const int t=5;
char buf[t+5];
When I compile this gives error in C but not in C++!!
Can anybody please explain me the reason?
Note: I know the const defaults to internal linkage in 'C++', where as in 'C' it defaults to external linkage. Does it has any relation to the above case??
This isn't valid in C89 C, though it may be valid in C99
As others explained, C is kept more simple than C++ and doesn't allow const variables to appear in integer constant expressions. But in both C89 and C++ declared arrays must have compile-time constant sizes.
You can use enumerations for this
enum {
BufSize = 5
};
char buf[BufSize + 5];
It doesn't have to do with internal linkage - external linkage variables are equally viable in integer constant expressions in C++. The internal linkage in C++ rather is a consequence, but not a neccessity, of allowing them to appear in constant expressions. The C++ Standard explains why they have internal linkage by default
Because const objects can be used as compile-time values in C++, this feature urges programmers to provide explicit initializer values for each const. This feature allows the user to put const objects in header files that are included in many compilation units
I think it's because the compiler can't evaluate t+5 to a constant expression. It looks like it should be OK, but:
One important point about array declarations is that they don't permit the use of varying subscripts. The numbers given must be constant expressions which can be evaluated at compile time, not run time.
In C the Size of an Array has to be an Constant Expression. Const Int is in C not an Constant Expression. It's meaning is more like "readonly". Use #define t 5 instead.
Yes, this is related to C's external linking of t.
You've declared an externally linked integer t. If you link this file with another that defines t, then your buffer's size would have to be determined after the file's compile-time, which is of course impossible in C.
extern const int t = 5; char buf[t + 5]; would work equally well in C++.
You have two issues here, dynamically sized arrays and consts. The concept of const is different in C and C++. For C it is just a variable that you don't have the right to change and thus is not a valid dimension for C-dialects that only allow for arrays with compile time fixed dimensions. The only way to define a `compile time constant' in C is to use an enumeration type
enum dummy { t=5 };
So such a thing would work with C89.
In contrast to that you code should also work in C99. There the array would be syntactically identified to be of dynamic size. Then any decent optimizer should be able to optimize this away. But beware also that
sizeof(buf)
would be the total size of the array (10) in C++ whereas with c99 it would be sizeof(char*)
sizeof(buf) would be evaluated at runtime and yield 10 too. :)
