C++ Program to Convert Number in Charact

Please learn to use code tags, they make reading and commenting on source code MUCH easier.

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HINT: you can edit your post and add code tags.

Some formatting & indentation would not hurt either

M'ok, you've written some code, now what?

You obviously have questions/concerns about your code, but you didn't let us know.

thmm (703)

Though the code looks quite clumsy and messy, it seems to work.
So what do you want ?

#include <iostream>
#include <cmath>

using namespace std;

int main() {
	int number = 1000;

	do {
		cout << "Enter an integer number< 1000:";
		cin >> number;
	} while (number < 1 || number >= 1000); // Check for number to be >=1 and < 1000

	int hundreds = number / 100;
	int tens = (number - (hundreds * 100)) / 10;
	int ones = number % 10;

	switch (hundreds) {
		case (9):
			cout << "Nine Hundred ";
			break;
		case (8):
			cout << "Eight Hundred ";
			break;
		case (7):
			cout << "Seven Hundred ";
			break;
		case (6):
			cout << "Six Hundred ";
			break;
		case (5):
			cout << "Five Hundred ";
			break;
		case (4):
			cout << "Four Hundred ";
			break;
		case (3):
			cout << "Three Hundred ";
			break;
		case (2):
			cout << "Two Hundred ";
			break;
		case (1):
			cout << "One Hundred ";
			break;
	}

	switch (tens) {
		case (9):
			cout << "Ninety ";
			break;
		case (8):
			cout << "Eighty ";
			break;
		case (7):
			cout << "Seventy ";
			break;
		case (6):
			cout << "Sixty ";
			break;
		case (5):
			cout << "Fifty ";
			break;
		case (4):
			cout << "Fourty ";
			break;
		case (3):
			cout << "Thirty ";
			break;
		case (2):
			cout << "Twenty ";
			break;

		case (1):
			{
				if (ones == 1) {
					cout << "Eleven";
					break;
				} else if (ones == 2) {
					cout << "Twelve";
					break;
				} else if (ones == 3) {
					cout << "Thirteen";
					break;
				} else if (ones == 4) {
					cout << "Fourteen";
					break;
				} else if (ones == 5) {
					cout << "Fifteen";
					break;
				} else if (ones == 6) {
					cout << "Sixteen";
					break;
				} else if (ones == 7) {
					cout << "Seventeen";
					break;
				} else if (ones == 8) {
					cout << "Eighteen";

					break;
				} else if (ones == 9) {
					cout << "Nineteen";
					break;
				} else
					break;
			}
	}

	if (tens > 1 || number < 10) // if number greater than 19 or number less than 10
	{
		switch (ones) {
			case (9):
				cout << "Nine";
				break;
			case (8):
				cout << "Eight";
				break;
			case (7):
				cout << "Seven";
				break;
			case (6):
				cout << "Six";
				break;
			case (5):
				cout << "Five";
				break;
			case (4):
				cout << "Four";
				break;
			case (3):
				cout << "Three";
				break;
			case (2):
				cout << "Two";
				break;
			case (1):
				cout << "One";
				break;
		}
	}

	cout << endl;
	return 0;
}

so what's the C++ question? Is it clumsy - yes, can it be improved - yes.

One way could be:

#include <iostream>
#include <array>

constexpr std::array Units { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };
constexpr std::array Tens { "zero", "tens", "twenty", "thirty", "fourty", "fifty", "sixty", "seventy", "eighty", "ninety" };
constexpr std::array Special { "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };

int main() {
	int number {};

	do {
		std::cout << "Enter an integer number (between 0 and 1000): ";
		std::cin >> number;
	} while (number < 1 || number >= 1000); // Check for number to be >=1 and < 1000

	const int hundreds { number / 100 };
	const int tens { (number - (hundreds * 100)) / 10 };
	const int ones { number % 10 };

	if (hundreds > 0)
		std::cout << Units[hundreds] << " hundred ";

	if (tens > 1)
		std::cout << Tens[tens] << ' ';
	else if (tens == 1)
		std::cout << Special[ones];

	if (tens != 1 && ones > 0)
		std::cout << Units[ones];

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

@srpathak18, the original code was C++, why have C code that doesn't really fulfill the original requirements?

A helpful reminder:

Your code, formatted:

#include<stdio.h>
void main()
{
   int num, i = 0, x, d;
   char* word_no[2000];
   printf("Enter an integer value: \n");
   scanf("%d", &num);
   while (num)
   {
      d = num % 10;
      num = num / 10;
      switch (d)
      {
      case 0: word_no[i++] = "zero";
         break;
      case 1: word_no[i++] = "one";
         break;
      case 2: word_no[i++] = "two";
         break;
      case 3: word_no[i++] = "three";
         break;
      case 4: word_no[i++] = "four";
         break;
      case 5: word_no[i++] = "five";
         break;
      case 6: word_no[i++] = "six";
         break;
      case 7: word_no[i++] = "seven";
         break;
      case 8: word_no[i++] = "eight";
         break;
      case 9: word_no[i++] = "nine";
         break;
      }
   }
   for (x = i - 1; x >= 0; x--)
   {
      printf("%s ", word_no[x]);
   }
}

Let me clarify, the requirement is to parse a number, say 235, into words. "two hundred thirty five".

What you are doing certainly doesn't create a valid C string (hint, hint, no terminating '\0').

void main is non-standard, it should be int. Even if the compiler accepts it, don't use it.

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Don't sweat over it George, they're gone the way of all things...

The string literals are null terminated. word_no is an array of char pointers. So for the number 123 then word_no[0] is pointer to "one", word_no[1] is pointer to "two" and word_no[2] is pointer to "three". So that the final loop will print one two three

Visual Studio doesn't like the code, complains mightily about assigning a constant char* to a char* 10 times, ending up in a nice big compile fail of 10 errors.

Code::Blocks' MinGW 8.1.0 complains about the void main as an error; only non-fatally warns about the assignments, it is forbidden to assign a string constant to char*.

But again, this isn't fulfilling the requirements of the assignment.

I ain't "sweating" anything other than it is crap code for different reasons with two compilers.

> only non-fatally warns about the assignments, it is forbidden to assign a string constant to char*.

With g++ (and compatibles), we need to explicitly insist on C++ conformance,
for example with: -std=C++11 -pedantic-errors
http://coliru.stacked-crooked.com/a/d06f435520c732f5

@JLBorges, I understand that, really. :)

The quickie project I used for temp testing the code in C::B didn't have pedantic errors set, the IDE's defaults. I was too lazy to set things properly.

I just find it a bit irritating that MinGW whinges about void main and VS does not.

With -Za (Disable Language Extensions), the Microsoft compiler generates a warning:

warning C4326: return type of 'main' should be 'int' instead of 'void'

It's c code - not C++ code. As c code with VS it compiles OK and runs as expected (with language extensions enabled!).

For C++ code 5, L5 should be:

const char* word_no[2000];

as in C++ "one" is of type const char*

and obviously void main() should be int main()

and don't try using Microsoft's Windows headers with -Za. They don't compile!

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Visual Studio's default warning level (3) is not set as high as it should when using language extensions, to get a "twitch" from VS about void main the warning level must be set to EnableAllWarnings (/Wall).

I'm not even going to get into "scanf is unsafe" religious debate, something C11 did by adding scanf_s long after MS "enhanced" C I/O.

No matter how ya slice it the code as was originally presented before disappearing into the aether isn't good C, let alone C++. Forget about whether it fulfilled the assignment's requirements, which it didn't.

On a side note VS doesn't like disabling language extensions when the C language standard is set to anything other than the default Legacy MSVC.

"D8016: '/Za' and '/std:c17' command-line options are incompatible." Same whinging with C11.

I think I'll stick to C++ code, I don't need to worry about legacy production code. Being a hobbyist at this.

> as in C++ "one" is of type const char*

In C++, the string literal "one" is of type array of 4 const char ie. const char[4].
It has a static storage duration.

In C, the string literal "one" is of type array of 4 char ie. char[4].
However, an attempt to modify an element of this array engenders undefined behaviour.

Also, in both C++ and C,
A string literal is not necessarily a null-terminated character sequence: if a string literal has embedded null characters, it represents an array which contains more than one (C-style) string.
https://en.cppreference.com/w/cpp/language/string_literal#Notes

This discussion and differing viewpoint is some good stuff to know! :)

Regarding the C format specifiers, I can find what %u is easily. But nothing "pings" when I do a search for %zu. Not even cppreference that I can see has it.*

I know what is used for, (https://stackoverflow.com/questions/2125845/platform-independent-size-t-format-specifiers-in-c), I'd just like to see some definitive documentation beyond a Linux man ref link buried in other pages. https://linux.die.net/man/3/printf

*Well, cppreference does have it, if one knows where to look and what to look for. z is a size modifier *ugh*

Last edited on

cppreference (see table 'The following format specifiers are available:')
Since C++11:
Length Modifier z, Conversion Specifier d ie. %zd : Expected Argument Type signed size_t
Length Modifier z, Conversion Specifier u ie. %zu : Expected Argument Type std::size_t
https://en.cppreference.com/w/cpp/io/c/fprintf

I got there about a minute before ya, JLBorges.

I guess my staff was too long, so I was looking in the wrong place.

https://youtu.be/Pk-B0s0jOwE?t=86