C++ Programming - References - Discussion
Discussion Forum : References - Programs (Q.No. 12)
12.
Which of the following statement is correct about the program given below?
#include<iostream.h>
int main()
{
int arr[] = {1, 2 ,3, 4, 5};
int &zarr = arr;
for(int i = 0; i <= 4; i++)
{
arr[i] += arr[i];
}
for(i = 0; i <= 4; i++)
cout<< zarr[i];
return 0;
}Discussion:
11 comments Page 1 of 2.
Harshad said:
5 years ago
@All.
#include<iostream>
using namespace std;
int main()
{
int arr[] = {1, 2 ,3, 4, 5};
int (&zarr)[5] = arr;
for(int i = 0; i <= 4; i++)
{
arr[i] += arr[i];
}
for(int i = 0; i <= 4; i++)
cout<< zarr[i];
return 0;
}
The above program gives correct Output.
#include<iostream>
using namespace std;
int main()
{
int arr[] = {1, 2 ,3, 4, 5};
int (&zarr)[5] = arr;
for(int i = 0; i <= 4; i++)
{
arr[i] += arr[i];
}
for(int i = 0; i <= 4; i++)
cout<< zarr[i];
return 0;
}
The above program gives correct Output.
(1)
Waqar Akram said:
5 years ago
@Manish maybe you are right. But if we use int &zarr=arr[5] it will show the next error (i is not initialized).
Ashish said:
5 years ago
Please explain someone correctly.
Manish said:
6 years ago
Here an array of reference is not acceptable, that's why the error occurs.
Kripa Mary Jose said:
6 years ago
This program will not cause any compilation error as;
int & zarr = arr; is not erroneous. Here the base address of the array is passed. I don't understand why its answer is a compilation error then?
Please, anyone explain in clear.
int & zarr = arr; is not erroneous. Here the base address of the array is passed. I don't understand why its answer is a compilation error then?
Please, anyone explain in clear.
DoanNguyen said:
7 years ago
for(i = 0; i <= 4; i++)
cout<< zarr[i];
return 0;
i no have in scope of for
And &zarr can't cast "int*" to 'int" need cast "&zarr = arr[0] "
- It can't print "zarr[i]" becase it is a pointer we need print " zarr + i".
cout<< zarr[i];
return 0;
i no have in scope of for
And &zarr can't cast "int*" to 'int" need cast "&zarr = arr[0] "
- It can't print "zarr[i]" becase it is a pointer we need print " zarr + i".
Chetan said:
7 years ago
Variable i is declared in loop 1.
After the end of the scope, it cannot be accessed.
But in loop 2 it is accessed without declaration will generate the error.
Correct me if I am wrong.
After the end of the scope, it cannot be accessed.
But in loop 2 it is accessed without declaration will generate the error.
Correct me if I am wrong.
(2)
Suman said:
10 years ago
No here it must be int (&zarr)[5] = arr.
Lavanyeswari motupalli said:
1 decade ago
A reference can be an int, char pointer but not array.
Atul Kumar said:
1 decade ago
Since an array of references is not acceptable.
Hence compile time error occur.
Hence compile time error occur.
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