Usage of the const keyword - c

Question

I know that using the const keyword on function arguments provides better performance, but I always forget to add it. Is the compiler (GCC in this case) smart enough to notice that the variabele never changes during the function, and compile it as if I would have added const explicitly?

Answer 1

You have a common misunderstanding about const. Only making an object const means that its value never changes, and then it's not just during a function, it never changes.
Making a parameter to a function const does not mean its value never changes, it just means that function cannot change the value through that const pointer. The value can change other ways.
For example, look at this function:
void f(const int* x, int* y)
{
cout << "x = " << *x << endl;
*y = 5;
cout << "x = " << *x << endl;
}
Notice that it takes a const pointer to x. However, what if you call it like this:
int x = 10;
f(&x, &x);
Now, f has a const pointer, but it's to a non-const object. So the value can change, and it does since y is a non-const pointer to the same object. All of this is perfectly legal code. There's no funny business here.
So there's really no answer to your question since it's based entirely on false premises.

Answer 2

Is the compiler (GCC in this case) smart enough to notice that the
variabele never changes during the function, and compile it as if I
would have added const explicitly?
Not necessarily. For example:
void some_function(int *ptr); // defined in another translation unit
int foo(int a) {
some_function(&a);
return a + 1;
}
The compiler can't see what some_function does, so it can't assume that it won't modify a.
Link-time optimization could perhaps see what some_function really does and act accordingly, but as far as this answer is concerned I'll consider only optimization for which the definition of some_function is unavailable.
int bar(const int a) {
some_function((int*)&a);
return a + 1;
}
The compiler can't see what some_function does, but it can assume that the value of a does not change anyway. Therefore it can make any optimizations that apply: maybe it can keep a in a callee-saves register across the call to some_function; maybe it computes the return value before making the call instead of after, and zaps a. The program has undefined behavior if some_function modifies a, and so from the compiler's POV once that happens it doesn't matter whether it uses the "right" or "wrong" value for a.
So, by in this example by marking a const you have told the compiler something that it cannot otherwise know -- that some_function will not modify *ptr. Or anyway that if it does modify it, then you don't care what your program's behavior is.
int baz(int a) {
some_function(NULL);
return a + 1;
}
Here the compiler can see all relevant code as far as the standard is concerned. It doesn't know what some_function does, but it does know that it doesn't have any standard means to access a. So it should make no difference whether a is marked const or not because the compiler knows it doesn't change anyway.
Debugger support can complicate this situation, of course -- I don't know how things stand with gcc and gdb, but in theory at least if the compiler wants to support you breaking in with the debugger and modifying a manually then it might not treat it as unmodifiable. The same applies to the possibility that some_function uses platform-specific functionality to walk up the stack and mess with a. Platforms don't have to provide such functionality, but if they do then it conflicts with optimization.
I've seen an old version of gcc (3.x, can't remember x) that failed to make certain optimizations where I failed to make a local int variable const, but in my case gcc 4 did make the optimization. Anyway, the case I'm thinking of wasn't a parameter, it was an automatic variable initialized with a constant value.
There's nothing special about a being a parameter in any of what I've said -- it could just as well be any automatic variable defined in the function. Mind you, the only way to for a parameter to get the effect of initialization with a constant value is to call the function with a constant value, and for the compiler to observe the value for that call. This tends to happen only when the function is inlined. So inlined calls to functions can have additional optimizations applied to them that the "out-of-line" function body isn't eligible for.

Answer 3

const, much like inline, is only a hint for a compiler and does not guarantee any performance gains. The more important task of const is to protect programmers from themselves so they do not unwilling modify variables where they shouldn’t be modified.

Answer 4

1) Really const is not affecting your performance directly. It may in some cases make simpler points-to analysis (so prefer const char* to char*), but const is more about semantics and readability of your code.
2) CV-qualified type forms different type in C and C++. So your compiler, even if it sees profit from making default const, will not do it, because it will change type and may lead to surprisingly odd things.

Answer 5

As part of the optimization the compiler is taking a deep look at when memory locations are read or written. So the compiler is quite good at detecting when a variable is not changed (const) and when it is changed. The optimizer does not need you to tell him when a variable is const.
Nevertheless you should always use const when appropriate. Why? Because it makes interfaces more clear and easier to understand. And it helps detect bugs when you are changing a variable that you did not want to change.