Temperature Conversion Error

Cecil · #1 27-May-2006, 16:29

Hello, I have just started my Introduction to C programming course out at UCF, and I was wondering if some of you would be able to give some help.

I'll paste the basic assignment details:

Problem A: Temperature
This program will convert between Fahrenheit and Celsius. The information you will collect from the user is:
1. Magnitude of the temperature
2. Type of measurement
Prompt the user to input these values.

You will output a single statement showing both original and converted temperature values (as rounded integers).

Input Specification
1. Magnitude of the temperature is an integer and the program will read it into an integer variable (even though it would be possible to read it into a floating point variable, why am I enforcing this?).

2. Type of measurement is a single lowercase character, which will be used to determine the type of the temperature entered. For example, if the user enters “123 f”, this means 123 is in Fahrenheit and it needs to be converted to Celsius.

Output Sample
You should test your program with different data than is shown here based on the specifications given. The user input is given in italics while the program output is in bold.

Sample Run 1
Enter the temperature and type: 26 c
Converting from Celsius to Fahrenheit…
The formula is F=(9/5)*C+32
26 Celsius is 79 Fahrenheit.

Sample Run 2
Enter the temperature and type: 78 f
Converting from Fahrenheit to Celsius…
The formula is C=(5/9)*(F-32)
78 Fahrenheit is 26 Celsius.

I'll start off with the code:

CPP / C++ / C Code:

/*
	Assignment #1
	5/25/06
	Problem A: Temperature Conversion */
	
#include <stdio.h>

int main(void)
{
	float temperature;
	char symbol;
	
	//Ask the user to input the temperature and type.
	printf("Enter the temperature and type:\n");
	
	//User enters information.
	scanf("%f %c", &temperature, &symbol);
	
	//Check the user's input
	if ((symbol == 'f') || (symbol == 'c'))
		printf("Correct input, continuing..\n");
		else
		printf("Invalid temperature symbol.  Please enter f for fahrenheit, or c for celsius\n");
		
	//Calculate the user's correct input
		if (symbol == 'f')
			{
				printf("\n%s\n\n%s",
				"converting from Fahrenheit to Celsius..",
				"the forumula is C=(5/9)*(F-32)");
				printf("The temperature in Celsius is: %d.\n", (5/9) * (temperature - 32));
			}
				
		if (symbol == 'c')
			{
				printf("\n%s\n\n%s",
				"converting from Celsius to Fahrenheit..",
				"the forumula is F=(9/5)*(C+32)");
				printf("The temperature in Fahrenheit is: %d.\n", (9/5) * (temperature + 32));
			}
		return 0;
}

The problem that I'm having is with the equations. The rest of this program seems to complie, link, and run perfectly fine. However, when I enter the input to be converted, it always returns '0' for the answer.

I'm sure that the problem lies in my equations. If you find anything else that could be a problem, please let me know. This is my first time learning C and I'm really avid on learning the ins and outs of it all.

Thanks in advance! Looking forward to your replies!

TurboPT · #2 27-May-2006, 16:51

Since you take temperature as a 'float' type, use %f in your printf's instead of %d.

Or if you keep %d, you'll have to cast your printf expressions to 'int', i.e. (int)( your expression ).

Cecil · #3 27-May-2006, 17:34

EDIT: I got it working now. Thanks! I had my math equations down wrong. I just misplaced the parenthesis is all.

Thanks for the help on the decimal placement for the answer by the way.

TurboPT · #4 27-May-2006, 17:53

Good,

I was going to ask about that calculation.

WaltP · #5 27-May-2006, 23:19

Hey Cecil, could you post your working code? I have a problem with what you posted that TurboPT didn't point out, and wondered if you fixed it.

Also, your formatting is a little horizontal heavy. Check out the Formatting code tutorial for different ideas.

Cecil · #6 28-May-2006, 00:48

Certainly, here is the working code:

CPP / C++ / C Code:

/*	
	Assignment #1
	5/25/06
	Problem A: Temperature Conversion */
	
#include <stdio.h>

int main(void)
{
	float temperature;
	char symbol;
	
	//Ask the user to input the temperature and type.
	printf("Enter the temperature and type:\n");
	
	//User enters information.
	scanf("%f %c", &temperature, &symbol);
	
	//Check the user's input
	if ((symbol == 'f') || (symbol == 'c'))
		printf("Correct input, continuing..\n");
		else
		printf("Invalid temperature symbol.  Please enter f for fahrenheit, or c for celsius\n");
		
	//Calculate the user's correct input
		if (symbol == 'f')
			{
				printf("\n%s\n\n%s",
				"converting from Fahrenheit to Celsius..",
				"the forumula is C=(5/9)*(F-32)");
				printf("The temperature in Celsius is: %.f.\n", temperature = .5555555 * (temperature - 32));
			}
				
		if (symbol == 'c')
			{
				printf("\n%s\n\n%s",
				"converting from Celsius to Fahrenheit..",
				"the forumula is F=(9/5)*(C+32)");
				printf("The temperature in Fahrenheit is: %.f.\n", temperature = (1.8 * temperature) + 32);
			}
		return 0;
}

WaltP · #7 28-May-2006, 04:00

OK, you fixed the problem I saw, but IMO you fixed it wrong.

CPP / C++ / C Code:

printf("\n%s\n\n%s",
"converting from Fahrenheit to Celsius..",
"the forumula is C=(5/9)*(F-32)");
printf("The temperature in Celsius is: %.f.\n", temperature = .5555555 * (temperature - 32));

#1) Yes, 5/9 is approx .5555555 but you really changed the equation. Not a big deal, but I feel it would be better keeping the equation intact:

CPP / C++ / C Code:

printf("The temperature in Celsius is: %.f.\n", 
      (5.0 / 9.0) * (temperature - 32));

When you had 5/9 you probably noticed the equation didn't work so you calculated the value of it. But if you simply add the .0 to the values the equation works.

#2) You shouldn't set the value of temperature in the printf()

Cecil · #8 28-May-2006, 18:42

Ah, very nice observation. Thanks for the help!

amneziac85 · #9 28-May-2006, 21:09

this is interesting, ive seen similar issues.

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