Swapping two variables using bitmasking

#1

I have not understood how the function in the code below worked.
We have made the addresses of variables a , b as the arguments. However in the driver program we have called the function by passing the values of a and b.

#2

manoher u are passing by reference that is when u make changes in the function it will reflect in the main function
now u are doing xor operation
and
a xor a=0
so when u are doing
a=a^b
now in 2nd step u have done
b=b^a put value of a here
b=b^a^b
so b^b=0 and then b=a^0 and a property of xor a^0=a
so here u get b=a;
now u do a=a^b put value of a here
a=a^b^a then u get b
That how its work
hope this help:)

#4

I have not understood the pass by reference part.
I am passing the address of the variables a and b in the function, however while calling the function in the driver program , I am calling the fucntion on the variables and not their addresses. How does this work?

#5

yes u are passing there address
when u use & with a variable then u are passing its address and without & u are passing a variable but u are making a another copy of it in the memory
so when u passed by reference the value changes in the address which is remain permanent and reflect in main memory
but when u r copy the variable then any change in the function will not affect in main as the new variable made during copying will be deleted when function is over

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#6

if ur doubt is resolved mark it resolved

#7

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