Question

literally can't make this program run on code blocks. help plz

Answer 1

```
//
// Conversion - Program to convert temperature from
// Celsius degrees into Fahrenheit:
// Farhrenheit = Celsius * (212 - 32)/100 + 32
//
//
#include <cstdio>
#include <cstdlib>
#include <iostream>

using namespace std;

int main(int nNumberofArgs, chars* pszArgs[])
{
// enter the temperature in Celcius
int celsius;
cout << "Enter the temperature in Celcius:";
cin >> celcius;

// calculate conversion factor for Celsius
//to Fahrenheit

int factor;
factor = 212 - 32;

//use conversion factor to convert Celsius
//into Fahrenheit values

int factor;
fahrenheit = factor * celsius/100 + 32;

//output the results (followed by a NewLine)
cout << "Fahrenheit value is:";
cout << fahrenheit << endl;

//wait until user is ready before terminating program
//to allow the user to see the pragram results

cout << "Press Enter to continue..." <<endl;
cin.ignore(10, '\n');
cin.get();
return 0;

}
```

Answer 2

https://onlinegdb.com/HJpc2JhlU

Answer 3

line 18 now

Answer 4

it says "Celsius was not declared in this scope." what does that mean?

Answer 5

nvm i got it fixed. just a bunch of misspelling and stuff.

Answer 6

Oof...

Answer 7

Yea, lotta mispelling, didn't realize this was already solved and just fixed it lol
```cpp
//
// Conversion - Program to convert temperature from
// Celsius degrees into Fahrenheit:
// Farhrenheit = Celsius * (212 - 32)/100 + 32
//
//
#include <cstdio>
#include <cstdlib>
#include <iostream>

using namespace std;

int main(int nNumberofArgs, char* pszArgs[])
{

//declare variables
int fahrenheit;
int celsius;
int factor;

//enter the temperature in Celcius
cout << "Enter the temperature in Celcius:";
cin >> celsius;

//calculate conversion factor for Celsius
//to Fahrenheit
factor = 212 - 32;

//use conversion factor to convert Celsius
//into Fahrenheit values
fahrenheit = factor * celsius/100 + 32;

//output the results (followed by a NewLine)
cout << "Fahrenheit value is:";
cout << fahrenheit << endl;

//wait until user is ready before terminating program
//to allow the user to see the pragram results

cout << "Press Enter to continue..." <<endl;
cin.ignore(10, '\n');
cin.get();
return 0;

}
```

Answer 8

dumb question, why was this in writing, LOL