Make this code work using these exact equations

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C++

#include<math.h>
#include<stdlib.h>
#include<stdio.h>

void main(void)
{
	printf("Name and Date: Matthew Adair, January 20, 2017 \n");
	printf("Course and Section: ENGR 19700 and Class #24743 \n");
	printf("Problem: Adair_temp5_M.c C Programming Homework 5 \n");
	printf("Statement: Program is to print information with \n");

	double TF_, TK_, TR_;


	printf("What is the temperature in Farenheit? \n");
	scanf("%Lf", &TF_);
	TF_ = TR_ - 459.67;
	TR_ = (9 / 5)*TK_;
	printf("The Kelvin equivalent to the given temperature in Farenheit is %Lf \n", TK_);


	

}

What I have tried:

I feel like it should work but I keep getting a runtime check failure #3 - T

Posted 6-Feb-17 9:18am

Member 12985896

Updated 6-Feb-17 12:24pm

PIEBALDconsult

v2

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[no name] 6-Feb-17 15:30pm

I don't see how you would think this would work. You use uninitialized variables, you read a value from the user and then overwrite it with a calculation with an uninitialized variable, perform another calculation using an uninitialized variable and then finally print the value of an uninitialized variable that you didn't even calculate. Learning to use the debugger is also a valuable skill that you will need often.

Member 12985896 6-Feb-17 15:55pm

In class we don't use the debugger, but anyways aren't the variables initialized when I use "double TF_, TK_, TR_;", same as they would be if I used "int"? The program works if I reverse the equations like "TF_ = TR_ - 459.67;" would be "TR_ = TF_ + 459.67;".

[no name] 6-Feb-17 17:14pm

"we don't use the debugger", then your teacher is not doing his job.
"variables initialized when I use "double ", not in C/C++ they are not. YOu have to initialize them.

Member 12985896 6-Feb-17 17:18pm

By initializing a variable, do you mean giving it a value? Such as using "double TF_ = 81.3;" instead of "double TF_;" and then later doing "scan_s("%Lf", &TF_);"? I'm appreciate the help, but i'm not sure what you mean. My professor isn't the best and programming is definitely not my strong suit as of now.

[no name] 6-Feb-17 18:07pm

"mean giving it a value", partially. When you declare a variable, it has a value but it could be anything and you won't know what it is. The point of initializing is giving your variable a known value.

Philippe Mori 6-Feb-17 21:11pm

Well, the problem is that you are confuse by how C++ interpret "equation". In C++, you have to put on the left the variable that get the result and on the right the equation itself. So you would have TR_ = TF_ + 459.67; for example.

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Jon McKee · Accepted Answer · 2017-02-06T12:24:00

*Updated after reading comment*

double TF_, TK_, TR_;
printf("What is the temperature in Farenheit? \n");
scanf("%Lf", &TF_); //Read TF
TF_ = TR_ - 459.67; //Overwrite TF?
TR_ = (9 / 5)*TK_;
printf("The Kelvin equivalent to the given temperature in Farenheit is %Lf \n", TK_);

In your code, you are reading in the temperature in Fahrenheit then overwriting that with TR_ - 459.67. Why even input TF in the first place? Also, these equations are still incorrect for what your printf statements say is happening. Fahrenheit to Rankine is R = F + 459.67 and Rankine to Kelvin is K = R * (5/9). Your equations are attempting to convert from Rankine to Fahrenheit, then from Kelvin to Rankine. This is the fixed code:

double TF_, TK_, TR_;
printf("What is the temperature in Farenheit? \n");
scanf("%Lf", &TF_);
TR_ = TF_ + 459.67;
TK_ = (5/9) * TR_;
printf("The Kelvin equivalent to the given temperature in Farenheit is %Lf \n", TK_);

I'd also recommend using useful variable names.

double fahrenheit, kelvin, rankine;
printf("What is the temperature in Farenheit? \n");
scanf("%Lf", &fahrenheit);
rankine = fahrenheit + 459.67;
kelvin = (5/9) * rankine;
printf("The Kelvin equivalent to the given temperature in Farenheit is %Lf \n", kelvin);

Quick note: double variableName; is a variable declaration. Now you have memory associated with a variable. That memory can hold anything right now. It could be 0; it could be random junk. You need to assign (initialize) it before using it. Example: variableName = 0; or more compactly double variableName = 0;.

The reason it often works anyways is because many operating system kernels as a security feature do not give out uninitialized memory. Also many language specifications have defaults (though whether the compiler respects this can vary). All of this can change and is out of your control. You shouldn't rely on it especially when avoiding semi-non-deterministic effects is so easy in this case. Just a simple = 0;.

Patrice T · Accepted Answer · 2017-02-06T11:18:00

Quote:
but anyways aren't the variables initialized when I use "double TF_, TK_, TR_;", same as they would be if I used "int"?

int and float are used to allocate memory, nothing else.
Initializing a variable means giving it a value.

Quote:
In class we don't use the debugger,

Major failure from your teacher.
-----
When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. It allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.

Debugger - Wikipedia, the free encyclopedia[^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]

The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.

Make this code work using these exact equations

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