[Umbrella]

How can i convert decimal to hexadecimal in c++?
The number can be any number.

[Amadeus]

There are a few ways...do you actually need it in a variable in it's changed form?

You can use the sprintf() function

http://www.cplusplus...io/sprintf.html

, or set a base format flag.

http://www.cplusplus...ec_hex_oct.html

Alternately, you can implement a conversion function of your own.

[Umbrella]

yea we haft to make are own algorithm that will convert a decimal into a hexadecimal.

cant figure that part out..

[NyeNye]

well make a template for that.....

[bluesuus]

Well, umbrella.if i understand problem.the i tnk this is wt ur professor wants.how do u convert from one base to another on paper to do is successive division and modulo.

take for e.g you want to convert 16 in base 10 to hexadeciaml number.

16/16 =1 remaider 0;

1/16 = 0 remainder 1

so ur anser is 10 in base 16....reading from bottom.

an appropriate c++ snippet can be.

while( number != 0 )
{
a[i] = number % 16;
number = number/16;
}

note: that u hv to put ur remainders in an array because, so that you can print the remainders from bottom

EDIT : Added Code Tags - born2c0de

[Umbrella]

ah wish i checked back i eventually figured it out and did it exactly like that . Thank you though!!

[born2c0de]

Quote

a[i] = number % 16;

Instead of changing 10 to 15 to A to F later in the code, we can implement it directly like this:

a[i] = (number%16) < 10 ? (number%16) : 'A' + (number%16) - 10;

[michaelangelo]

born2c0de, on 16 Oct, 2006 - 04:53 AM, said:

Quote

a[i] = number % 16;

Instead of changing 10 to 15 to A to F later in the code, we can implement it directly like this:

a[i] = (number%16) < 10 ? (number%16) : 'A' + (number%16) - 10;

so if that number is bigger than 10 it can be print as a hexadecimal with this code

for example

cout<<number;

and answer could be 2B or we should do something else to print that number as hex, thank you

[sssmartie]

hmmm don't the components of an arrays always have to be the same type...like I mean can we place both a number and a character in an array?!...if so what type of array should it be defined as in the begining of the program?[font=Comic Sans Ms]

[born2c0de]

A character is stored in memory as a number as per its ASCII code.
So although integer and character sound like two completely different data types, they are actually the same except that an integer takes 2/4 bytes on 16/32-bit systems while char takes up only 1 byte.

[sssmartie]

So as your algorithm when we want to print out the hexadecimal number ,if one of the digits would be 10 for example,is it going to be shown as 65 (the ascii code of 'A') or as A itself?
Thank you

[born2c0de]

It will be stored as 65 in the Array.
However, since the data type of the array is char, cout will interpret it as a set of characters and print it as a character i.e. 'A'

The same applies to printf().

Try this out to have a better idea.

char c = 'A';
int i = 66;

printf("\n%d",c);
printf("\n%c",i);

[h3X]

I wrote this in C++ before I saw this thread, and i just thought I'd share it with you:

#include <cstdio>
#include <cstdlib>
#include <stack>
#include <string>

using std::string;
using std::stack;

string dec2hex(int dec)
{
	int i = 0;
	stack <int>remainder;
	string hex, temp;
	char hexDigits[] = { "0123456789ABCDEF" };

	while (dec != 0)
	{
		remainder.push(dec % 16);
		dec /= 16;
		++i;
	}
	
	while (i != 0)
	{
		if (remainder.top() > 15)
		{
			temp = dec2hex(remainder.top());
			hex += temp;
		}
		hex.push_back(hexDigits[remainder.top()]);
		remainder.pop();
		--i;
	}
	return hex;
}

int main(int argc, char *argv[])
{
	if (argc != 2)
		exit(printf("Error: expected 1 param\nUsage: %s <integer>\n", argv[0]));

	int dec = atoi(argv[1]);
	
	string hex = dec2hex(dec);
	printf("%s\n", hex.c_str());
	
	return 0;
}

I hope someone finds this useful.