What will be the output of the following C++ code?#include <iostream>using namespace std;void square (int *x, int *y){*x = (*x) * -(*y);}int main ( ){int number = 30;square(&number, &number);cout << number;return 0;}

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What will be the output of the following C++ code?
#include <iostream>
using namespace std;
void square (int x, int y)
{
x = (x) * –(*y);
}
int main ( )
{
int number = 30;
square(&number, &number);
cout << number;
return 0;
}

What will be the output of the following C++ code?
#include <iostream>
using namespace std;
void square (int *x, int *y)
{
*x = (*x) * –(*y);
}
int main ( )
{
int number = 30;
square(&number, &number);
cout << number;
return 0;
}
a) 870
b) 30
c) error
d) Segmentation fault

Answer: a
Explanation: As we are passing value by reference therefore the change in the value is reflected back to the passed variable number hence value of number is changed to 870.

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What will be the output of the following C++ code?#include <iostream>using namespace std;void square (int *x, int *y){*x = (*x) * –(*y);}int main ( ){int number = 30;square(&number, &number);cout << number;return 0;}

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What will be the output of the following C++ code?
#include <iostream>
using namespace std;
void square (int x, int y)
{
x = (x) * –(*y);
}
int main ( )
{
int number = 30;
square(&number, &number);
cout << number;
return 0;
}