CODEWITHSUNDEEP
c++syntax
rtuner
rtuner

Reputation: 2410

Is "struct type variablename" valid as a declaration?

My lecturer just astounded me by insisting that the following is a valid C++ statement:

struct charlistrecord *next

The complete condition was found in declaring a recursive datatype in C/C++ as follows:

typedef struct charlistrecord
{
     char data;
     struct charlistrecord *next;
} *charlist;

I want to know whether maybe there has been a time when this was accepted syntax or whether I've just been dumb for months of thinking I know C++.

Upvotes: 2

Views: 147

Answers (4)

WhozCraig
WhozCraig

Reputation: 66194

Prior to C++ and the assumption of typedefs, the only way to get a typed structure was via this syntax, and it was the only way to get a self-type-linked node.

struct Foo
{
    int val;
    struct Foo *link;
};

In C, this required any usage of Foo to be:

struct Foo foo;
struct Foo *foo_ptr;

etc..

A typedef helped for this by doing this:

typedef struct Foo
{
   int val;
   struct Foo *link;
} Foo;

Since now you could do this:

Foo foo;       // same as struct Foo
Foo *foo_ptr;  // same as struct Foo *

Note: Using a typedef to alias a struct Name is not restricted to Name as the alias. Yo were perfectly valid to do this as well:

typedef struct Foo
{
    int val;
    struct Foo *link;
} Bar;

and now the following are doable;

struct Foo foo;
Bar bar;
struct Foo *fooptr = &bar;
Bar *barptr = &foo;

Really makes you wish you programmed in C back in the day, doesn't it ? Probably doesn't clear things up, but hopefully a little less gray.

Upvotes: 2

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171127

It's perfectly valid, and it can even be useful if your struct's name is shadowed by a variable, like here:

struct C {};
int C = 0;
int a;

void foo() {
  C *a; // multiplies int C by int a
  struct C *a; //defines 'a' as pointer to struct C
}

Upvotes: 0

SingerOfTheFall
SingerOfTheFall

Reputation: 29966

Yes it's totally valid. The word struct here just explicitly states that next is a pointer to a structure type named charlistrecord.

It's a reminiscence from C, and you might as well omit the keyword in C++, but if you want to use it, you can.

Upvotes: 1

john
john

Reputation: 8027

It's a hangover from C. In C this is how you have to declare struct variables and members, so it's legal in C++ for reasons of backwards compatibility. Sounds like your lecturer is a bit 'old school'.

Upvotes: 1

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