'Clipping a binary number to required length C/C++
I have written a short function to convert an input decimal number to a binary output. However, at a much higher level of the code, the end user should toggle an option as to whether or not they desire a 5B or 10B value. For the sake of some other low level maths, I have to clip the data here.
So I need some help figuring out how to clip the output to a desired length and stuff the required number of leading zeros.
The incomplete C code:
long dec2bin(int x_dec,int res)
{
long x_bin = 0;
int x_bin_len;
int x_rem, i = 1;
while (x_dec != 0)
{
x_rem = x_dec % 2;
x_dec /= 2;
x_bin += x_rem * i;
i *= 10;
}
return x_bin;
}
I had completed a working proof of concept using python. The end application however, requires I write this in C.
The working python script:
def dec2bin(x_dec,x_res):
x_bin = bin(x_dec)[2:] #Convert to Binary (Remove 0B Prefix)
x_len = len(x_bin)
if x_len < x_res: #If Smaller than desired resolution
x_bin = '0' * (x_res-x_len) + x_bin #Stuff with leading 0s
if x_len > x_res: #If larger than desired resolution
x_bin = x_bin[x_len-x_res:x_len] #Display desired no. LSBs
return x_bin
I'm sure this has been done before, Indeed, my python script proves it should be relatively straightforward, but I'm not as experienced with C.
Any help is greatly appreciated.
- Mark.
Solution 1:[1]
As @yano suggested, I think you have to return an ascii string to the caller, rather than a long. Below's the short function I wrote for my own purposes, for any base...
char *itoa ( int i, int base, int ndigits ) {
static char a[999], digits[99] = /* up to base 65 */
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz#$*";
int n=ndigits;
memset(a,'0',ndigits); a[ndigits]='\000';
while ( --n >= 0) {
a[n] = digits[i%base];
if ( (i/=base) < 1 ) break; }
return ( a );
} /* --- end-of-function itoa() --- */
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | eigengrau |
