Project to calculate money and time to spend all of it.

[SmokeyBear]

Hello, I'm trying to make a program that would calculate a certain sum of money and how long it would take you to spend it. This is a very sloppy rough draft. I'm trying to figure after the "if statment" how to return back to the beginning of the code. Also I would appreciate any other criticism on how I can better my project. Thank you.

 #include <iostream>
#include <math.h>
using namespace std; 
int main()
{
	double days = 0.0;
	double hours = 0.0;
	double minutes = 0.0;
	double seconds = 0.0;
	//double money = 0.0;
	double spending = 0.0;
	double amount = 0.0;

	long long money = 0.0; 
	long long day = 0.0;
	long long hrs = 0.0;
	long long min = 0.0;
	long long sec = 0.0;



	cout << "This project is to determine" << endl;
	cout << "how long it would take to spend a certain amount of money" << endl;
	cout << "in a specified amount of time" << endl << endl;

	cout << "Enter the desired amount of money" << endl;
	cin >> money;

	if (money > 9223372036854775807 || money <= 0)
	{
		cout << "\nInvalid amount - Must be greater than $0 or less than $9,223,372,036,854,775,807 \n";
	}

	cout << "Enter how much would you want to spend - $" << endl;
	cin >> spending; 

	day = money / spending;
	hrs = money / spending * 24;
	min = hrs * 60;
	sec = min * 60;


	cout << "It would take this many days - " << day << endl;
	cout << "It would take this many hours - " << hrs << endl;
	cout << "It would take this many minutes - " << min << endl;
	cout << "It would take this many seconds - " << sec << endl;

	system("Pause");
	return 0;
}

[Thomas1965]

It's not clear how it should work.
Lets say I have 1000£ and want to spend 500£
What should be the outcome?

[SmokeyBear]

Well lets say I have a $1,000. I chose to spend $1 a day.
It would take 1000 days. X amount of hours. X amount of seconds. etc.

[Ganado]

Side note:
money > 9223372036854775807
If you're trying to detect overflow here, don't bother, because if that value can't be stored in a 64-bit integer, there's no way for money to be greater than the greatest number representable by a 64-bit integer.

I'm trying to figure after the "if statment" how to return back to the beginning of the code.

Use a while loop.
http://www.cplusplus.com/doc/tutorial/control/ --> See Loops section

#include <iostream>

int main()
{
	int n = 0;
	while (n != 42)
	{
		std::cout << "Enter 42: ";
		std::cin >> n;
	}

	return 0;
}

[SmokeyBear]

Something like this?

	while (money > 9223372036854775807 || money <= 0)
	{
		cout << "\nInvalid amount - Must be greater than $0 or less than $9,223,372,036,854,775,807 \n";

	}

[Thomas1965]

You mean like this?

double available_money = 1000;
double amount_per_day = 500;

double days = available_amount / amount_per_day;
double hours = days * 24;
double minutes = hours * 60;
double seconds = minutes * 60;

cout << "It would take this many days - " << days << endl;
cout << "It would take this many hours - " << hours << endl;
cout << "It would take this many minutes - " << minutes << endl;
cout << "It would take this many seconds - " << seconds << endl;

Output:

It would take this many days - 2
It would take this many hours - 48
It would take this many minutes - 2880
It would take this many seconds - 172800
Press any key to continue . . .

[jlb]

Something like this?

No, you would be better off testing the stream state if you're worried about numeric overflow or non-digit values. If the value inputted is larger than the type can hold or a non-digit character is entered the stream will enter a fail state.

[SmokeyBear]

Yes @Thomas1965 exactly.

I tried my best to clean it up. Although its not perfect I hope you can see my intent with the code. Criticism and advice would be much appreciated.

#include <iostream>
#include <math.h>
#include <limits.h>
using namespace std; 
int main()
{
	long long days = 0.0;
	long long hours = 0.0;
	long long minutes = 0.0;
	long long seconds = 0.0;
	long long money = 0.0;
	long long spending = 0.0;
	long long amount = 0.0;


	cout << "This project is to estimate" << endl;
	cout << "how long it would take to spend a certain amount of money" << endl;
	cout << "in a specified amount of time" << endl << endl; 

	
	bool continueLoop = true;
		do
	{
	
	cout << "Enter the desired amount of money" << endl;
		cin >> money;


		if (money > 900,000,000,000,000,000 || money <= 0)
					
		{
			cout << "\nInvalid amount - Must be greater than 0 or less than 900,000,000,000,000,000 \n";
			system("Pause");
			return 0;
		};


		cout << "Enter how much would you want to spend daily - $" << endl;
			cin >> spending; 


			if (spending > money)

			{
				cout << "\nCan't spend more than whats available \n";
				system("Pause");
				return 0;
			};


			days = money / spending;
			hours = money / spending * 24;
			minutes = hours * 60;
			seconds = minutes * 60;


			cout << "It would take this many days - " << days << endl;
			cout << "It would take this many hours - " << hours << endl;
			cout << "It would take this many minutes - " << minutes << endl;
			cout << "It would take this many seconds - " << seconds << endl;

			system("Pause");
			return 0;
		} while (continueLoop == true);


	}
 

[icy1]

- return returns from the function.
- The function in question is main(). You'll be exiting the program.
- For the first two returns, were you instead looking for continue ?
- There are commas in your giant number -- does that even compile?
- semicolons after the 'if' statement brackets, probably not legal
- nothing ever changes your continueLoop flag; why even have a variable? while(true) { ... } prob simpler, or if you want a certain amount of complete runthroughs, while(count<3) , initialize count before while loop and increment that at the bottom after outputting the seconds.

[SmokeyBear]

#include iostream
#include math.h
#include limits.h
using namespace std; 
int main()
{
	long double days = 0.0;
	long long hours = 0.0;
	long long minutes = 0.0;
	long long seconds = 0.0;
	long double money = 0.0;
	long double spending = 0.0;
	long long amount = 0.0;


	cout  This project is to estimate  endl;
	cout  how long it would take to spend a certain amount of money  endl;
	cout  in a specified amount of time  endl  endl; 

	
	bool continueLoop = true;
		do
	{
	
	cout  Enter the desired amount of money  endl;
		cin  money;


		if (money  900000000000000000  money = 0)
					
		{
			cout  nInvalid amount - Must be greater than 0 or less than 900,000,000,000,000,000nn;
			continue;
		}


		cout  Enter how much would you want to spend daily - $  endl;
			cin  spending; 


			if (spending  money)

			{
				cout  nCan't spend more than whats available n;
				continue;
			}


			days = money  spending;
			hours = money  spending  24;
			minutes = hours  60;
			seconds = minutes  60;

			cout  It would take this many days -   days  endl;
			cout  It would take this many hours -   hours  endl;
			cout  It would take this many minutes -   minutes  endl;
			cout  It would take this many seconds -   seconds  endl;

			system(Pause);
			return 0;
		} while (continueLoop == true);


	} 

I've done this. I have two issues.
- I am not totally sure on my loop and would appreciate a more detailed example
- I am trying to find an integer for "days" that would handle large numbers and decimals

[Ganado]

Something went wrong when you pasted your code, half of your symbols don't show up. Try re-posting.

As was mentioned by jlb, one option to detect valid input is to check the state of cin.

long long n;
if (std::cin >> n)
{
    std::cout << "Valid input.\n";
}
else
{
    std::cout << "Invalid input!\n";
}

[E0p9LyTq]

I would appreciate any other criticism on how I can better my project.

Don't use the using namespace std; directive.

Starting out it seems easier to save a few keystrokes, but later you will run into problems with name clashes.

// why using namespace std is bad

#include <iostream>
#include <algorithm>

using namespace std;

unsigned int count;

int main()
{
   count++; // conflicts with std::count() in <algorithm>

   std::cout << count << '\n';
}

count in main() confuses the compiler, there is an ambiguous "match" between the global variable count and std::count in the <algorithm> header.
https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice

Using global variables can also cause hard to detect runtime problems, if used without careful consideration.

[icy1]

Added a few tweaks:
- made all vars the same type (long double)
- added var for the upper limit
- to avoid scientific notation, manipulated cout to print with fixed notation and an arbitrary decimal precision
- bunched input pieces together
- bunched error-handling pieces together
- ask user for a single character to repeat and eat the newline after it
- made example bool variable 'repeat' to show how the choice can influence the loop (can also just move the logic over to the while loop condition)
- aligned some of the output for consistency

#include <iostream>
#include <iomanip>
#include <limits>
#include <locale>

using namespace std; 

struct ForceCommas : std::numpunct<char>
{
    char do_thousands_sep() const { return ','; }
    string do_grouping() const { return "\03"; }
};

int main()
{
    long double days = 0.0;
    long double hours = 0.0;
    long double minutes = 0.0;
    long double seconds = 0.0;
    long double money = 0.0;
    long double upper_limit = 900000000000000000;
    long double spending = 0.0;

    cout << "This project is to estimate how long it would " << endl << 
            "  take to spend a certain amount of money in a" << endl <<
            "  specified amount of time." << endl << endl; 
            
    // If you want fixed notation with two decimal points
    cout << fixed << setprecision(2);
    cout.imbue(locale(cout.getloc(), new ForceCommas));
    
    bool repeat;
    char redo_choice;
    do
    {
        cout << "Enter desired amount of money: $";
        cin >> money;
        cout << "         Enter daily spending: $";
        cin >> spending; 

        if (money <= 0 || money > upper_limit)
        {
            cout << "\nInvalid amount - Must be greater than 0 or less than "<< upper_limit << endl;
            continue;
        }
        if (spending > money)
        {
            cout << "\nCan't spend more than whats available \n";
            continue;
        }

        days = money / spending;
        hours = money / spending * 24;
        minutes = hours * 60;
        seconds = minutes * 60;
        
        cout << endl;
        cout << "   It would take this many days  " << days << endl;
        cout << "  It would take this many hours  " << hours << endl;
        cout << "It would take this many minutes  " << minutes << endl;
        cout << "It would take this many seconds  " << seconds << endl;

        cout << "\nAgain (y/n)? ";
        cin >> redo_choice;
        cin.ignore(numeric_limits<int>::max(), '\n');
        repeat = redo_choice=='y' || redo_choice=='Y';
    } while (repeat);
    
    return 0;
}

can test here https://repl.it/repls/PaleStarryKeyboard

example output:

This project is to estimate how long it would 
  take to spend a certain amount of money in a
  specified amount of time.

Enter desired amount of money: $ 900000000000000000
         Enter daily spending: $ 7

   It would take this many days  128,571,428,571,428,571.43
  It would take this many hours  3,085,714,285,714,285,714.25
It would take this many minutes  185,142,857,142,857,142,848.00
It would take this many seconds  11,108,571,428,571,428,571,136.00

Again (y/n)?  n

Edit: added commas to the locale for the lulz.
I Also realized the site i like to use, repl.it, adds an extra space after every 'cin', unlike cpp.sh. Weird; your output may vary slightly.

[SmokeyBear]

Thank you guys very much you all were a big help!
I really need to work on my loops and decimal formatting.

[SmokeyBear]

struct ForceCommas : std::numpunct<char>
{
    char do_thousands_sep() const { return ','; }
    string do_grouping() const { return "\03"; }
};

    // If you want fixed notation with two decimal points
    cout << fixed << setprecision(2);
    cout.imbue(locale(cout.getloc(), new ForceCommas));

What do these two parts of the code exactly do?
The second one sets it to two decimal points but what is the other code used for?