#include<iostream> using namespace std; int main() { int t_in_fah, t_in_cel; cout << "Enter a temperature in Fahrenheit \n"; cin >> t_in_fah; t_in_cel = 5/9*(t_in_fah - 32); cout << "The temperature in Celsius is: " << t_in_cel << endl; return 0; }

          #include<iostream>
using namespace std;
int main()
{
int t_in_fah, t_in_cel;
cout << "Enter a temperature in Fahrenheit \n";
cin >> t_in_fah;
t_in_cel = 5/9*(t_in_fah - 32);
cout << "The temperature in Celsius is: " << t_in_cel << endl;
return 0;
}

#include<iostream>
using namespace std;
int main()

int tinfah, tincel;
cout << "Enter a temperature in Fahrenheit ";
cin >> tinfah;
tincel = 5/9*(tinfah - 32);
cout << "The temperature in Celsius is: " << tincel << endl;
return 0;

Added by Cristina N.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Nick Johnson

Step 1

The given program is intended to convert a temperature from Fahrenheit to Celsius. The formula for conversion is correct: Celsius = (5.0/9.0) * (Fahrenheit - 32). However, the program does not produce the correct results because of how it handles the data types Show more…

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The following program is supposed to convert a temperature in degrees Fahrenheit to degrees Celsius, but it will not produce the correct results. Use only TYPE CASTING to fix the problem and run the program for the test values: (1) 32 F is 0 C; and (2) 212 F is 100 C. #include<iostream> using namespace std; int main() { int t_in_fah, t_in_cel; cout << "Enter a temperature in Fahrenheit: "; cin >> t_in_fah; t_in_cel = static_cast<int>(5.0/9.0 * (t_in_fah - 32)); cout << "The temperature in Celsius is: " << t_in_cel << endl; return 0; }