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So I have got a binary tree where each node (say v) has 2 children, one called 'atom' and the other 'cell'. In the lval_pop function, I would like to return the 'atom' child but shift the 'cell' node to the address of v, so the when I dereference the pointer to v, I would get the cell node (before shifting). However, while the function returns the 'atom' children correctly, the shifting does not happen. Why is that?

Thanks for your help in advance!

What I have tried:

lval* lval_pop(lval* v) { 
	lval* x = v->atom;	
		v = v->cell;
	return x;  
Updated 31-May-17 13:17pm
Afzaal Ahmad Zeeshan 31-May-17 18:15pm
How does it fail? Did you try to check it line-by-line by debugging it?

1 solution

Pointers are 1 of the most difficult things for a beginner, only exprerience can teach you how they work, and the debugger is the only tools that can help you by showing you what your code is really doing.
There is a tool that allow you to see what your code is doing, its name is debugger. Mastering the debugger is not optional, it is mandatory for any programmer, no exception.
It is also a great learning tool because it show you reality and you can see which expectation match reality.
When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.

Debugger - Wikipedia, the free encyclopedia[^]

Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.

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