Re: Quick basic C question!
Hello,
I thankyou all for your feedback on my post... much appreciated....
As for my homework which Allen Carre has assigned to me, here it is.
Even though the results are precise and the snippet is functional, since I
have compared the results with Windows Calculator, I beleive this is *not*
the best way of doing it. :-)
I am sure there is a much easier way! Sorry If I disapointed you Allen, but
this is all I could come up with right now:
////////////////////////////////////////////////////////////////////
#include <stdio.h>
#include <iostream>
#include <math.h>
using namespace std;
void main(void)
{
char x[9];
int n[8],bin[8],Stop,i,decimal=0,z=0,h,BitPos,Totalize=0;
bool k;
double base = 2, ii=0, NUM=0;
printf("Enter a BCD\HEX # !");
cin >> x;
for(i=0;i<8;i++)
n[i] = x[i];
for(i=0;i<8;i++)
{
if(n[i] >= 0x30 && n[i] <= 0x39)
{bin[i] = n[i] - 48;}
else if (n[i] >= 0x41 && n[i] <= 0x46)
{bin[i] = n[i] - 55;}
}
for(i=9;i>1;i--)
{
NUM = 0;
if(bin[i-2])
{
for(h=0;h<4;h++)
{
BitPos=pow(2,h);
k = BitPos & bin[i-2];
if(k)
NUM = NUM + pow(base,z);
z++;
}
}
Totalize = Totalize+NUM;
}
printf("The Decimal equivalent is: %u\n", Totalize);
cin >> Stop;
}
////////////////////////////////////////////////////////////////////////
I based myself on using the binary weight of every bit and assigning it its
respective power of two calculation!
I don't know of any other way, I am sure C must have a quick function to do
all this, however I did not find one!
When I needed to do stuff like this I always look at chart which shows the
the Decimal equivalents.. I don't know, maybe its a bad habit....
Anyhow, its been fun....
Thankyou all for you information regarding the (n/16) stuff, it makes sence
and understand it now....
You all have a nice day!
--
Best regards
Robert
"Alan Carre" wrote:
"Robby" <Robby@discussions.microsoft.com> wrote in message
news:1944D700-226A-4C49-AAD1-4C47A91025C1@microsoft.com...
Hello,
I am reviewing some C code when I have come across the following piece of
code, please take a look:
/////////////////////////////////////////////////////////////
void main()
{
int hour;
display_bcd(hour);
}
void display_bcd( byte n)
{
putc(n/16+'0');
putc(n%16)+'0');
}
///////////////////////////////////////////////////////////////////////////////
Regarding the second line of code of the display_bcd function, If I
understand right, say n = 10, so,
10%16 = 10 + (ascii value of 0, which is 48) sent through putc we get
the following character ":" right?
However I don't understand the first line: putc(n/16+'0'); ???
Why is he dividing a byte value by 16?
What happens to a byte value when divide by a number anyways?
I looked around in a few C books but could not find any samples resembling
this.
All feedback sincerely appreciated!
Thanks
BCD stands for "Binary Coded Decimal" which really means "Hexadecimal that
looks like decimal". So, for instance, a number like 0x12 (hex) 'looks like'
the decimal number "12" decimal. So it is called the BCD value of 12
(decimal). So how do we do the conversion?
First we want the '10's' value which in hexadecimal is really the '16's'
value (the number of 16's in the number). To get that we divide by 16 (using
integer division the remainder is thrown away).
Tens value: 0x12 / 16 == 0x12 / 0x10 == 1.
To get the 1's value we simply take the remainder when dividing by 16
(0x10):
One's value: 0x12 % 16 == "remainder of" 0x12 / 0x10 == 2.
Summing up:
putc (1 + '0'); // prints a 1
putc (2 + '0') ; // prints a 2
Total Printout == "12".
- Alan Carre.
Note that the given procedure ONLY works for BYTE values. It will not work
for shorts or longs etc.
Homework: Make a function that prints out the decimal value for any BCD
number up to 0x99999999 (hint: no divisions or mods ('%'s) are required!).