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This is supposed to print 3,5,-1. I don't understand what is happening under-the-hood. What is happening when the function foo is called? I am having trouble understanding the last 4 lines in the function foo. I understand everything else.

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HaveNoDisplayName
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mistertwister
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    Please post both the question, the code, what you _do_ understand and what exactly you are confused about. – Aasmund Eldhuset Oct 13 '15 at 01:48
  • Obviously the line `y++` increments the pointer, not the value where it points. You would need to make it `(*y)++`. In this exercise, they are showing the difference between pass by reference `&var` and pass by pointer `*var`. In passing a pointer, explicit dereference is needed. – alvits Oct 13 '15 at 02:20
  • "The last four lines"? You mean starting from `*y = z;`? – Beta Oct 13 '15 at 02:31
  • Oh you need explanation of the last 4 lines of `foo()`. The first line increments the local copy `z` by `2` giving it a new value `5`. The second line assigns to the memory pointed by `x` the difference by subtracting the value of local variable `z` from the value in the memory pointed by `y`. The next line increments the local pointer `y` by 1. As I have mentioned in my previous comment, you probably want to increment the value contained in the memory pointed by `y`. – alvits Oct 13 '15 at 02:38
  • [What are the barriers to understanding pointers and what can be done to overcome them?](http://stackoverflow.com/q/5727/995714) – phuclv Oct 13 '15 at 04:04

1 Answers1

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function essentially becomes

foo(&x, arr[0], 3)
z = z+2 // z=3+2=5
x = *y -z; // x=arr[0]-5=4-5=-1
*y = z; //arr[0]=5

notice a is still 3 because it is passed by value. b points to the start of the array which is 5 after the execution of the function. and c is -1 because pass by reference.

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