What will be the output of the following C++ code?
#include <iostream>
#include <complex>
using namespace std;
int main()
{
complex <double> cn(3.0, 5.0);
cout<<"Complex number is: "<<real(cn)<<" + "<<imag(cn)<<"i"<<endl;
return 0;
}
#include <iostream>
#include <complex>
using namespace std;
int main()
{
complex <double> cn(3.0, 5.0);
cout<<"Complex number is: "<<real(cn)<<" + "<<imag(cn)<<"i"<<endl;
return 0;
}A. Complex number is: 3 + 5i
B. Complex number is: 5 + 3i
C. Complex number is: 9 + 25i
D. Complex number is: 3 – 5i
Answer: Option A
Solution (By Examveda Team)
In the given C++ code, the complex class from the C++ Standard Library is used to represent a complex number.In the statement complex
The function real(cn) returns the real part of the complex number, which is 3.
The function imag(cn) returns the imaginary part, which is 5.
Hence, the output will be: Complex number is: 3 + 5i.
This Question Belongs to C Plus Plus Programming >> C Plus Plus Miscellaneous
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Comments (1)
What does the 'sizeof' operator return in C++?
A. Size of a data type in bits
B. Size of a data type in bytes
C. Size of a variable in bytes
D. Size of a variable in bits
What is the purpose of the 'static' keyword in C++?
A. To declare a variable with dynamic storage duration
B. To declare a constant
C. To declare a variable with external linkage
D. To declare a variable with static storage duration
What is the difference between '++i' and 'i++' in C++?
A. None of the above
B. They both have the same effect
C. '++i' increments the value of 'i' before returning it, while 'i++' increments the value of 'i' after returning it
D. '++i' increments the value of 'i' after returning it, while 'i++' increments the value of 'i' before returning it
Your answer is incorrect check it
Answer will be -
Complex number is :5 + 3i