Thread: Issue with pointer to a const char using typedef

I have declared a pointer to const char and incremented the pointer, it is not giving any issue.

Code:

int main()
{
        const char * ptr;
        char *c = "ABC";
        ptr = c;
        ptr++;


        return 0;
}

I have used typedef for declaring a pointer to const char as shown below, but it is giving error during compilation. It seems it is treating pointer as constant pointer.

Code:

typedef char *P;
int main()
{


        const P ptr;
        char *c = "ABC";
        ptr = c;
        ptr++;


        return 0;
}

Error during compilation:

Code:

test.c: In function ‘main’:
test.c:10:2: error: assignment of read-only variable ‘ptr’
test.c:11:2: error: increment of read-only variable ‘ptr’

I think typedef is used for defining typedefs, in this case ptr is treated as constant pointer, not a pointer to constant char.

Code:

typedef char *P;
const P ptr;

Is there any limitations with typedef in C.

First of all, 'c' is assigned the address of a CONSTANT string, "ABC", which cannot be altered, so for the examples below, I will use a different string example:

Code:

char str[] = "ABC";

This defines a non-const char array, initialized with a CONSTANT string, "ABC". You now have two copies of the string, one non-const (the array), and the other const.

Code:

typedef char * P;

const P ptr = str;

This declares a const pointer to a non-const char. The const pointer would need to be initialized!

Code:

typedef char const * P;
P ptr = str;

In this case, it declares a non-const pointer to a const char.

And to be thorough:

Code:

typedef char const * P;
const P ptr = str;

ptr is a const pointer to a const char. neither can be altered.

I hope this has help answered your question, rather than confusing you even more! ;^)

There are many tutorials on interpreting complex pointer definitions, this is but one of them.

Thanks for the info.