Reference and Pointer ( Lecture 10)

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[@subhabhaskaran1827]

Sir!! As X+1 is same as ++x ..how it is correct to return(x+1)...it is also a change being made to reference parameter. Please explain sir

[@brijrajkishore8913]

x+1 is not same as ++x. x+1 does not increase the x itself but ++x does change the x's value

[@shahbazkhan1850]

++x is equivalent to x = (x+1); which is an assignment statement, whereas (x+1) is simply an expression and doesn't affect any variable's existing value

[@ROHITKUMARIN]

Sir !! Which software I need to use for compilation. Please suggest because I am facing some problems

[@anilsarode6164]

very nice thanks a lot sir

[@snehasishmondal677]

Sir, I got two doubt from this lecture :
1. int a=10;
int &b = a;
If we print b we will get address of a.
Now if i increment a value of a will increment.
Then if i increment b means ++b or (b+1) then value of b means address of a variable should increment.
how value of a is increment?
2. If i use reference in function call then actual and formal both variable will update. if i dont want to change actual parameter then we can use call by value like C.
Why const reference is required?

[@shahbazkhan1850]

1. because a and b point to the same memory location, so any update in either of the variable will be reflected in the other variable. So if you increment b, the value at address pointed by b(as well as a) will be incremented, hence a will also refer to the same incremented value.
2. We still use it so that we can avoid the copy operation (copy of values from actual parameters to formal parameters) which can be compute heavy if the parameters are of complex types like struct.