Pointer in C, don't understand how they got this result

Question

here is code snippet

void F (int a, int *b)
{
 a = 7 ;
 *b = a ;
 b = &a ;
 *b = 4 ;
 printf("%d, %d\n", a, *b) ;
}
int main()
{
 int m = 3, n = 5;
 F(m, &n) ;
 printf("%d, %d\n", m, n) ;
 return 0;
}

answer

4 4 
3 7

I see how 4 4 was computed, I don't get how they got 3 7 (I do understand how 3 is computed, it is not changed since it was not passed by reference)

Thanks !

Answer 1

I've annotated the F function with comments to explain what's going on:

a = 7 ;  // a is now 7
*b = a ; // b points to n, so n is now 7
b = &a ; // b now points to a and no longer to n
*b = 4 ; // a is now 4. n is unaffected

Answer 2

Take a close look at this line

b = &a ;
*b = 4 ;

b gets the reference of a (it's memory address). When you now access *b it points to the memory of the variable a, not to n anymore

Answer 3

Quite subtle indeed. :)

In F a lot of nonsense happens: whichever value is passed via a, it's discarded, and instead 7 is assigned to it. Then, the value of a (which is 7) is assigned to the variable to which b points (which is n, so it's here that n becomes 7).

At the next line, the object to which b points is changed, so that b now points to the local parameter a, and at the following line the object pointed by b (=>a) is set to 4.

Thus, now we have a that is 4, b that points to a (=>so *b==4, since *b==a), m is still 3 (it was passed by value, so it couldn't have been changed by F) and n was set to 7.

By the way, this is exactly the mess you shouldn't to at all in your applications to avoid confusing any people who have the disgrace to work on them. :)

Answer 4

At the start of main, we have

m=3   n=5  // int m = 3, n = 5;

then we call F(m, &n), passing m by value, and n by pointer such that

m = 3   n = 5
a = 3   b->n   // F(m, &n);

Now, inside F(), we assign 7 to a:

m = 3   n = 5
a = 7   b->n     // a = 7

then we assign a (=7) to the memory address pointed by b (->n)

m = 3   n = 7
a = 7   b->n     // *b = a;

next we change b, so that now b points to a:

m = 3   n = 7
a = 7   b->a     // b = &a;

and then we assign 4 to the memory address pointed by b (->a)

m = 3   n = 7
a = 4   b->a     // *b = 4;

printing a (=4) and *b (->a =4)

and printing m (=3) and n (=7) outside the function