Thread: Confused with const in function parameters

My teacher posted sample code with a function with parameters:
void funcName ( char const * const init )
What I understand is that both the char and the pointer are constants.

I'm confused as to what type of pointer init is from the declaration alone.
Usually if I want to pass a pointer I just do
void myFunc (int * pointer)
So the type of the pointer is an int because it's to the left of "*".

Looking at the code I know that init is really a string.
So shouldn't the function instead be?:
void funcName ( char const char* const init)

Thanks

Init may be a string. There's other uses for a char pointer. Strings are simply the most common.

Originally Posted by seal308

My teacher posted sample code with a function with parameters:
void funcName ( char const * const init )
What I understand is that both the char and the pointer are constants.

That is true. However, note that this is only known to be true of the formal parameter; the actual argument could be a non-const pointer to non-const char.

Originally Posted by seal308

Usually if I want to pass a pointer I just do
void myFunc (int * pointer)
So the type of the pointer is an int because it's to the left of "*".

No, the type of the pointer is int*, not int. Since the type of the pointer is int*, the type of what the pointer points to is int.

Originally Posted by seal308

Looking at the code I know that init is really a char*
So shouldn't the function instead be?:
void funcName ( char const char* const init)

This:

Code:

void funcName ( char const* const init)

is exactly the same as this:

Code:

void funcName ( const char* const init)

except for spelling. That is, when the const qualifier is at the left-most, it applies to what is on its immediate right, whereas it would otherwise apply to what is on its immediate left. Consequently, what you have in mind would be redundant: you would essentially be trying to const qualify the char twice.

Thank you, I understand it perfectly now.
It clicked when you said that const applies to what is on its immediate left.
It's easier for me to first think of it as empty of const so:

Code:

void funcName (char* init)

If I want to make the pointer constant I write const with the * on its immediate left.

Code:

void funcName (char* const init)

If I want to make the data constant I write const with the char on its immediate left.

Code:

void funcName (char const* init)

So when reading the parameters, I should read them from right to left.

Is the advantage of this to prevent memory leaks?

Originally Posted by laserlight

That is true. However, note that this is only known to be true of the formal parameter; the actual argument could be a non-const pointer to non-const char.


No, the type of the pointer is int*, not int. Since the type of the pointer is int*, the type of what the pointer points to is int.


This:

Code:

void funcName ( char const* const init)

is exactly the same as this:

Code:

void funcName ( const char* const init)

except for spelling. That is, when the const qualifier is at the left-most, it applies to what is on its immediate right, whereas it would otherwise apply to what is on its immediate left. Consequently, what you have in mind would be redundant: you would essentially be trying to const qualify the char twice.