Can an array be used in a case expression?
I would like to use a const int array in a switch / case statement in C ++. Is it possible? So far I have tried something like:
int main()
{
int const tab[3] = {1,2,3};
int value(2);
switch(value)
{
case tab[1]:
cout << "result is: " << tab[0]<< endl;
}
return 0;
}
However, the compiler keeps telling me:
.../main.cpp|11|error: the value of ‘tab’ is not usable in a constant expression
Well, I declared my array as "int const", right?
source to share
Each statement case
must accept a constant expression, which is defined as:
A conditional expression e is a constant constant expression if evaluating e, following the rules of the abstract machine (1.9), evaluates one of the following expressions:
- [...]
- lvalue-to-rvalue conversion (4.1) if not applied to
- non-volatile glvalue of an integral or enumerated type that refers to a non-volatile const object with a previous initialization, initialized with a constant expression [Note: a string literal (2.14.5) matches an array of such objects. -end note] or
- a volatile glvalue that refers to a non-volatile object defined with
constexpr
, or which refers to a non-volatile sub-object of such an object, or- a nonvolatile literal glvalue that refers to a nonvolatile object that began its lifespan as part of the e score;
- [...]
Your case is an lvalue-to-rvalue conversion, but none of these three bullet points apply, so tab[1]
it is not a basic constant expression. However, this second point of the subpool gives us a clue: what if the object was defined with constexpr
! This would make it a tab[1]
constant expression, and thus this would compile:
constexpr int tab[3] = {1,2,3};
int value(2);
switch(value)
{
case tab[1]:
cout << "result is: " << tab[0]<< endl;
}
const
does not make the object a constant expression. It just makes it non-mutable. Please note that the code below is fully valid:
int x;
cin >> x;
const int y = x; // obviously, y can't be a constant expression
source to share
No, you don't need const
, you need constexpr
. Rewrite your code with:
int main()
{
constexpr int tab[3] = {1,2,3};
int value(2);
switch(value)
{
case tab[1]:
cout << "result is: " << tab[0]<< endl;
}
return 0;
}
Since you need tab
the compile-time value , it must be an expression constexpr
. Also, each operator case
accepts a constant expression not a constant value. const
will make it a constant value, whereas it constexpr
will make it a constant expression.
source to share