+ 1
Please explain output.
void main() { int a[]={1,2,6,4}; int *p,**q,*t; p=a; t=a+1; q=&t; cout<<*++p<<**q<<*t++; }
5 Respuestas
+ 4
Before cout:
==> p points to first element of a
==> t points to second element of a
==> q points to t since it is a pointer-to-pointer
At cout:
==> *++p means "increment p first and then derefence it" or simply *(p + 1). It will print the second element of a
==> since q is a pointer to a pointer, you need to dereference it twice: *q returns t pointer, **q returns what t is pointing to, which is 6.
==> *t++ also means *(t + 1) but since it's the post-increment operator it will derefence t first 'then' do the incrementing. So cout will print the second element of a which is 2.
After cout:
==> t will point to the third element of a because of the t++ part.
Hope this helps.
+ 4
Oh, I didn't noticed it when I wrote it. You're right, how q really shows 6 when t is pointing to 2?
Now I'm confused too :D
+ 4
Ah, I see. So that's why. Thanks.
+ 1
I got answer today as post increment operator has highest precedence thus t++ will execute first thats why q is printing 6 and t value is printing 2 bcoz compiler printing the value we asked in current statement.
0
how q is pointing to 6 when t is point to 2. Not getting that point please try to explain that.