What is a value known at compile time ?

[pieroborrelli]

Hello, I'm studying the C++ programming language and I'm in a chapter where thebook introduces the constexpr qualifier, to initialize a constexpr variable we need of a constant expression that can be evaluated at compile time, what does evaluated at compile time mean ? What's the difference between a variable and a constant for the compiler ?

int a = 10; 
constexpr int b = 15; // what's the difference ? 

what happens to the variable knwon at compile when the programs run ? the variable a for example is simply created in memory but what about b ?
Thank you for your help I'm just a beginner

[SECMoG1T]

evaluated at compile time simply means that the compiler is able to evaluate solve for the value of an expression when converting it to machine instructions

A simple example would be that of the size of a static array, in such an arry the size can't change when the program is running hence a rule of thumb is to provide the size of such an arry as a const integral, a simple integral literal or a constant expression so that the compiler will be able to determine it size before the program runs.

For a dynamic array whose size is determined at runtime it won't require that it size be know at compile time

int ar [5]; //this size must be know when the compiler is translating the code to machine instructions
                 /// the size is an integral literal 5;

constexpr int f ()/// this is a constexpr function compiler can evaluate this expression at compile
   {                         /// time since it will always return a constant value of 4;
       return 2+2;
    }

constexpr int f2 () /// this function is an error it violates rules of a const exp function
  {
     int x;    /// not that most constexpr function I av come across only contains a return
     cout <<"enter a value "; cin>> x;  /// statement
     return x+2;
  }

  constexpr int k=f ()//correct k is a constexpr intialized bu a constexpr function

 constexpr int funct (int x)  /// this is a constexpr function compiler can evaluate it at compile time 
   {                     ///providef x is a const or constexpr statement; 
      return x*x;
   }

  E.g. constexpr int z= funct (6); // this is ok

         int val;
         cout <<" enter a value "; cin>> val;
         constexpr int z1=funct (val);///  this is an error val is not a const or constexpr statement

 compiler can always resolve a constexpr statement at compile time ; note that a constexpr statement or variable is implicitly const and should be initialized on declaration

  z=6;// this is an error z is a constexpr variable. 

 integral are all const 
   eg 5/// can be used in a cconstexpr statement 5 will always be 5; 5=2 this is an error; 

 Other variables can be subject to change hence cant be used in constexpr statements

  E.g. int h=7; /// h can't be used to intialize a constexpr because h isn't a const variable
  
     constexpr int expr=h; /// ! Error h isn't a const variable 

[dhayden]

"Compile time" means "when you compile the program from source code. "Run time" is when you are actually running the program. In a constexpr, the compiler must be able to reduce the expression to a constant at compile time.

int a = 10; 
constexpr int b = 15; // what's the difference ?  

a is a variable. You can assign a new value to it. b is a constant - basically a synonym for 15. You can't assign a new value to b any more than you can "assign" 22 to 10.

what happens to the variable known at compile when the programs run ?

A constexpr stays the same. That's the whole point.

[pieroborrelli]

@dhayden So a constexpr variable will be treated as a constant value at compile time, but because this does not happen with the variables that do not use constexpr? for example :

int a = 10; // this is not a constant expression 

in this case I'm initializing a with constant expression, however, the variable is not a constant expression right? why ? .

[dhayden]

the variable is not a constant expression right? why ?

Right. It is a variable because you defined it as one when you said it was an int. The main thing about the variable is that you can change its values:

int a=10;
int b = 20;
a = 82;
a = b-17;
a = a+b;
// etc.