Made Easy Mock Test

Question

Answer 1

When you call fun(p) you are passing the copy of the pointer p and then changing its value. So when the function returns the actual p remains unchanged.

If you wish to make the change persist, then you will have to pass the address of the pointer. This will also mean that you will have to change the signature of the function fun to

 

void fun(int **p){…..*p = &q;}

 

Comments

okay got it...I forgot about that *p = q

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A silly question but fun(p) share the value at p right ? Like the address of r? Which will be then stored in another local variable ?

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No fun(p) is sharing the the value of p that happens to be lets say 1000...And to store the address we declare a local ptr variable stated as the arg of the func.

But the maincatch which I missed the modification statement was p = &x which is 1000 changed to 10….

If it would have been *p = &x then the value at 1000 that is lets say 5 would get changed to 10….

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Have a look at this question and answer, this will clear things up nicely Yuuchan :)

https://gateoverflow.in/374988/go-classes-test-series-2024-data-structures-test-1-question