Functions returning pointers

[JS]

If I have a function that looks like this:



char *alloc(int n)
{
....
return allocp - n;

}

I guess the "*" in the first line of the function belongs to "char" and not
the name of the function. "char *"....not "*alloc". Right?

JS

 

[Ioannis Vranos]

If I have a function that looks like this:



char *alloc(int n)
{
....
return allocp - n;

}

I guess the "*" in the first line of the function belongs to "char" and not
the name of the function. "char *"....not "*alloc". Right?

Right.

 

[Alf P. Steinbach]

* JS:

If I have a function that looks like this:



char *alloc(int n)
{
....
return allocp - n;

}

I guess the "*" in the first line of the function belongs to "char" and not
the name of the function. "char *"....not "*alloc". Right?

Yes.

 

[Victor Bazarov]

If I have a function that looks like this:



char *alloc(int n)
{
....
return allocp - n;

}

I guess the "*" in the first line of the function belongs to "char" and not
the name of the function. "char *"....not "*alloc". Right?

Right. An asterisk cannot be part of a name. It says that the function
'alloc' returns a pointer to char.

V

 

[HappyHippy]

If I have a function that looks like this:



char *alloc(int n)
{
....
return allocp - n;

}

I guess the "*" in the first line of the function belongs to "char" and
not
the name of the function. "char *"....not "*alloc". Right?

JS

Yes, you are right, * belongs to the return type (char).

 

[JS]

Right. An asterisk cannot be part of a name. It says that the function
'alloc' returns a pointer to char.

V

Just a bit misleading because of the missing blank followed by the "*".
Would seem a bit more clear if one wrote:

char * alloc(int n)

 

[Karl Heinz Buchegger]

Just a bit misleading because of the missing blank followed by the "*".
Would seem a bit more clear if one wrote:

char * alloc(int n)

It is the same as
2+3

you can write it any way you want
2+ 3
2 +3
2 + 3

it always means the same thing: the addition of 2 and 3

There is a possible pitfall with that pointer-'*' however:
Actually the situation is this:
The '*' belongs more to the thing on its right then to the thing
on its left. In case of function return types this makes no difference.
But consider:

int* i, j;

What does this define?
Well. It defines a pointer variable named 'i' and (surprise) it defines
an int variable named 'j'. Note: 'j' is an int (!) not a pointer to int (!).
That's why most C programmers prefer to write

int *i, j;

to make this clear without any doubt.

In C++ much less pointers are used, and most programmers agree, that there
should be only one variable defined in each definition, thus a C++ programmer
would write:

int *i;
int j;

and now the whitespace can easily be moved to the data type without fooling
anybody:

int* i;
int j;

 

[JS]

It is the same as
2+3

you can write it any way you want
2+ 3
2 +3
2 + 3

it always means the same thing: the addition of 2 and 3

There is a possible pitfall with that pointer-'*' however:
Actually the situation is this:
The '*' belongs more to the thing on its right then to the thing
on its left. In case of function return types this makes no difference.

but in this function:

char *alloc(int n)

it belongs to the left.....right?

 

[Ioannis Vranos]

but in this function:

char *alloc(int n)

it belongs to the left.....right?

Yes. There is no such ambiguity in the language. No identifier (name) is
permitted to contain character '*'.

 

[Karl Heinz Buchegger]

but in this function:

char *alloc(int n)

it belongs to the left.....right?

If it is your understanding that alloc() returns a pointer, in particlar
a character pointer, then yes.