimization is being done by the compiler, or something to do with IEEE floating point representation internally, but I am curious as to exactly how this difference occurs. I'm hoping to avoid looking at the assembler if at all possible.

The above was compiled in C++, but I presume the same would hold if it was converted to ANSI-C code using printfs.

The above discrepancy occurs on 32-bit Linux, but not on 64-bit Linux.

Thanks bg

Answer 1

On x86, with optimizations on, the results of subexpressions are not necessarily stored into a 64-bit memory location before being used as part of a larger expression.

Because x86's standard floating-point registers are 80 bits, this means that in such cases, extra precision is available. IF you then divide (or multiply) that especially-precise value by another, the effects of the increased precision can magnify to the point where they can be perceived by the naked eye.

Intel's 64-bit processors use SSE registers for floating point math, and those registers don't have the extra precision.

You can play around with g++ flags to fix this if you really care.