Code to convert a decimal number to hexadecimal

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Code to take a decimal number as input and prints its Hexadecimal equivalent, the code must have recursive function and no arrays in C language ??

I just wrote the following code, but sure about nothing in this, just tried to write something.

You can tell a new code instead of modifying this one.

What I have tried:

C++

#include<stdio.h>

int convert(int n){
	int x;
	if(n==0) {
		printf("0");
		return;
	}

	else{
		x = convert(n/16)%16;
		if(x>9){
			x = x+55;
			printf("%c", x);
		}
		else{
			printf("%d", x);
		}

		return n;
	}
}



int main(void){
	int n;
	scanf("%d", &n);
	convert(n);

}</stdio.h>

Posted 8-Feb-16 10:00am

Dishank Bansal

Updated 8-Feb-16 10:51am

v4

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Comments

Patrice T 8-Feb-16 16:15pm

What is the problem with this code ?
What is going wrong when you run it ?

Dishank Bansal 8-Feb-16 16:53pm

It's simply not giving the correct answer.
Suppose for input 15, it must give output F, but is giving 00.

3 solutions

Solution 1

There is no such thing as conversion between decimal and hexadecimal. This is because a number cannot be decimal or hexadecimal, it is "binary"; more exactly, calculation is agnostic of the internal representation of the number, except for some special cases.

So, "conversion" function profile cannot be int convert(int n). Note that you don't return anything at all, which is a plain bug. If you call some function "conversion", it should not print anything, but should only return some value. But you have no "another" integer value to return. All numeric values, such as int and other ones, represent mathematical value, which represent the value, amount, not the way of showing this value in a string.

The only relevant "conversion" would be a "conversion" to string. One of the way to do so is to create a function expecting a number and returning a string. But with C-style strings, you should not return such string (because it would have a side effect, allocation of memory, which is bad). Therefore, you should fill in some memory provided by the called. Such functions already exist in the standard library; this is, in particular, sprintf or snprintf:
sprintf — C++ Reference[^],
snprintf — C++ Reference[^].

All your "hexadecimal vs decimal" stuff is in the format specifiers which define how the data is "printed": printf — C++ Reference[^].
See also this tutorial: C Tutorial: printf, Format Specifiers, Format Conversions and Formatted Output | CodingUnit Programming Tutorials[^].

—SA

Posted 8-Feb-16 10:18am

Sergey Alexandrovich Kryukov

Updated 8-Feb-16 10:38am

v2

Comments

CPallini 8-Feb-16 16:35pm

5.

Sergey Alexandrovich Kryukov 8-Feb-16 16:36pm

Thank you, Carlo.
—SA

Solution 3

You should learn to use the debugger as soon as possible. Rather than guessing what your code is doing, It is time to see your code executing and ensuring that it does what you expect.

The debugger allow you to follow the execution line by line, inspect variables and you will see that there is a point where it stop doing what you expect.
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
http://docs.oracle.com/javase/7/docs/technotes/tools/windows/jdb.html[^]
https://www.jetbrains.com/idea/help/debugging-your-first-java-application.html[

Pay attention to the value returned by convert.

Posted 8-Feb-16 12:10pm

Patrice T

Updated 8-Feb-16 12:16pm

v2

Comments

Dishank Bansal 9-Feb-16 2:08am

I know how to use debugger.
But here i'm unable to make the algorithm.
this one is just a random try.

jeron1 9-Feb-16 10:26am

A 'random' try, really? At least map it out and go through it on paper, if you can't do it there, then you certainly can't write code for it.

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CPallini · Accepted Answer · 2016-02-08T11:07:00

Sergey already showed you the right way of thinking about it, that is the hexadecimal representation of a number. He also suggested you the proper way to handle it in C.
Now, let me try to go a bit into the actual implementation:

C

  #include <stdio.h>
  #include <string.h>
  #include <assert.h>

void step( int k, int l, char * p)
{
  if ( l )
  {
    *p = (k & 0xF);
    *p += *p < 0xA ? '0' : ('A' - 0xA);
    step(k>>=4, --l, --p);
  }
}

void hexstr(int n, char buf[], int bufsize)
{
  int len = 2 * sizeof(n);
  assert(buf && bufsize > len);
  buf[len] = '\0';
  step(n, len, &buf[len-1]);
}


int main()
{
  int n;
  char buf[2*sizeof(n)+1];
  printf("please enter a number\n");
  if ( scanf("%d", &n) == 1)
  {
    hexstr(n, buf, sizeof(buf));
    printf("the hexadecimal representation of %d is %s\n", n, buf);
  }
  return 0;
}

[update]
This one recursively prints the hexadecimal representation of the number without using arrays.

C++

#include <stdio.h>
#include <string.h>
#include <assert.h>


void hexstep( unsigned int k, int l)
{
  if ( l )
  {
    hexstep (k >> 4, --l);
  }

  char c = (char) (k & 0xF);
  printf("%c", c < 0xA  ? c + '0' : c + 'A' - 0xA);
}

int main()
{
  unsigned int n;
  printf("please enter a number\n");
  if ( scanf("%u", &n) == 1)
  {
    printf("the hexadecimal representation of %u is ", n);
    hexstep(n, sizeof(n)*2-1);
    printf("\n");
  }
  return 0;
}

[/update]

Code to convert a decimal number to hexadecimal

3 solutions

Solution 1

Solution 2

Solution 3

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