Can cout alter variables somehow?

Question

So I have a function that looks something like this:

float function(){
    float x = SomeValue;
    return x / SomeOtherValue;
}

At some point,

Answer 1

Welcome to the wonderful world of floating point. The answer you get will likely depend on the floating point model you compiled the code with.

This happens because of the difference between the IEEE spec and the hardware the code is running on. Your CPU likely has 80 bit floating point registers that get use to hold the 32-bit float value. This means that there is far more precision while the value stays in a register than when it is forced to a memory address (also known as 'homing' the register).

When you passed the value to cout the compiler had to write the floating point to memory, and this results in a lost of precision and interesting behaviour WRT overflow cases.

See the MSDN documentation on VC++ floating point switches. You could try compiling with /fp:strict and seeing what happens.

Answer 2

Printing a value to cout should not change the value of the paramter in any way at all.

However, I have seen similar behaviour, adding debugging statements causes a change in the value. In those cases, and probably this one as well my guess was that the additional statements were causing the compiler's optimizer to behave differently, so generate different code for your function.

Adding the cout statement means that the vaue of x is used directly. Without it the optimizer could remove the variable, so changing the order of the calculation and therefore changing the answer.