Output question (this question was asked in instagram post of coding blocks)

#1

int main()
{
int arr[]={10,20};
int *p=arr;
(p++)++;
cout<<ar[0]<<" “<<1[arr]<<” "<<–p;
return 0;
}

#2

in cpp, arr[i] is resolved as *(arr+i). where arr is base address of array.
now arr[i] = i[arr] = *(arr+i)… same thing.
(p++)++ is syntactically incorrect.
it should be p++; p++; and then
p is incremented twice and decremented once, so it will point to second element of array.
–p will print address of second element.

thanks

#3

That was the wrong question
can u please explain this

#4

aa[i] and i[arr] are same as explained.
(*p++)++ can be rewritten as
(*p++); => *p = *p+1 =. it means , arr[0] =11, since p is pointing to arr[0]
p++; => p start pointing to arr[1]

cout<<*–p; => this is *(–p) by associativity. p will point to arr[0] again and thus print 11.

these kind of questions are mainly bases on operator precedence and associativity.
thanks

#5

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#6

Thanks for the help , but I always get confused in question regarding output .
Could you please help me with some learning resources and practice Questions .

For example in this question , What will be associativity and precedence ?