CODEWITHSUNDEEP
cfunctionprototypereturn-type
user1005284
user1005284

Reputation:

what does return; mean?

The return data-type of a function,whose prototype is declared in main(), is void. It cointains an instruction return; as in 

main()
{
    void create(int *p);
    *some code*
}
void create(node *list)
{
     *some code*
     return;
}

What will it return,and where will it return??

Upvotes: 3

Views: 18809

Answers (4)

Salvatore Previti
Salvatore Previti

Reputation: 9050

In this case doesn't mean much.

return; means exit suddenly from this function returning void.

int a()
{
    return 10;
}

void b()
{
     return; // we have nothing to return, this is a void function.
}

void c()
{
    // we don't need return; it is optional.
}

return; for a void function is not useful, it can be omitted, is optional. However sometime it is useful, for example, for exiting a loop or a switch.

void xxx()
{
    int i = 0;
    while (1)
    {
        if (++i >= 100)
            return; // we exit from the function when we get 100.
    }
}

Upvotes: 4

AusCBloke
AusCBloke

Reputation: 18492

It's not going to return anything, you might have return statements in a void function to kind of alter the flow and exit from the function. ie rather than:

void do_something(int i)
{
   if (i > 1) {
      /* do something */
   }
   /* otherwise do nothing */
}

you might have:

void do_something(int i)
{
   if (i <= 1)
      return;

   /* if haven't returned, do something */
}

Upvotes: 9

thumbmunkeys
thumbmunkeys

Reputation: 20764

it will return from the executing function with a void return value (which means no value).

Upvotes: 1

Fred Foo
Fred Foo

Reputation: 363567

return; won't return anything, which matches the declared void return type of the function create. It will return to the caller of the function it appears in (not shown in your example), just like return EXPRESSION; will.

In this particular piece of code, return; is redundant since it appears at the very end of create, but it's useful when you want to exit a function prematurely:

void print_error(char const *errmsg)
{
    if (errmsg == NULL)
        // nothing to print
        return;
    perror(errmsg);
}

Upvotes: 2

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