Compiler: Does the compiler evaluate constant expressions at compile time?

Euan Andrew1

Tool/software: TI C/C++ Compiler

I'm working on an F28379D with Compiler v20.2.2.LTS. Let's say I have the following code:

outputA = (float)2/3*inputA;

The compiler reference manual states that:

"For optimal evaluation, the compiler simplifies expressions into equivalent forms, requiring fewer instructions or registers. Operations between constants are folded into single constants. For example, a = (b + 4) - (c + 1) becomes a = b - c + 3."

My question is this, does the compiler recognise that (float)2/3 is always 0.66666666667 and make a subsitution at compile time, or does it instead ask the microprocessor to actually compute 2/3 using its FPU every time? I am considering making this substitution myself with #define preprocessor statements but I am curious if the compiler is actually doing it for me.

over 5 years ago

0 George Mock over 5 years ago

TI__Guru**** 249260 points

Euan Andrew1 said:
does the compiler recognise that (float)2/3 is always 0.66666666667

Yes.

Thanks and regards,

-George

0 Euan Andrew1 over 5 years ago in reply to George Mock

Intellectual 410 points

Thanks for the quick response. If I may ask a follow-up question...

If the example were slightly more complex, and the constant expression contained a function call, would the compiler still recognise that the result was a constant and replace it with its actual value. For example:

outputA = (float)1/sqrt(3)*inputA;

Does the compiler realise that this is always 0.57735026919, or is time spent on a function call?

0 George Mock over 5 years ago in reply to Euan Andrew1

TI__Guru**** 249260 points

Euan Andrew1 said:
Does the compiler realise that this is always 0.57735026919, or is time spent on a function call?

The sqrt function is called.

Thanks and regards,

-George

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Compiler: Does the compiler evaluate constant expressions at compile time?