c++

Mac1ek · 07-06-2024, 04:41 PM

I have question - if i have loop - for, when in compilation time index begin to end is knowed, i cant use constexpr for this evaluation like this:

Code:

for(int i = 0; i<10; ++i) {
   constexpr foo(i);
}

? Thanks.

astrogeek · 07-07-2024, 11:49 AM

As no one else has stepped up...

I have never actually used c++ constexpr but from looking over the docs and an article here, I think you cannot use it as in your example.

First, foo(int) would have to have been declared as constexpr which is not shown in your example. but even if it were it's parameters must be known at compile time and in your example that is not possible. Even though you know the range of i within the loop, it is still a range, not a constant value that can be known at compile time.

NevemTeve · 07-07-2024, 12:01 PM

I guess it is not impossible to actually try it. In my case it didn't work

Code:

/* Mac1ek.cc */

extern int foo(int);

int main ()
{
    for(int i = 0; i<10; ++i) {
        constexpr foo(i);
    }
    return 0;
}

compilation:

Code:

g++ -std=c++23 -o Mac1ek Mac1ek.cc Mac1ek2.cc
Mac1ek.cc: In function ‘int main()’:
Mac1ek.cc:8:19: error: ISO C++ forbids declaration of ‘foo’ with no type [-fpermissive]
    8 |         constexpr foo(i);
      |                   ^~~
Mac1ek.cc:8:23: error: the value of ‘i’ is not usable in a constant expression
    8 |         constexpr foo(i);
      |                       ^
Mac1ek.cc:7:13: note: ‘int i’ is not const
    7 |     for(int i = 0; i<10; ++i) {
      |             ^

smallpond · 07-07-2024, 03:21 PM

How can it be a constexpr if you are calling a function with a variable argument?

sundialsvcs · 07-07-2024, 05:13 PM

To me, constexpr is yet-another "over-engineered thing" that someone added to C++. The formal description says:

Quote:

The constexpr specifier declares that it is possible to evaluate the value of the function or variable at compile time. Such variables and functions can then be used where only compile time constant expressions are allowed [...]

So it functions like a "constant," because it can be evaluated at compile time, but it is an "expression." The compiler now has to do all the heavy-lifting to figure out what the "constant value" is, in any particular case. But, having done so, the result is "a constant value."

Clearly, in your case this cannot be done: you cannot possibly "evaluate foo(i) at compile time," simply because of "i"!

Incidentally, I'd be interested to hear from any C++ programmers out there who think that this "fee-chur" is "the bee's knees ..."

If a use-case exists to justify this level of compiler complexity, I'd like to hear what it is.

slackerz · 07-09-2024, 07:53 AM

Ideally what you want to do is constexpr pretty much everything in your code. In practice the use of constexpr ultimately reduces to a handful of constants in the program.

EdGr · 07-09-2024, 08:23 AM

Gcc is very good at inlining functions and evaluating constant expressions at compile time. Here is an example:

Code:

// compile with g++ -O3 -Wall -S test.cpp

namespace {
  long i=4;
  long foo () { return i; }
}

long bar () { return foo (); }

The assembly code for bar () simply returns 4. No code is generated for i or foo ().

I use const but have not found a need for constexpr.
Ed

sundialsvcs · 07-09-2024, 09:45 AM

Of course, I was not wondering about consts, but specifically: constexpr ...