Can you help me with this code

[tomisrb]

When i compile this code, i get that c is 40, but I dont understand why is that.. If anyone can help me it would be good..

    int a,b,*pa,*pb,c,d;
    pa=&a;
    pb=&b;
    scanf("%d%d",pb,pa);
c= *(++pa);
    d=++(*pb);
    printf("a=%d,b=%d,c=%d,d=%d",*pa,*pb,c,d); 
return 0;

[Ganado]

What is your user input?

Well, regardless of what your user input is, your program invokes undefined behavior. On line 5 you are incrementing the pa pointer, and then dereferencing it. But there is no guarantee that the pa pointer now points to valid memory.

[George P]

I certainly don't get c=40 when I run the code using Visual Studio 2019:

#include <cstdio>

int main()
{
   int a, b, * pa, * pb, c, d;
   pa = &a;
   pb = &b;
   scanf("%d%d", pb, pa);
   c = *(++pa);
   d = ++(*pb);
   printf("a=%d,b=%d,c=%d,d=%d", *pa, *pb, c, d);
}

5 6
a=9216000,b=6,c=9216000,d=6

pa and c point to a memory address your program doesn't own, never initialized. That is undefined behavior.

[seeplus]

This is 'c hacking at its worst'. The result is undefined as it depends upon how the compiler lays out's its variables in memory.

If we consider:

#include <stdio.h>

int main()
{

	int a, b, * pa, * pb, c, d;
	pa = &a;
	pb = &b;
	scanf("%d%d", pb, pa);
	printf("&a: %p, &b: %p, &pa: %p, &pb: %p, &c: %p, &d: %p\n", pa, pb, &pa, &pb, &c, &d);
	c = *(++pa);
	d = ++(*pb);
	printf("a=%d,b=%d,c=%d,d=%d", *pa, *pb, c, d);
	return 0;
}

Then for input

 2 31 

I get for VS2019 64bit:

&a: 00000000001FF994, &b: 00000000001FF998, &pa: 00000000001FF980, &pb: 00000000001FF988, &c: 00000000001FF990, &d: 00000000001FF99C
a=3,b=3,c=2,d=3

OK. The address of a is 1FF994 and the address of b is 1ff998. b has the value 2 and a has the value 31 (scanf is b then a, not a then b as might be expected!).

pa is incremented so is now 1ff998. This is the address of b. *(1ff998) is the value of b so c is now 2.

pb is 1ff998 so *1ff998 is 2. So d is 2 + 1 which is 3

The final output 'lies' as a is not 3, but is still 2. *pa is 3 as pa is now the same as pb.

This all assumes that &b has a higher address than &a - but is probably true for some compilers/systems but shouldn't be assumed.

But notice that the address of c is 1ff990 - which is the address before that of a!

So the memory layout is:

1ff980 - pa (64 bits as compiled as 64 bit)
1ff988 - pb (64 bits)
1ff990 - c (32 bits)
1ff994 - a (32 bits)
1ff998 - b (32 bits)
1ff99c - d (32 bits)

These do not correspond to the order of definition! So really, no assumption should be made about any memory ordering of variables!