Defined type not recognised by pointer initialization

Question

Trying to define a user-type I stumbled upon an error I don't quite understand: the following code works fine

const int a[5] = { 1, 2, 3, 4, 5 };

typedef const int costant;

costant b = 5;
costant *c = a;

void func( const int item )
{
    printf("Item: %i\n", item );
}

int main( void )
{
    func( a[0] );
    func( b );
    func( *c );
}

Nonethe less, initializing the pointer

costant *c = a[0];

fails stating that Initializer element is not costant. Trying to access the location a[0] to alter its content is refused with error assignment of read-only location.

Why is considered the array address a as const but not its element a[0], even though it is "read-only"?

Replacing the typedef with a #define costant const int doesn't change the behaviour.

The typedef line doesn't compile. Please do not present uncompilable code. — Jonathan Leffler, Jan 04 '17 at 15:46
`costant * c = &a[0];` would work (assuming a fixed typedef) -- because a[0] is a `const int` but `c` is a `pointer to a const int` — John Hascall, Jan 04 '17 at 15:47
@JonathanLeffler Sorry, I corrected the typedef line. It does compile now — Matteo Furlan, Jan 04 '17 at 16:09

score 1 · Answer 1 · answered Jan 04 '17 at 15:47

1

Unlike #define, which has the syntax #define NEW_SYMBOL [old_symbol], the syntax of typedef is:

typedef existing_type new_type;

So:

typedef const int constant;

Is valid C, whereas

typedef constant const int;

Is not.

answered Jan 04 '17 at 15:47

Govind Parmar

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You are right, I corrected the mistake (I just copied wrongly from my editor in another pc) – Matteo Furlan Jan 04 '17 at 16:10

Defined type not recognised by pointer initialization

1 Answers1