Please explain how it's output is 100? #include <iostream> using namespace std; void myFunc(int *x) { *x = 100; } int main() { int var = 20; myFunc(&var); cout << var; }

Answer 1

Actually any changes to x itself is dereferencing.
But *x receives values for var.
add "cout <<x;" in myFunc after *x = 100 and
add "cout <<&var" in main after calling myFunc
both will give you the same address.

Answer 2

The myFunc receives address of var in pointer x. So, x is nothing but a pointer to var. Thus, whatever value assigned to printer's content is directly assigned to var.
I hope you understand the concept of pointers.

Answer 3

*(& var) is var

Answer 4

consider, let address of var be 0x2345.
in myFunc(&var) yu are sending address of var,ie. address of 20. which can be written for understanding purpose as myFunc(address of value 20),ie myFunc(0x2345). so in function body, since yu have sent address any change made to X changes its value. ie (int *x) takes address 0x2345.
now in the body.. yu have derefrenced the pointer.so it says "value@0x2345"=100. it overwrites before-value 20 to after-value 100.since yu have made changes directly through address.. "count<<var" gives O/p as 100..

Answer 5

*x=100 doesn't do the dereferencing?