Write a program that prints messages for weather. The program will get Fahrenhei
ID: 3797870 • Letter: W
Question
Write a program that prints messages for weather. The program will get Fahrenheit degree from user input and print out following messages on the screen:
If temperature in Fahrenheit is over 100, print out message “Give us Ice! we are almost dead!!”.
If temperature in Fahrenheit is over 95, print out message “It's too hot!”.
If temperature in Fahrenheit is over 86, print out message “It's still hot!”.
If temperature in Fahrenheit is over 77, print out message “It's O.K…!”.
If temperature in Fahrenheit is over 68, print out message “It's good to do exercise!”.
If temperature in Fahrenheit is over 50, print out message “ It'scool !”.
If temperature in Fahrenheit is over 32, print out message “It's chilly!”.
If temperature in Fahrenheit is less than 32, print out message “It’s cold!”.
use if... else if... else structure what we studied in the class.
Explanation / Answer
solution in c language(in any programming language these statements remains same)
#include<stdio.h>
int main()
{
int temperature;
printf("please enter the temperature in fahrenheit ");
scanf("%d",&temperature);
if(temperature<32)
{
printf("it's cold");
}else if(temperature>=32&temperature<50)
{
printf("it's chilly'");
}else if(temperature>=50&temperature<68)
{
printf("it's cool");
}else if(temperature>=68&temperature<77)
{
printf("it's good to do exercise'");
}else if(temperature>=77&temperature<86)
{
printf("it's ok'");
}else if(temperature>=86&temperature<95)
{
printf("it's still hot");
}else if(temperature>=95&temperature<100)
{
printf("it's too hot'");
}else
printf("give us Ice! we are almost dead!!");
return 0;
}
output
please enter the temperature in fahrenheit
45
it's chilly'
--------------------------------
Process exited after 7.632 seconds with return value 0
Press any key to continue . . .
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.