Pass by value and pointers - c

Question

Can you explain why the output is 3?
I was trying to trace the answer and it shows that the line
i+=(a==b?1:0)
gives 1, but doesn't fun1() pass by value so q and p were copied to different variables?
int fun1(int* a, int* b)
{
int i = 0;
i += (&a == &b ? 1 : 0);
i += (a == b ? 1 : 0);
i += (*a == *b ? 1 : 0);
return i;
}
int fun2(int** a, int* b)
{
int i = 0;
i += (a == &b ? 1 : 0);
i += (*a == b ? 1 : 0);
return i;
}
int main(void)
{
int i = 0;
int* p = &i;
int* q = &i;
printf("%d\n", fun1(p, q) + fun2(&p, q));
return 0;
}

Answer 1

In fun1:
(&a==&b?1:0)
This gives 0. They are two different arguments to fun1, basically two different local variables, so they cannot have the same address.
(a==b?1:0)
This gives 1. The values are both &i from main.
(*a==*b?1:0)
This gives 1. Since a and b are equal, they point to the same thing.
In fun2:
(a==&b?1:0)
This gives 0. a and b are both arguments, so a can't equal the address of b (and in fact, it equals &p from main).
(*a==b?1:0)
This gives 1. a is equal to &p from main, so *a is equal to p from main, which is &i from main. And b is equal to q from main, which is again &i from main.
The total is therefore 3.

Answer 2

int fun1(int *a, int*b)
{
int i=0;
i+=(&a==&b?1:0); // 0. &a is the memory where a stands. it's different from &b.
i+=(a==b?1:0); // 1. Both pointers point to the same memory.
i+=(*a==*b?1:0); // 1. *a is the value where it points to, and that's the same as *b.
return i; // returns 2
}
int fun2(int **a, int*b)
{
int i=0;
i+=(a==&b?1:0); // 0
i+=(*a==b?1:0); // 1. *a is the value where it points to, and this is a memory value that is the same as b.
return i; // returns 1
}
That's why it returns 3.
Good question, by the way.

Answer 3

a==b is actually testing whether a and b points to same memory location or not. In function fun1, both a and b points to the same memory location and therefore a==b comes out to be true.

Answer 4

Let's consider at first the first function
int fun1(int *a, int*b)
{
int i=0;
i+=(&a==&b?1:0);
i+=(a==b?1:0);
i+=(*a==*b?1:0);
return i;
}
It was called like
fun1(p,q)
where the both pointers point to the same variable
int *p=&i;
int *q=&i;
So the function got two equal values as its argument. It is the address of variable i.
Function parameters are its local variable. You can imagine the called function like
int fun1( /*int *a, int*b */)
{
int *a = p;
int *b = q;
//...
}
These local variables occupies different extents of memory. So &a is not equal to &b.
As result the value of expression (&a==&b?1:0) will be equal to 0 and variable i in the statement below
i+=(&a==&b?1:0);
will not be changed
The values stored in variables a and b id the address of variable i in main. As it was said early the two variables contain the same value.
So expression (a==b?1:0) will yield 1 and variable i in the statement below
i+=(a==b?1:0);
will be increased.
As the both pointers points to the same object then expression (*a==*b?1:0) also will yield 1. As result variable i will be increased/
i+=(*a==*b?1:0);
The function will return value 2.
Now let's consider the second function
int fun2(int **a, int*b)
{
int i=0;
i+=(a==&b?1:0);
i+=(*a==b?1:0);
return i;
}
As it was said above parameter b is a local variable of the function. Its address does not equal to the address of argument p
So expression (a==&b?1:0) yields 0.
Expression *a is the value stored in argument p The same value is stored in parameter b
So expression (*a==b?1:0) yields 1.
In total the sum of the return values of the functions will be equal to 3.