C++ Programming - References - Discussion
Discussion Forum : References - Programs (Q.No. 6)
6.
Which of the following statement is correct about the program given below?
#include<iostream.h>
int main()
{
int x = 10, y = 20;
int *ptr = &x;
int &ref = y;
*ptr++;
ref++;
cout<< x << " " << y;
return 0;
}Discussion:
11 comments Page 1 of 2.
Aishwarya said:
4 years ago
Because the priority is to the increment:
*ptr++; // == *(ptr++);
So it's the value if the POINTER wich is incremented, not the value of 'x'.
But if you write this :
(*ptr)++; // The value of 'x'is incremented.
*ptr++; // == *(ptr++);
So it's the value if the POINTER wich is incremented, not the value of 'x'.
But if you write this :
(*ptr)++; // The value of 'x'is incremented.
(3)
Ghanshyam said:
1 decade ago
*ptr++;
this statement will not increase value of x by 1
* and ++ have same precedence..
associativity of both of them is from right to left.
first ++ evaluated then * , so it doesn't increase value, instead after execution of this statement ptr will point to next location..
this statement will not increase value of x by 1
* and ++ have same precedence..
associativity of both of them is from right to left.
first ++ evaluated then * , so it doesn't increase value, instead after execution of this statement ptr will point to next location..
(1)
Sherin said:
10 years ago
How it come 10 20?
(1)
Manisha said:
8 years ago
Hello, anyone please help me to get this.
(1)
Sangeetha said:
7 years ago
Anyone can give the clean explanation, about in which case cout will start from right to left?
(1)
SomebodY said:
1 decade ago
++ operator have associativity L to R.
Swapnil said:
1 decade ago
How it happens ?
Ilango said:
1 decade ago
If we can perform arithmetic operation on pointers then why we can't increment the value of the pointer. Please anyone explain this.
Pawar Manisha said:
8 years ago
How? Explain its output.
Hoang Pham said:
8 years ago
The key is the precedence of the two operators, postfix ++ and indirection *
++ has a higher precedence than *.This mean *p++ is going to be grouped as: *(p++).
++ has a higher precedence than *.This mean *p++ is going to be grouped as: *(p++).
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