programming

Question

main()

{

int * ptr =(int *) malloc (5*sizeof(int));

for(i=0;i<5;i++)

*(ptr+i)=i;

pf("%d",*ptr++);

pf("%d",(*ptr)++);

pf("%d",*ptr1);

pf("%d",*++ptr);

pf("%d",++*ptr);

}

 

Answer 1

Text in Yellow shows current ptr position.

Even though the array is dynamically allocated let me name it as A[5]={ ,  ,  ,  ,  } //Initially

Loop line for(..) { *(p+i)=i }                    
Array will look like A={0,1,2,3,4}

Before printing output points to be noted:-

1. Precedence of post (++/--) > dereference (*) >= pre(++/--)
2. when precedence is same like in case of dereference (*) and pre(++/--) Rigth to left associativity is used.

• printf(*ptr++) 
  • it will be parentheses as *(p++) but, post-increment effect comes in the picture, so it will dereference current position address value and then increase the pointer.
  • OUTPUT 0
  • A={0,1,2,3,4}
• printf( (*ptr)++) )
  • easy one, dereference the value and increment the value.
  • OUTPUT = 1
  • but due to post increment side effect value will be updated after print.
  • A={0,2,2,3,4,}
• don't know what does that mean.
• print( *++ptr )
  • precedence same but Right to left associativity so it will be treated as *(++ptr)
  • increment pointer then dereference it.
  • OUTPUT = 2
  • A={0,2,2,3,4}
• print(++*ptr)
  • precedence same but Right to left associativity so it will be treated as ++(*ptr) 
  • dereference first then increment value.
  • OUTPUT = 3
  • A={0,2,3,3,4}