Function returning pointer to struct

Question

I am having some issues with code that is returning a pointer to a struct declared inside a class. Here is my code so far:

SortedList.h

#ifndef SORTEDLIST_H
#define SORTEDLIST_H

class SortedList{

 public:

    SortedList();

 ...

 private:

    struct Listnode {    

      Student *student;

      Listnode *next;

    };

    static Listnode *copyList (Listnode *L);

};

#endif

SortedList.cpp

#include "SortedList.h"

...

// Here is where the problem lies

Listnode SortedList::*copyList(Listnode *L)

{

    return 0; // for NULL

}

Apparently, the copy list method wont compile. I am using Microsoft Visual Studio and the compiler tells me that "Listnode" is unidentified. When I try to compile, here is whhat I get:

1>------ Build started: Project: Program3, Configuration: Debug Win32 ------

1>  SortedList.cpp

sortedlist.cpp(159): error C2657: 'SortedList::*' found at the start of a statement (did you forget to specify a type?)

sortedlist.cpp(159): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int

sortedlist.cpp(159): error C2065: 'L' : undeclared identifier

sortedlist.cpp(159): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int

sortedlist.cpp(159): fatal error C1903: unable to recover from previous error(s); stopping compilation

========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

Help would be greatly appreciated...ASAP

The format you should use for the post is Markdown: daringfireball.net/projects/markdown/basics — jogojapan
– jogojapan, Commented Nov 13, 2012 at 4:55

sgarizvi · Accepted Answer · 2012-11-13 07:57:00Z

2

Inside the cpp file, the function should be defined as:

SortedList::Listnode* SortedList::copyList(ListNode* L)
{
    return 0; //For NULL
}

Also, the struct Listnode should be declared either public or outside the class SortedList.

edited Nov 13, 2012 at 7:57

answered Nov 13, 2012 at 4:56

sgarizvi

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1 Comment

Jarryd Over a year ago

Note that inside the parentheses, you don't need the SortedList:: because at that point you are inside a SortedList function so you are in its scope, which means that Listnode is in scope.

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