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int arr[] = {1,2,3,4}
int *ptr=arr;
*(arr+3) = *++ptr + *ptr++;
Final contents of arr[]
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Answer / jai
{1,2,3,4}
++ has higher precedence over *, assigment will resolve to
*(arr+3) = *(++ptr) + *(ptr++);
*(arr+3) = 2 + 2;
=> Though ptr is pointing to address of 3rd element after
post increment.
| Is This Answer Correct ? | 13 Yes | 1 No |
Answer / anag
*(arr+3)------>arr[0][3] that means the there is any chnage
in the last value of an array
{1,2,3,--}
we know ++ has higher prededence than * so
*++ptr---->*(++ptr)
*(++ptr)----> increment in the location after that it point
to the value
it represent the second location of an array
* represent the value at this address
the value at the second location is 2.
in the second expression first it refer the value after
that it increment in the location
ptr currently points to the second location . ptr holds
that location for the second expression * represent the
value at that location that is 2.
so 2+2->4
{1,2,3,4} ----------->ans
suppose if we add a another expression after this that *ptr
then it print the value 3
because previous expression increment the location of the
value
Thank you
| Is This Answer Correct ? | 5 Yes | 0 No |
Answer / vijaisankar
In this statement
first ptr holds base address of the array(4000),
then as per precedence operators ptr gets post incremented
(4002)though it points the value 1(4000)(ptr is post
incremented) and then ptr gets preincrement so (4004) the
value in that one is 3 then 3+1=4.
*(arr+3)=3;
| Is This Answer Correct ? | 2 Yes | 7 No |
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