C++ Programming - References - Discussion
Discussion Forum : References - Programs (Q.No. 2)
2.
Which of the following statement is correct about the program given below?
#include<iostream.h>
int main()
{
int x = 80;
int &y = x;
x++;
cout << x << " " << --y;
return 0;
}Discussion:
17 comments Page 1 of 2.
Bhavesh Garg said:
2 years ago
The Correct Answer is (B) 81 80.
(15)
Blake said:
2 years ago
@All.
Here, the correct answer is D since you're using "cout" and you didn't declare beforehand that you are "using namespace std;".
Here, the correct answer is D since you're using "cout" and you didn't declare beforehand that you are "using namespace std;".
(5)
Sid said:
4 years ago
In C++17 answer is 81 80 and in C++ 14 ans is 80 80.
(4)
Anusha said:
5 years ago
The correct answer is 81 80.
(4)
Anant Supekar said:
5 years ago
The correct answer will be 81 80.
(3)
Hatos said:
8 years ago
My output is;
81 80
Code::Blocks
Release 17.12 rev 11256
g++.exe (GCC) 7.2.0
std = c++1z
81 80
Code::Blocks
Release 17.12 rev 11256
g++.exe (GCC) 7.2.0
std = c++1z
(1)
Ak7 said:
6 years ago
1st: cout << x will work after that cout address is returned again and cout<<--y will be printed.
Correct ans is 81 80.
Correct ans is 81 80.
(1)
Srdjan said:
8 years ago
Operator << is left-to-right associative, but operator --(in --y) has greater precedence hence it will be executed sooner. Standard representation of this expression is:
<<(cout, <<(x, <<(" ", --(y)))) //inermost is always executed sooner.
ie) for typical a*b+c:
+(*(a,b), c).
<<(cout, <<(x, <<(" ", --(y)))) //inermost is always executed sooner.
ie) for typical a*b+c:
+(*(a,b), c).
(1)
Sharath said:
1 decade ago
&y means alias of x , so here now x & y are same.
Then cout execution start right to left, So first x++ is 81 then --y is 80
output is 80 80
Then cout execution start right to left, So first x++ is 81 then --y is 80
output is 80 80
Serg said:
7 years ago
Of course, it is 81 80.
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