Casting from pointer to int

Posted on 2000-04-04
Last Modified: 2010-04-02
This code works:
  if ((int) pNode()->pUserField() == 2) .....

but am not sure how to do the opposite ie. plug a value back into the object pNode.   I have tried

  pNode()->pUserField() = (void *) 2;

but this doesn't compile.  How do I set the pUserField to an integer value?

Question by:abulka
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Expert Comment

ID: 2686170
Short answer?  Use another function (not pUserField()).  i.e., pUserField() is a function that, if the second line you show doesn't compile, does not return something that can show up on the left hand side of an assignment operator (an "l-value").  Maybe you can rewrite the function pUserField() (and note, it is a function, not a variable, as the name would lead you to believe) to return by reference.  Then you'd be all set!

Author Comment

ID: 2686199
I am using a 3rd party library and cannot rewrite or add new functions.  It should be possible to do with the existing function.  Here is more info on the UserField property:

// This method sets a void pointer to a user field. This pointer is
// originally initialized to NULL.

void * TSDigraph::pUserField(void * pUserField);

// This method gets a void pointer to a user field. This pointer is
// originally initialized to NULL.

void * TSDigraph::pUserField();


Accepted Solution

GlennDean earned 5 total points
ID: 2686239
Instead of writing
pNode()->pUserField() = (void *) 2;
pNode()->pUserField((void *)2);

Expert Comment

ID: 2686254
I work for Tom Sawyer and I can tell you that Glenn is definitely right. He fully deserves the enormous 5 point reward :)

Author Comment

ID: 2686327
Seems to work.  Much appreciated!  Thanks to mnewton, too.

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