C Programming - Pointers - Discussion

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

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..

Here the output when i compiled is is x=31,y=-10,z=-10.++ has higher precedence than *,

So address is incremented to 504 and the value at that address is fetched.....
Is this correct??

@Kishore Mylavarapu
Yes explanation is very good I agree

Thanks.

*y++ and *z++ means 1st assign the add. of(z=500) then increase it. If it is written like (*y)++ then we put the value of that given ponter then increment will perform.

Simple concept I have commented the flow just check out.

int main()
{
int x=30, *y, *z;

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

z=y; //both pointers r pointing to same value

*y++=*z++; /* z is post incremented hence original value adress 500 is assigned then incremented but, here even we r incrementing y hence it changes from 500 to 504(it is integer type). */

x++; //incremented in the next line it is printed

printf("x=%d, y=%d, z=%d\n", x, y, z);

return 0;
}

In a pointer char stores 2 bytes, int stores 4 bytes, long stores 8 bytes.

Hence in above example x=504 and y=504.

Since Z and y have same values in them y.

The statment *y++=*z++ actually assigns the value in z (address of x)to ythus Y++ makes it 500, similarly z++.

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

y=&x; /* Assume address of x is 500 and integer is 4 byte size */
z=y;
*y++=(*z)++;
x++;
printf("x=%d,x add=%u, y=%u, z=%u\n", x,&x, y, z);
/run dis code u.ll get to kw d diff b/w (*Z)++ n *(Z++)