Convertion from Decimal to Hexadecimal

Hi,
Can you please tell me the way to covert a given decimal number into hexadecimal (for example 4 -> 04, 20 -> 14 etc) and also the vice versa.
thank you.

Hi
Try this function module:
" CRM_EI_KB_CONV_DEC_TO_HEX "
This fm converts any decimal to hexadecimal, but not vice-versa.........
Hope it helps....

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Luciana T. Angeli wrote:
One more question, I see that negative numbers can't be converted to Hexadecimal this way.
It is possible to convert negetive decimal numbers to hexadecimal format ?Negative numbers need to be displayed as 2's complement to be shown in hexadecimal, so you would have to specify (in effect) how many bytes you are dealing with.
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Wrote file afiedt.buf
1 with t as (select &num as num from dual)
2* select to_char(case when num<0 then 65536+num else num end, 'fmXXXX') as hx from t
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old   1: with t as (select &num as num from dual)
new   1: with t as (select 123 as num from dual)
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SQL> /
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old   1: with t as (select &num as num from dual)
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old   1: with t as (select &num as num from dual)
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old   1: with t as (select &num as num from dual)
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8000
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Edited by: Srinivas on Feb 22, 2008 10:48 PM

use.
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Subbu S.

In the user exit of Outbound Idoc, they might have used some custom logic to implement the given logic.
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I have a question.
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6kyAngel wrote:
actually it convert the string value into decimal value.In fact what it does is convert the (binary) string into an integer value. It's the String representation of integer quantities that have a base (two, ten, hex etc): the quantities themselves don't have a base.
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Hi all,
I am working on a C++ program that converts a decimal to a fraction, finds the lowest common multiple, multiplies the numerator by the denominator, then reduces the fraction again so the fraction will be 'numerator/1'.
I am using this code in a larger program to change a floating point slope value to a whole value for slope-intercept equations and standard form equations.
The program will run but I tested almost 500 numbers and 20 numbers made the program crash (it stalled and did not finish the equation, I had to close the terminal and open it up again... )
These 20 numbers I had problems with:
2.36, 7.36, 1.11, 1.001, 1.66, 1.88, 2.22, 2.13, 3.24, 3,26, 3.32, 3.43, 3.49, 3.51, 3.57, 3.68, 3.74, 3.76, 3.82
(I am sure there are more...)
I have tried c-style casting and dynamic and static casting to try to resolve this issue. Can someone help me out and point out why these certain numbers make the program crash and offer a possible sloution?
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(gdb) break 16
No line 16 in file "init.c".
(gdb)
Here is the program's code:
//example.cc
#include <iostream>
#include "reduce.h"
#include "decimal-to-fraction.h"
using namespace std;
int main()
double deci;
double n, d;
while(1) {
cout << "Enter a decimal to convert to a fraction ('0' to quit): ";
cin >> deci;
cin.ignore(INT_MAX, '\n');
if(!deci) exit(0);
dec_to_frac(deci, n, d);
if (n * d != n){
cout << '\n' << deci << " = " << n << '/' << d << "\n\n";
cout<<'\n' << d << " x " << n;
n = d * n;
cout<< " = " << n << "\n\n";
cout<<'\n' << n << '/' << d;
reduce(n, d);
cout << " = " << n << '/' << d << "\n\n";
cout<<'\n' << n << "\n\n";
else
cout<<'\n' << deci<< "\n\n";
return 0;
#ifndef _REDUCE_H_
#error You must include "reduce.h" before this file
#endif /* def _REDUCE_H_ */
#ifndef _DECIMAL_TO_FRACTION_H_
#define _DECIMAL_TO_FRACTION_H_
void dec_to_frac(double decimal, double &numerator, double &denominator)
//numerator = static_cast<int >(decimal);
//numerator = (int )decimal;
numerator = decimal;
denominator = 1;
while(numerator != (int)numerator) {
numerator *= 10;
denominator *= 10;
reduce(numerator, denominator);
#endif /* def _DECIMAL_TO_FRACTION_H_ */
#ifndef _REDUCE_H_
#define _REDUCE_H_
void reduce(double &numer, double &denom)
int i;
for(i=2; i<=numer; ++i) {
if( ((numer/i) == ((int )(numer/i))) && ((denom/i) == ((int)(denom/i))) ) {
numer /= i;
denom /= i;
--i;
#endif /* def _REDUCE_H_ */

Okay, a few things ('cause I'm picky too): LCM == "least common multiple" or "lowest common multiple", it's really interchangable. LCD == lowest common denominator, which is when you're adding two fractions and want to find the lowest denom. Now that we have that out of the way...
Using ++i or i++ as the third expression in your for loop will make zero difference.
for (int i=2; i<=numer; ++i)
will have exactly the same effect as
for (int i=2; i<=numer; i++)
, though they may have slightly different performances. The reason for this is that the third expression (i++ or ++i) is evaluated after every iteration of the loop, meaning it doesn't happen until the end of the loop body is reached, and then it's evaluated independently, meaning all you end up with is 'i' incrementing.
As for this:
if( ((numer/i) == ((int )(numer/i))) && ((denom/i) == ((int)(denom/i))) ) {
becomes
if (numer % i == denom % i == 0) {
That's about right, it would look more like
if (numer %i == 0 && denom % i == 0)
- the other way would probably not behave the way you expect.
Some other tips:
numer and denom should be 'int' or 'long' - not double. You don't expect your numerator and denominator to have decimal points, so use integral types - using floating point 'double's is likely what's getting you into this trouble -- floating point arithmetic is inherently imprecise. You might have better luck parsing the input yourself and constructing the numerator from that instead of reading directly into a floating-point number.
I'm hacking around with your code right now for fun; I'll let you know if I find anything.
==edit==
Oh, something else I wanted to mention; to break on a line in gdb you type:
break filename:linenumber
ie.
break example.cc:16
Last edited by Cerebral (2007-06-03 02:43:42)

Conversión decimal a hexadecimal

¡Hola!
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hi,
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Convertion from Decimal to Hexadecimal

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