Constant Expression Clarification Required

Question

K & R c 2nd edition (section 2.3) mentions

A constant expression is an expression that includes only constants. Such expressions can be evaluated at compile time, rather than at run time, and, accordingly, can be used anywhere where a constant may occur.

however, I have a few doubts about this:

• Will this expression be considered a constant expression?

  const int x=5;
  const int y=6;
  int z=x+y;
  

  ie is the use of a keyword constconsidered a constant expression or not?

• Is there any method by which we can check if the expression was evaluated at compile time or at run time?

• Are there cases where estimating compilation time gives an excellent result than estimating runtime?

• ? (, )

Answer 1

• . , , , , .
• () , .
• , , . : " , " (§6.6 11 C11).
• , .:) const !

Answer 2

const ?

→ , . , const, const , .

- , , ?

→ ( - Unwind). .

, , ?

→ , . . §6.6 11, C11.

^{FWIW, sizeof ( , ), NULL . . NULL undefined .}

? (, )

→ , .

Answer 3

• x y - const, z - . , , x y, z. , , compiller 5 + 6 z.

• , , , .

• . , .

• :), , .

Answer 4

• C const - , , . , ++. .

• gcc builtin (int __builtin_constant_p (exp)), , .

• , - undefined, -. [1]

• Because constant expressions are evaluated at compile time, they provide a safe processing time and often use code space and possibly data space. In addition, in some places (for example, global initializers), only constant expressions are allowed. See Standard.

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[1]: One example is the right shift of a sign constant with a constant sign, for example. -1 >> 24. Since this implementation is defined, the compiler may give a different result from running the program using a variable that has the same value:

int i = -1;
(-1 >> 24) == (i >> 24)
^             ^--- run-time evaluated by target
+--- compile-time evaluated by compiler

Comparison may fail.