C Programming - Pointers - Discussion

y=&x; /* Assume address of x is 500 and integer is 4 byte size */
z=y;
*y++=*z++;
x++;

here remember 3 points..
initially v knw d value of x = 30, which is incremented, then x=31.
the address of( *z) z is 500 bcoz z=y and x=&y , when incrementing this the value of integer added to it then 504. similarly yalso 504..

I'm gonna explain *y++=*z++
here ++ is the post increment . when u use ++ operator with assignment operator(=), first assignment takes place then the value will be incremented by one. see here y++=z++ is the two step process 1) y=z
2) increment the value of z by one and increment the value of y by one

In all the operating system pointers will be incremented by 4 bytes. so the answer is D(504)

y=&x so y=500;
z=y so y=z=500;
so *y++=*z++=504
and x++=31
so ans is: x=31, y=504, z=504

just try the following program and u will get the difference between *ptr++ and ++*ptr

# include<stdio.h>
int main()
{
int arr[]={15,6,9,22,8,21};
int i,*ptr;
ptr=arr;
for(i=0;i<6;i++)
printf("%d\t",*ptr++);
return 0;
}

Now replace printf() statement with:
printf("%d\t",++*ptr);

In *ptr++,the pointer is incrememted whereas in ++*ptr,the vaue is incremented.

500 is memory address and space occupied is 4bytes.

If both *p++ & ++*p used together in one program than *p++ increment the value & ++*p increment the address.

If it is used in the program alone than it is only increase the value.*p++ & (*p)++ are not same.*p++ increase the address array value else (*p)++ increase the array value.

*y++=*z++ is explicitly written as

*y = *z;
y++;
z++;

Compile the code in the C Compiler provided and add a printf() before incrementing the values, you'll get the difference:

/* Note: GCC Compiler (32 Bit Linux Platform). */

#include<stdio.h>
int main()
{
int x=30, *y, *z;
y=&x; /* Assume address of x is 500 and integer is 4 byte size */
z=y;
printf("x=%d, y=%d, z=%d\n", x, y, z);
*y++=*z++;
x++;
printf("x=%d, y=%d, z=%d\n", x, y, z);
return 0;
}

Output:
x=30, y=500, z=500
x=31, y=496, z=496

The value of x increases by 1, which is normal variable
But the values of *y and *z are seen to decrease by 4 (bytes)
This happens because the memory addresses are arranges in descending order from top to bottom, so when 500 becomes 496, it actually means that the pointer values has INCREASED!!!

The memory addresses look something like this:
|10|
|09|
|08|
|07|
|06|
|05|
|04|
|03|
|02|
|01|
|00|

That is why the answer option provided is wrong as it shows the changed values of *y and *z to be 504, which is actually DECREASING of the memory address!

Thank all
Finally I understood!
Follow condition of this exercise
y = &x; It mean y point to x that allocate at address 500
z = y; mean z point to address 500 too.
*y++ = *z++; => *(y++) = *(z++) It means increase the address of y and z by 4. So when we printf them, we have result are y = z = 504