Pass By Reference In Go {DEMYSTIFIED!}

I started using go a few weeks back and still exploring go concepts, trust me when they say “Go Will Make You Fall In Love With Programming” this is actually true, you will love writing go code :)

Okay, So about this article, As I said I am still a beginner, A few days back I was reading a few blogs and came across this concept of pass by reference in go , I came across an example to implement the concept and by looking at it a programmer with an experience of ‘C’ will say it is right but to my surprise the output was completely opposite to what I thought of, Below is the code:

https://play.golang.org/p/K5eMNtkwGWA

Here one can see that output is not changed whereas to our knowledge output should be “akshit”, So this was something kind of disturbing for me, Because we are passing the address (known as ‘pass by reference’) still the output is not what we thought of it to be, What Is Going At The Back?

After giving few hours for this I just realized that whatever we do ‘Go’ will still remain a pass by value language, so this is what happens at the back:

So, When change() is called, the variable which is changed is u (copy) address ie. addr4 is written in addr3 (u copy) thus addr2 (u original) remains intact.

What is the solution?

Here it is,

https://play.golang.org/p/SyzMK0TdqIG

This code works the way we want it to be, so let’s dive into it, How is this working?

The above code, takes addr3 ie. copy of u (variable in change()) which holds the address addr1 (value of original u) and points to it using dereference operator and assigns the value to it, thus the contents of addr1 are changed because addr2 (original u) points to addr1, thus the output.

Conclusion:

After going through various blogs, hours of thinking and writing this article, I have a solid image that whatever we do, Go will remain a pass by value language. So, what now, If you are beginner like me, Remember one thing,
Go is more of thinking than to code.

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