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What will be the output of the following C++ code?
#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
int main(int argc, char const *argv[])
{
int &q = 5;
cout<<q;
return 0;
}
#include <string>
#include <cstdlib>
using namespace std;
int main(int argc, char const *argv[])
{
int &q = 5;
cout<<q;
return 0;
}
- a)5
- b)Run-time error
- c)Segmentation fault
- d)Compile-time error
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
What will be the output of the following C++ code?#include <iostrea...
References require are other names for variables not for a constant literal. No such assignment are allowed in C++.
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What will be the output of the following C++ code?#include <iostrea...
Understanding the Code
The provided C++ code attempts to create a reference to an integer literal. Here’s a breakdown of why this leads to a compile-time error.
Code Analysis
- The line `int &q = 5;` tries to bind a reference variable `q` to the integer literal `5`.
- In C++, a reference must bind to an existing variable, not to a temporary value or literal.
Why Compile-time Error Occurs
- References Must Bind to Variables: References in C++ are meant to provide an alias for existing variables. Since `5` is a constant and not a variable, the compiler cannot create a reference to it.
- Temporary Values: The integer literal `5` is a temporary value, and C++ does not allow references to bind to temporaries.
Expected Output
- Since the code attempts to perform an invalid operation, the compiler will generate an error during compilation, not allowing the program to run.
Conclusion
The correct answer is option 'D' (Compile-time error) due to the invalid reference binding in the code. This highlights the importance of understanding how references work in C++.
The provided C++ code attempts to create a reference to an integer literal. Here’s a breakdown of why this leads to a compile-time error.
Code Analysis
- The line `int &q = 5;` tries to bind a reference variable `q` to the integer literal `5`.
- In C++, a reference must bind to an existing variable, not to a temporary value or literal.
Why Compile-time Error Occurs
- References Must Bind to Variables: References in C++ are meant to provide an alias for existing variables. Since `5` is a constant and not a variable, the compiler cannot create a reference to it.
- Temporary Values: The integer literal `5` is a temporary value, and C++ does not allow references to bind to temporaries.
Expected Output
- Since the code attempts to perform an invalid operation, the compiler will generate an error during compilation, not allowing the program to run.
Conclusion
The correct answer is option 'D' (Compile-time error) due to the invalid reference binding in the code. This highlights the importance of understanding how references work in C++.
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What will be the output of the following C++ code?#include <iostream>#include <string>#include <cstdlib>using namespace std;int main(int argc, char const *argv[]){ int &q = 5; cout<<q; return 0;}a)5b)Run-time errorc)Segmentation faultd)Compile-time errorCorrect answer is option 'D'. Can you explain this answer?
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