I tend to use math functions of constant expressions for convinience and coherence (i.e log(x)/log(2)
instead of log(x)/0.3...
). Since these functions aren't actually a part of the language itself, neither are they defined in math.h
(only declared), will the constant ones get precalculated at compile time, or will they be wastefully calculated at runtime?
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478answers:
7I think they would be calculated at runtime, since you are calling a library function. If it was, say (2+3), I guess a compiler optimization would take care of it. I may be wrong, however.
They'll usually get calculated at runtime (see other answers for how to inline them), though I wouldn't necessarily use the word "wastefully" unless there were a lot of them and/or the code was executed lots of times.
Rather than just put in the constant value, create a #define
or const
variable that expresses what the value means (PI
, LOG_2
, etc.) and use that instead.
For example:
#define LOG_2 0.30102999566
This doesn't do any calculation and you can specify the value to what ever precision you desire or can manage given your environment (32 bit v 64 bit).
This happens at runtime since the code for the function only becomes available during the linking step (which happens after the compile step).
To optimize the step, use a global variable which you initialize before you use it.
It depends. If the compiler can do the math exactly as it would be done at runtime, and if link time optimizations are performed, and if the library is kept in some sort of intermediate form, then yes it could be done.
Most compilers don't do all of that.
Some compilers will evaluate them at compile time, but this behavior is not guaranteed (and doing so may also cause problems). You'll need to check your compiler and see what it does.
Note that on many systems, log(2)
is available from the macro M_LN2
in <math.h>
.
In case of library functions, some compiler might implement these functions as intrinsics, meaning that the compiler knows enough about the functions to replace the call with a constant at compile time. Whether it will do this depends on the compiler. In practice, I often notice that some compilers are reluctant to pre-calculate floating-point expressions at compile time, even if they don't involve any function calls.
In general case, normally they won't and can't get calculated at compile time, assuming the compiler simply doesn't know enough about these functions to be able to calculate them at compile time. Maybe they have some prominent run-time side-effects?
Some compilers might have non-standard compiler-dependent extensions that allow user to provide the additional information about the functions to the compiler, so that the compiler can optimize function calls better and even figure out whether a given call can be replaced with a compile-time pre-calculation. For example, GCC compiler supports such function attributes (GCC-specific extension) as const
and pure
. If the arguments are known at compile time, a call to function with the const
attribute can be theoretically replaced with compile-time pre-calculation. Although I can't say whether GCC can actually do this.
In C++ (even though your question is tagged C) language a planned new feature is a constexpr
declaration specifier that serves a similar purpose and supposed to have the effect you describe.
It depends on the compiler and the optimization flags. As @AndrewyT points out, GCC has the ability to specify which functions are constant and pure via attributes, and in this case the answer is positive, it will inline the result, as you can easily check:
$ cat constant_call_opt.c
#include <math.h>
float foo() {
return log(2);
}
$ gcc -c constant_call_opt.c -S
$ cat constant_call_opt.s
.file "constant_call_opt.c"
.text
.globl foo
.type foo, @function
foo:
pushl %ebp
movl %esp, %ebp
subl $4, %esp
movl $0x3f317218, %eax
movl %eax, -4(%ebp)
flds -4(%ebp)
leave
ret
.size foo, .-foo
.ident "GCC: (Ubuntu 4.3.3-5ubuntu4) 4.3.3"
.section .note.GNU-stack,"",@progbits
no function call there, just loads a constant (0x3f317218 == 0.69314718246459961 == log(2)
)
Althought I have not any other compiler at hand to try now, I think you could expect the same behaviour in all the major C compilers out there, as it's a trivial optimization.