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Lets assume I have an input array like below
int input_arr[100] = {10,20,1255,1200,50,55,1,5,6,1000};
Here to store each elements of array it took 32 bits even though value of array elements is very small i.e 1255 is the maximum elements in array and to store that I need only 11 bit that means in 11 bit I can fit all other elements of array.
So my task to compress 32-bit elements of array into 11-bit array elements ? Expected compressed array looks like
int output_arr[] = {00000001010 00000010100 .... 10011111111 ... }
| | |
11 bits(1) 11 bits(2) 11 bits( 1255)
To do the above task what I did is here
find the maximum elements in given array
find the bits required to store maximum elements(previous step)
find bytes required to store no of bits for e.g to store 11 bits I need equivalent 2 bytes(in below code new_size contains this). Here is I need your help. Here is the memory wastage as told by my manager because to store 11 bits my new_size is 2 bytes i.e 5 bits are still extra or wastage. How can I avoid this.
Here is what I tried
int my_pow(int input_num,int p) {
int temp = 1;
for(int iter = 0;iter < p; iter++) {
temp = temp * input_num;
}
return temp;
}
int main() {
#if 0
int input_array[53069] = {1,2,2,3,4,1,2,4,6,1255,1,2,5,1233};
#endif
int input_array[] = {1,2,3,4,6,1255,1,2,5,1233};
int max = input_array[0], ele = sizeof(input_array)/sizeof(input_array[0]);
/* finding max elements in a array */
for(int i = 0;i < ele; i++) {
if(input_array[i] > max) {
max = input_array[i];
}
}
/* finding no of bits required to store highest elements of array */
int bit_required = 0;
while(1) {
if(max < my_pow(2,bit_required))
break;
bit_required+=1;
}
/* when above loop fails bit_required is nothing
but no of bit required to store the highest element of array */
/* finding size of new/compressed array */
int new_size = 0;
if(bit_required % 8 == 0) {
new_size = bit_required/8;
}
else {
new_size = (bit_required/8) + 1;
}
/* construct the new array again */
typedef struct array_task {
unsigned char new_array[new_size];/* in each cmp_arr, can store new_size char
now for each B[] I'm not using 32 bits , its new_size bits */
}cmp_arr;/* creating new array of ele elements */
cmp_arr cmpressed[ele];
/* store elements of input_array[] into output_array[] */
for(int row = 0 ; row < ele ;row++) {
for(int col = bit_required - 1; col >= 0; col-- ) {
cmpressed[row].new_array[col] = ((input_array[row] >> col & 1) + 48) ;
printf("%d",(cmpressed[row].new_array[col]) - 48);
}
printf("\n");
}
#if 0
printf("Size of A before %d\n",sizeof(input_array)); /* 40 bytes */
printf("size of compressed array %d\n",sizeof(cmp_arr));/* same task, it perform in 2 bytes,
each elements won't take 32 bits */
#endif
return 0;
}
Is there any other way to do the same task efficiently ? All suggestion are most welcome ?
To put values shifted by 11 bits instead of 8, 16 or 32 will require manipulations with bits. You will basically have to emulate an array of bits in an array of (say 32 bits) integers. In this case if a value is stored at a bit offset X it will be (possibly) stored in your array somewhere on indexes X/32 and X/32+1 (if it is crossing border of 32 bits). Each time when you have to set a value into the array you have to load those two values and "place" your number there. The implementation is a bit technical, try the following code:
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#define MASK32 ((uint64_t)0xffffffff)
void setValue(uint32_t *a, int bits, int i, int n) {
int bitoffset = i * bits;
int index = bitoffset / 32;
int shift = bitoffset % 32;
uint64_t maskbits = (~(uint64_t)0) >> (64-bits);
uint64_t val = ((uint64_t)a[index+1]<<32) + a[index];
val = val & ~(maskbits << shift) | ((n & maskbits) << shift);
a[index] = (val & MASK32);
a[index+1] = (val >> 32) & MASK32;
}
int getValue(const uint32_t *a, int bits, int i) {
int bitoffset = i * bits;
int index = bitoffset / 32;
int shift = bitoffset % 32;
uint64_t maskbits = (~(uint64_t)0) >> (64-bits);
int val = ((((uint64_t)a[index+1]<<32) + a[index]) >> shift) & maskbits;
return(val);
}
int input_arr[100] = {10,20,1255,1200,50,55,1,5,6,1000};
int main() {
int i, j;
uint32_t a[100*11/32+2];
for(i=0; i<100; i++) setValue(a,11,i,input_arr[i]);
for(j=0; j<100; j++) printf("a[%d/11] == %d\n", j, getValue(a,11,j));
}
Another approach that I find "interesting" is allocating an array of chars and doing a cast to an type that fits the maximum value. Something like this:
NumBytesMaxValue = ...;
void* pointers = malloc(NumBytesMaxValue * NumValues);
if (NumBytesMaxValue == 1)
cast_pointer_to_char_and_fill_it();
else if (NumBytesMaxValue == 2)
cast_pointer_to_short_and_fill_it();
...
Data compression is a vast subject, an active area of research... compressing your data can be done in so many different ways as to make it off topic.
Finding the smallest type for the array can however be done by a utility program or a preliminary phase:
#include <limits.h>
#include <stdio.h>
int main() {
int input_array[] = { 1, 2, 2, 3, 4, 1, 2, 4, 6, 1255, 1, 2, 5, 1233 };
size_t i, count = sizeof(input_array) / sizeof(input_array[0]);
int min, max;
int nc = 0;
min = max = input_array[0];
for (i = 1; i < count; i++) {
if (min > input_array[i]) min = intput_array[i];
if (max < input_array[i]) max = intput_array[i];
}
printf("min value is %d, max value is %d\n", min, max);
if (min >= SCHAR_MIN && max <= SCHAR_MAX)
nc += printf("type signed char is appropriate\n");
if (min >= 0 && max <= UCHAR_MAX)
nc += printf("type unsigned char is appropriate\n");
if (min >= SHRT_MIN && max <= SHRT_MAX)
nc += printf("type short is appropriate\n");
if (min >= 0 && max <= USHRT_MAX)
nc += printf("type unsigned short is appropriate\n");
if (nc == 0)
printf("no type smaller than int is appropriate\n");
return 0;
}
You can use the same approach for a set of numbers with values unknown at compile time with these steps:
start with an allocated array of a small type such as signed char.
read the next value: if it fits in the current type, add it to the array and continue.
if not, allocate an array of a larger type such as short, copy the values parsed so far into it, free the previous array, store the new value and continue.
if the new value does not fit in a short, use a larger type such as int.
you could write code for even larger types such as long and long long, but you need specific code for each type.
at the end of the read phase, you have an array of the smallest type that handles all the values in the dataset. Handle this array with code for its specific type. This means you have to duplicate the processing code for each type, which can be tricky.
Suppose I have an array of bytes from a secure PRNG, and I need to generate a number between 1 and 10 using that data, how would I do that correctly?
Think of the array as one big unsigned integer. Then the answer is simple:
(Big_Number % 10) + 1
So all that is needed is a method to find the modulus 10 of big integers. Using modular exponentiation:
#include <limits.h>
#include <stdlib.h>
int ArrayMod10(const unsigned char *a, size_t n) {
int mod10 = 0;
int base = (UCHAR_MAX + 1) % 10;
for (size_t i = n; i-- > 0; ) {
mod10 = (base*mod10 + a[i]) % 10;
base = (base * base) % 10;
}
return mod10;
}
void test10(size_t n) {
unsigned char a[n];
// fill array with your secure PRNG
for (size_t i = 0; i<n; i++) a[i] = rand();
return ArrayMod10(a, n) + 1;
}
There will be a slight bias as 256^n is not a power of 10. With large n, this will rapidly decrease in significance.
Untested code: Detect if a biased result occurred. Calling code could repeatedly call this function with new a array values to get an unbiased result on the rare occasions when bias occurs.
int ArrayMod10BiasDetect(const unsigned char *a, size_t n, bool *biasptr) {
bool bias = true;
int mod10 = 0;
int base = (UCHAR_MAX + 1) % 10; // Note base is usually 6: 256%10, 65536%10, etc.
for (size_t i = n; i-- > 0; ) {
mod10 = (base*mod10 + a[i]) % 10;
if (n > 0) {
if (a[i] < UCHAR_MAX) bias = false;
} else {
if (a[i] < UCHAR_MAX + 1 - base) bias = false;
}
base = (base * base) % 10;
}
*biaseptr = bias;
return mod10;
}
As per the comments follow-up, it seems what you need is modulus operator [%].
You may also need to check the related wiki.
Note: Every time we use the modulo operator on a random number, there is a probability that we'll be running into modulo bias, which ends up in disbalancing the fair distribution of random numbers. You've to take care of that.
For a detailed discussion on this, please see this question and related answers.
It depends on a bunch of things. Secure PRNG sometimes makes long byte arrays instead of integers, let's say it is 16 bytes long array, then extract 32 bit integer like so: buf[0]*0x1000000+buf[1]*0x10000+buf[2]*0x100+buf[3] or use shift operator. This is random so big-endian/little-endian doesn't matter.
char randbytes[16];
//...
const char *p = randbytes;
//assumes size of int is 4
unsigned int rand1 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand2 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand3 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3]; p += 4;
unsigned int rand4 = p[0] << 24 + p[1] << 16 + p[2] << 8 + p[3];
Then use % on the integer
ps, I think that's a long answer. If you want number between 1 and 10 then just use % on first byte.
OK, so this answer is in Java until I get to my Eclipse C/C++ IDE:
public final static int simpleBound(Random rbg, int n) {
final int BYTE_VALUES = 256;
// sanity check, only return positive numbers
if (n <= 0) {
throw new IllegalArgumentException("Oops");
}
// sanity check: choice of value 0 or 0...
if (n == 1) {
return 0;
}
// sanity check: does not fit in byte
if (n > BYTE_VALUES) {
throw new IllegalArgumentException("Oops");
}
// optimization for n = 2^y
if (Integer.bitCount(n) == 1) {
final int mask = n - 1;
return retrieveRandomByte(rbg) & mask;
}
// you can skip to this if you are sure n = 10
// z is upper bound, and contains floor(z / n) blocks of n values
final int z = (BYTE_VALUES / n) * n;
int x;
do {
x = retrieveRandomByte(rbg);
} while (x >= z);
return x % n;
}
So n is the maximum value in a range [0..n), i.e. n is exclusive. For a range [1..10] simply increase the result with 1.
suppose I have n1 and n2 I want to multiply them
for example I have array
n1={1,2,3};
and in
n2={5,6}
they are two integers in n1 we have the 123 and in n2 56
123*56=6888
then in result I should have
result = {6,8,8,8}
here is the incomplete algorithm which I thought
for(i in n1 bigger array)
for(j in n2 smaller one)
{
mult=n1[i]*n2[j]
mult+= carry;
if(mult>=10)
{
carry = (mult/10);
mult-= (carry*10);
}
}
}
How can I write it? I don't know the place of store
after finishing the insider loop I should store num in array and then compute again and...
How should I write it? I searched the whole of overflow here but I didn't find about it in c code
The Goal is to Compute the Large numbers Integer Numbers has 8 Bytes,in other words 64 bits so they can store 2pow64-1 which is 19 digits now this will help to compute very larger than 19 digits
It would be slightly easier if your digit-arrays were little-endian. Then your example multiplication would look
3 2 1 * 6 5
---------------
18 12 6
15 10 5
---------------
18 27 16 5 // now propagate carries
8 28 16 5
8 8 18 5
8 8 8 6
============
The product of n1[i] and n2[j] would contribute to result[i+j]. The main loop could roughly look like
for (i = 0; i < l1; ++i) // l1 is length of n1
{
for (j = 0; j < l2; ++j) // l2 is length of n2
{
result[i+j] += n1[i]*n2[j];
}
}
// now carry propagation
You see that the result must be at least (l1-1) + (l2-1) + 1 long, since the product of the most significant digits goes int result[(l1-1) + (l2-1)]. On the other hand, n1 < 10^l1 and n2 < 10^l2, so the product is < 10^(l1+l2) and you need at most l1+l2 digits.
But if you're working with char (signed or unsigned), that will quickly overflow in each digit, since (for k <= min(l1-1,l2-1)) k+1 products of two digits (each can be as large as 81) contribute to digit k of the product.
So it's better to perform the multiplication grouped according to the result digit, accumulating in a larger type, and doing carry propagation on writing the result digit. With little-endian numbers
char *mult(char *n1, size_t l1, char *n2, size_t l2, size_t *rl)
{
// allocate and zero-initialise, may be one more digit than needed
char *result = calloc(l1+l2+1,1);
*rl = l1 + l2;
size_t k, i, lim = l1+l2-1;
for (k = 0; k < lim; ++k)
{
unsigned long accum = result[k];
for (i = (k < l2) ? 0 : k-(l2-1); i <= k && i < l1; ++i)
{
accum += (n1[i] - '0') * (n2[k-i] - '0');
}
result[k] = accum % 10 + '0';
accum /= 10;
i = k+1;
while(accum > 0)
{
result[i] += accum % 10;
accum /= 10;
++i;
}
}
if (result[l1+l2-1] == 0)
{
*rl -= 1;
char *real_result = calloc(l1+l2,1);
for (i = 0; i < l1+l2-1; ++i)
{
real_result[i] = result[i];
}
free(result);
return real_result;
}
else
{
result[l1+l2-1] += '0';
return result;
}
}
For big-endian numbers, the indexing has to be modified - you can figure that out yourself, hopefully - but the principle remains the same.
Indeed, the result isn't much different after tracking indices with pencil and paper:
char *mult(char *n1, size_t l1, char *n2, size_t l2, size_t *rl)
{
// allocate and zero-initialise, may be one more digit than needed
// we need (l1+l2-1) or (l1+l2) digits for the product and a 0-terminator
char *result = calloc(l1+l2+1,1);
*rl = l1 + l2;
size_t k, i, lim = l1+l2-1;
// calculate the product from least significant digit to
// most significant, least significant goes into result[l1+l2-1],
// the digit result[0] can only be nonzero by carry propagation.
for (k = lim; k > 0; --k)
{
unsigned long accum = result[k]; // start with carry
for (i = (k < l2) ? 0 : k-l2; i < k && i < l1; ++i)
{
accum += (n1[i] - '0') * (n2[k-1-i] - '0');
}
result[k] = accum % 10 + '0';
accum /= 10;
i = k-1;
while(accum > 0)
{
result[i] += accum % 10;
accum /= 10;
--i;
}
}
if (result[0] == 0) // no carry in digit 0, we allocated too much
{
*rl -= 1;
char *real_result = calloc(l1+l2,1);
for (i = 0; i < l1+l2-1; ++i)
{
real_result[i] = result[i+1];
}
free(result);
return real_result;
}
else
{
result[0] += '0'; // make it an ASCII digit
return result;
}
}
Edit: added 0-terminators
Note: these are not NUL-terminated (unsigned) char arrays, so we need to keep length information (that's good to do anyway), hence it would be better to store that info together with the digit array in a struct. Also, as written it only works for positive numbers. Dealing with negative numbers is awkward if you only have raw arrays, so another point for storing additional info.
Keeping the digits as '0' + value doesn't make sense for the computations, it is only convenient for printing, but that only if they were NUL-terminated arrays. You may want to add a slot for the NUL-terminator then. In that case, the parameter rl in which we store the length of the product is not strictly necessary.
Definitely an interesting problem.
Here was my thought:
For the given array, append each value to the end of a string. Thus you construct a string of the numbers in order. {1,2,3} = "123"
Then, you use a "ToInteger" method that you can find in one of the C libraries. Now you have your number to multiply with.
With this logic, you can probably look up how the "ToInteger" or "ToString" methods work with numbers, which would lead to an answer.
Think how you would do it on paper, since you are simulating multiplying two decimal numbers. For starters, I think you'd go from least significant to most significant digit, so you'd be counting down the indexes (2, 1, 0 for the larger array; 1, 0 for the smaller). Also, you'd somehow have to arrange that when you multiply by n2[0] (the 5 in 56), you start adding at the tens place, not the units.
You won't find complete C code for your problem at SO. Your first approach isn't that bad. You could do the following:
Multiply n1 and n2, conversion is done by mulitplication and addition, i. e. a{1,2,3} -> 1*100 + 2*10 + 3*1, easy to implement
Count the digits of your multiplication result (use division inside a loop)
While looping through the digits you can store them back into another array
If you can't or if you don't want to deal with dynamic array allocation, then think about how big your array for storage must be beforehand and perform a static allocation.
Edit
Based on the discussion another approach:
Suppose, that r = n1 * n2
Create a n*m 2D array, where
n = number of digits in n2
m = number of digits in n1 + 1
Within a loop multiply each digit of n1 with one of the elements of n2, store the result in the array, store the result per-digit in the 2D-array, don't forget to add the carry to each digit
Repeat 2 with all other digits of n2
Now the array is filled and you'll have to add each digits like you would do it on paper, store each result within a target array, take care of the carry again
There is one thing left in the algorithm: Determine the size of the target array, based on the informations within the intermediate array, you can think about this by using pencil and paper ;)
This code isn't optimized, nor does it account for generic lengths of arrays/numbers, but it should give you the general idea of how to implement the algorithm:
(This is similar to string-to-int or int-to-string algorithms, just add the ASCII offset to each item of the array and you have it.)
#include <stdio.h>
#include <stdint.h>
#define N1_N 3
#define N2_N 2
#define MAX_N 4 /* maximum array length allowed */
void print_array (const uint8_t* array, size_t size);
uint32_t array_to_ulong (const uint8_t* array, size_t size);
size_t ulong_to_array (uint8_t* array, size_t size, uint32_t val);
int main()
{
uint8_t n1[N1_N] = {1,2,3};
uint8_t n2[N2_N] = {5,6};
uint8_t n3[MAX_N];
size_t n3_size = MAX_N;
uint32_t n1_int;
uint32_t n2_int;
uint32_t result;
print_array(n1, N1_N);
printf(" * ");
print_array(n2, N2_N);
n1_int = array_to_ulong (n1, N1_N);
n2_int = array_to_ulong (n2, N2_N);
result = n1_int * n2_int;
printf(" = %d = ", result);
n3_size = ulong_to_array (n3, n3_size, result);
print_array(n3, n3_size);
getchar();
return 0;
}
void print_array (const uint8_t* array, size_t size)
{
size_t i;
printf("{");
for(i=0; i<size; i++)
{
printf("%d", array[i]);
if(i != size-1)
{
printf(", ");
}
}
printf("}");
}
uint32_t array_to_ulong (const uint8_t* array, size_t size)
{
uint32_t result = 0;
uint32_t multiplier = 1;
size_t i;
for(i=1; i<=size; i++)
{
result += array[size-i] * multiplier;
multiplier *= 10;
}
return result;
}
size_t ulong_to_array (uint8_t* array, size_t size, uint32_t val)
{
size_t i;
for(i=1; i<=size && val!=0; i++)
{
array[size-i] = val % 10;
val /= 10;
}
return i-1;
}
12345 * 6789 is:
12345 * 6 * 1000 +
12345 * 7 * 100 +
12345 * 8 * 10 +
12345 * 9 * 1
and that is:
1 * 6*1000 * 10000 + 2 * 6*1000 * 1000 + 3 * 6*1000 * 100 + 4 * 6*1000 * 10 + 5 * 6*1000 * 1 +
1 * 7*100 * 10000 + 2 * 7*100 * 1000 + 3 * 7*100 * 100 + 4 * 7*100 * 10 + 5 * 7*100 * 1 +
1 * 8*10 * 10000 + 2 * 8*10 * 1000 + 3 * 8*10 * 100 + 4 * 8*10 * 10 + 5 * 8*10 * 1 +
1 * 9*1 * 10000 + 2 * 9*1 * 1000 + 3 * 9*1 * 100 + 4 * 9*1 * 10 + 5 * 9*1 * 1
so the algorith is multiply each value by each value and add (cumulate) it to the appropriate result array element (1000 is 10^3 so array element 3 (array starting by zero)).
then move thru the result array and shift for results bigger than 10 the div by ten to the left (starting by the far right)
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#define MAX 10000
char * multiply(char [],char[]);
int main(){
char a[MAX];
char b[MAX];
char *c;
int la,lb;
int i;
printf("Enter the first number : ");
scanf("%s",a);
printf("Enter the second number : ");
scanf("%s",b);
printf("Multiplication of two numbers : ");
c = multiply(a,b);
printf("%s",c);
return 0;
}
char * multiply(char a[],char b[]){
static char mul[MAX];
char c[MAX];
char temp[MAX];
int la,lb;
int i,j,k=0,x=0,y;
long int r=0;
long sum = 0;
la=strlen(a)-1;
lb=strlen(b)-1;
for(i=0;i<=la;i++){
a[i] = a[i] - 48;
}
for(i=0;i<=lb;i++){
b[i] = b[i] - 48;
}
for(i=lb;i>=0;i--){
r=0;
for(j=la;j>=0;j--){
temp[k++] = (b[i]*a[j] + r)%10;
r = (b[i]*a[j]+r)/10;
}
temp[k++] = r;
x++;
for(y = 0;y<x;y++){
temp[k++] = 0;
}
}
k=0;
r=0;
for(i=0;i<la+lb+2;i++){
sum =0;
y=0;
for(j=1;j<=lb+1;j++){
if(i <= la+j){
sum = sum + temp[y+i];
}
y += j + la + 1;
}
c[k++] = (sum+r) %10;
r = (sum+r)/10;
}
c[k] = r;
j=0;
for(i=k-1;i>=0;i--){
mul[j++]=c[i] + 48;
}
mul[j]='\0';
return mul;
}
I've just started learning C and I'm having some problems with some code I want to write.
Basically I have this struct that is a bit array, with the number of bits in the array, and a pointer to a buffer of chars, that stores the bits.
My strategy for rotating the bit array is simply taking the number of rotations (mod the length to avoid full rotations) and using a simple reversal algorithm to rotate the array.
EDIT:
However, my problem is that I want to rotate the bits in the actual buffer.
I also want to be able to rotate a subsequence of bits within the entire bit array. So for 1101101, I might want to rotate (0-indexed from the left) the subsequence starting at index 2 and ending at index 5. I'm not entirely sure how to use my char buffer to do this.
Thanks for the help!
struct arrayBits{
size_t numBits;
char *buf;
}
The buf array holds 8-bit integers, not bools as I previously mentioned.
The way that I can access and set an individual bit is just by indexing into the byte that holds the bit I want (so for an array ab, ab->buf[index_of_desired_bit/8] and then performing some bitwise operations on it to change the value, for performance reasons.
EDIT: Thanks to everyone for all the suggestions. I've looked at all of them and I believe I understand the code better. Here's the code I ended up writing, however, I think there are some problems with it.
While it passes some of my basic test cases, it seems to run a little too fast on an bitarray of size 98775 bits, randomly filled. By this I mean, is there some case in which my code just outright fails and crashes? The test cases do three rotations, in a row, on the full 98775-bit array. One rotation of -98775/4 (<--this is a size_t, so wrap around?), one rotation of 98775/4, and then a final rotation of 98775/2.
Is there something I'm missing or some problem I'm not seeing?
/*Reverse a bit array*/
/*v1.1: basic bit reversal w/o temp variable*/
static void arrayReversal(bitarray_t *ba, size_t begin, size_t end){
while(begin < end)
{
bitarray_set(ba, begin, (bitarray_get(ba, begin) ^ bitarray_get(ba, end))); /*x = x ^ y*/
bitarray_set(ba, end, (bitarray_get(ba, begin) ^ bitarray_get(ba, end))); /*y = x ^ y*/
bitarray_set(ba, begin, (bitarray_get(ba, begin) ^ bitarray_get(ba, end))); /*x = x ^ y*/
begin++;
end--;
}
}
/*Main Rotation Routine*/
void bitarray_rotate(bitarray_t *ba, size_t bit_off, size_t bit_len, ssize_t bit_right_amount) {
assert(bit_off + bit_len <= ba->bit_sz);
assert(bit_off + bit_len > 0);
if(bit_off + bit_len > ba->bit_sz || bit_off + bit_len < 0)
{
printf("\nError: Indices out of bounds\n");
return;
}
/*Find index to split bitarray on*/
if(bit_len == 0) return; //Rotate only 1 bit i.e. no rotation
size_t reversal_index;
reversal_index = modulo(-bit_right_amount, bit_len);
if(reversal_index == 0) return; //No rotation to do
/*3 bit string reversals*/
assert(reversal_index - 1 + bit_off < ba->bit_sz);
/* Reverse A*/
arrayReversal(ba, bit_off, reversal_index - 1 + bit_off);
assert(reversal_index + bit_off < ba->bit_sz);
/*Reverse B*/
arrayReversal(ba, reversal_index + bit_off, (bit_off + bit_len - 1));
/*Reverse ArBr*/
arrayReversal(ba, bit_off, (bit_off + bit_len -1));
}
Well the easy way to start is to consider how to rotate the bits in a single value. Let's say that you have x, which is an N-bit value and you want to rotate it by k places. (I'm only going to look at rotating upwards/left, it is easy to convert to downwards/right). The first thing to observe is that if k=N then x is unchanged. So before rotating we want to reduce k modulo N to throw away complete rotations.
Next we should observe that during the rotation the k upper-bits will move to the bottom of the value, and the lower N-k bits will move up k places. This is the same as saying that the top k-bits move down N-k places. The reason that we phrase it this way is that C has shift operators, but not rotation.
In psuedo-C we can say:
#define N sizeof(type)*8
type rotate(type x, int k) {
type lower = x & ((1 << (N-k)) - 1);
type upper = x >> (N-k) & ((1 <<k)-1);
return upper | lower;
}
This takes care of the simple atomic case, simply replace type with char or int as appropriate. If type is unsigned then the mask on the value of upper is unnecessary.
The next thing to consider is rotating in an array of values. If you think of the above code as glueing together two halves of a value then for the more complicated case we need to glue together upper and lower parts from different places in the array. If k is small then these places are adjacent in the array, but when k>N we are rotating through more than one intermediate word.
In particular if we are rotating up k places then we are moving bits from k/N words away in the array, and the N bits can span floor(k/N) and ceil(k/N) locations away in the array. Ok, so now we're ready to put it all together. For each word in the array the new upper N-(k mod N) bits will be the lower bits of floor(k/N) words away, and the new lower (k mod N) bits will be the upper bits of ceil(k/N) words away.
In the same psuedo-C (i.e replace type with what you are using) we can say:
#define N sizeof(type)*8
#define ARR_SIZE ...
type rotate(type *x, int k,type *out) {
int r = k % N;
int upperOff = k/N;
int lowerOff = (k+N-1)/N;
for(int i=0; i<ARR_SIZE; i++) {
int lowerPos = (i + ARR_SIZE - lowerOff) % ARR_SIZE
int upperPos = (i + ARR_SIZE - upperOff) % ARR_SIZE
type lower = x[lowerPos] & ((1 << (N-k)) - 1)
type upper = x[upperPos] >> (N-k) & ((1 <<k)-1)
out[i] = upper | lower;
}
}
Anyway, that's a lot more than I was intending to write so I'll quit now. It should be easy enough to convert this to a form that works inplace on a single array, but you'll probably want to fix the types and the range of k first in order to bound the temporary storage.
If you have any more problems in this area then one place to look is bitmap sprite graphics. For example this rotation problem was used to implement scrolling many, many moons ago in 8-bit games.
I would suggest a pointer/offset to a starting point of a bit in the buffer instead of rotating. Feel free to overload any operator that might be useful, operator[] comes to mind.
A rotate(n) would simply be a offset+=n operation. But I find the purpose of your comment about -"However, my problem is that I want to rotate the actual buffer" confusing.
You dont need an extra buffer for rotate (only for output).
You should implement a function for one rotate and loop this, eg: (right-shift variation)
char *itoa2(char *s,size_t i)
{
*s=0;
do {
memmove(s+1,s,strlen(s)+1);
*s='0'+(i&1);
} while( i>>=1 );
return s;
}
size_t bitrotateOne(size_t i)
{
return i>>1 | (i&1) << (sizeof i<<3)-1;
}
...
size_t i=12,num=17;
char b[129];
while( num-- )
{
i = bitrotateOne(i);
puts( itoa2(b,i) );
}
Since your criteria is so complex, I think the easiest way to do it would be to step through each bit and set where it would be in your new array. You could speed it up for some operations by copying a whole character if it is outside the shifted bits, but I can't think of how to reliably do shifting taking into account all the variables because the start and end of the shifted sequence can be in the middle of bytes and so can the end of the entire bits. The key is to get the new bit position for a bit in the old array:
j = (i < startBit || i >= startBit + length) ? i :
((i - startBit + shiftRightCount) % length) + startBit;
Code:
#include "stdafx.h"
#include <stdlib.h>
#include <string.h>
typedef struct {
size_t numBits;
unsigned char *buf;
} ARRAYBITS;
// format is big endian, shiftint left 8 bits will shift all bytes to a lower index
ARRAYBITS rotateBits(ARRAYBITS *pOriginalBits, int startBit, int length, int shiftRightCount);
void setBit(unsigned char *buf, int bit, bool isSet);
bool checkBit(unsigned char *buf, int bit);
ARRAYBITS fromString(char *onesAndZeros);
char *toString(ARRAYBITS *pBits);
int _tmain(int argc, _TCHAR* argv[])
{
char input[1024];
ARRAYBITS bits = fromString("11110000110010101110"); // 20 bits
ARRAYBITS bitsA = rotateBits(&bits, 0, bits.numBits, 1);
ARRAYBITS bitsB = rotateBits(&bits, 0, bits.numBits, -1);
ARRAYBITS bitsC = rotateBits(&bits, 6, 8, 4);
ARRAYBITS bitsD = rotateBits(&bits, 6, 8, -2);
ARRAYBITS bitsE = rotateBits(&bits, 6, 8, 31);
ARRAYBITS bitsF = rotateBits(&bits, 6, 8, -31);
printf("Starting : %s\n", toString(&bits));
printf("All right 1: %s\n", toString(&bitsA));
printf("All left 1 : %s\n", toString(&bitsB));
printf("\n");
printf(" : ********\n");
printf("Starting : %s\n", toString(&bits));
printf("6,8,4 : %s\n", toString(&bitsC));
printf("6,8,-2 : %s\n", toString(&bitsD));
printf("6,8,31 : %s\n", toString(&bitsE));
printf("6,8,-31 : %s\n", toString(&bitsF));
gets(input);
}
ARRAYBITS rotateBits(ARRAYBITS *pOriginalBits, int startBit, int length, int shiftRightCount)
{
// 0-8 == 1, 9-16 == 2, 17-24 == 3
ARRAYBITS newBits;
int i = 0, j = 0;
int bytes = 0;
while (shiftRightCount < 0)
shiftRightCount += length;
shiftRightCount = shiftRightCount % length;
newBits.numBits = pOriginalBits->numBits;
if (pOriginalBits->numBits <= 0)
return newBits;
bytes = ((pOriginalBits->numBits -1) / 8) + 1;
newBits.buf = (unsigned char *)malloc(bytes);
memset(newBits.buf, 0, bytes);
for (i = 0; i < pOriginalBits->numBits; i++) {
j = (i < startBit || i >= startBit + length) ? i : ((i - startBit + shiftRightCount) % length) + startBit;
if (checkBit(pOriginalBits->buf, i))
{
setBit(newBits.buf, j, true);
}
}
return newBits;
}
void setBit(unsigned char *buf, int bit, bool isSet)
{
int charIndex = bit / 8;
unsigned char c = 1 << (bit & 0x07);
if (isSet)
buf[charIndex] |= c;
else
buf[charIndex] &= (c ^ 255);
}
bool checkBit(unsigned char *buf, int bit)
{
// address of char is (bit / 8), bit within char is (bit & 7)
int index = bit / 8;
int b = bit & 7;
int value = 1 << b;
return ((buf[index] & value) > 0);
}
ARRAYBITS fromString(char *onesAndZeros)
{
int i;
ARRAYBITS bits;
int charCount;
bits.numBits = strlen(onesAndZeros);
charCount = ((bits.numBits -1) / 8) + 1;
bits.buf = (unsigned char *)malloc(charCount);
memset(bits.buf, 0, charCount);
for (i = 0; i < bits.numBits; i++)
{
if (onesAndZeros[i] != '0')
setBit(bits.buf, i, true);
}
return bits;
}
char *toString(ARRAYBITS *pBits)
{
char *buf = (char *)malloc(pBits->numBits + 1);
int i;
for (i = 0; i < pBits->numBits; i++)
{
buf[i] = checkBit(pBits->buf, i) ? '1' : '0';
}
buf[i] = 0;
return buf;
}
I suggest you use bit-level operations (>>,<<,~,&,|) rather than wasting space using int. Even so, using an int array, to rotate, pass the left & right index of substring:
void rotate ( struct arrayBits a, int left , int right )
{
int i;
int first_bit;
if(*( a.buf + right ) == 1) first_bit = 1;
else first_bit = 0;
for( i = left+1 ; i <= right ; i++ )
{
*( a.buf + i )=*( a.buf + i - 1 );
}
*a.buf = first_bit;
}
Example:
If struct_array is 010101,
rotate (struct_array,0,5); => rotates whole string 1 int to right
o/p: 101010
rotate (struct_array,2,4); => rotates substring 1 int to right
o/p: 01 001 1
To reverse the bit array call the rotate() function on the substring, size_of_substring times.
I have an array of unsigned chars in c I am trying to print in base 10, and I am stuck. I think this will be better explained in code, so, given:
unsigned char n[3];
char[0] = 1;
char[1] = 2;
char[2] = 3;
I would like to print 197121.
This is trivial with small base 256 arrays. One can simply 1 * 256 ^ 0 + 2 * 256 ^ 1 + 3 * 256 ^ 2.
However, if my array was 100 bytes large, then this quickly becomes a problem. There is no integral type in C that is 100 bytes large, which is why I'm storing numbers in unsigned char arrays to begin with.
How am I supposed to efficiently print this number out in base 10?
I am a bit lost.
There's no easy way to do it using only the standard C library. You'll either have to write the function yourself (not recommended), or use an external library such as GMP.
For example, using GMP, you could do:
unsigned char n[100]; // number to print
mpz_t num;
mpz_import(num, 100, -1, 1, 0, 0, n); // convert byte array into GMP format
mpz_out_str(stdout, 10, num); // print num to stdout in base 10
mpz_clear(num); // free memory for num
When I saw this question, I purpose to solve it, but at that moment I was very busy.
This last weekend I've could gain some prize hours of free time so I considered my pending challenge.
First of all, I suggest you to considered above response. I never use GMP library but I'm sure that it's better solution than a handmade code.
Also, you could be interest to analyze code of bc calculator; it can works with big numbers and I used to test my own code.
Ok, if you are still interested in a code do it by yourself (only with support C language and Standard C library) may be I can give you something.
Before all, a little bit theory. In basic numeric theory (modular arithmetic level) theres is an algorithm that inspire me to arrive at one solution; Multiply and Power algorithm to solve a^N module m:
Result := 1;
for i := k until i = 0
if n_i = 1 then Result := (Result * a) mod m;
if i != 0 then Result := (Result * Result) mod m;
end for;
Where k is number of digits less one of N in binary representation, and n_i is i binary digit. For instance (N is exponent):
N = 44 -> 1 0 1 1 0 0
k = 5
n_5 = 1
n_4 = 0
n_3 = 1
n_2 = 1
n_1 = 0
n_0 = 0
When we make a module operation, as an integer division, we can lose part of the number, so we only have to modify algorithm to don't miss relevant data.
Here is my code (take care that it is an adhoc code, strong dependency of may computer arch. Basically I play with data length of C language so, be carefully because my data length could not be the same):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
enum { SHF = 31, BMASK = 0x1 << SHF, MODULE = 1000000000UL, LIMIT = 1024 };
unsigned int scaleBigNum(const unsigned short scale, const unsigned int lim, unsigned int *num);
unsigned int pow2BigNum(const unsigned int lim, unsigned int *nsrc, unsigned int *ndst);
unsigned int addBigNum(const unsigned int lim1, unsigned int *num1, const unsigned int lim2, unsigned int *num2);
unsigned int bigNum(const unsigned short int base, const unsigned int exp, unsigned int **num);
int main(void)
{
unsigned int *num, lim;
unsigned int *np, nplim;
int i, j;
for(i = 1; i < LIMIT; ++i)
{
lim = bigNum(i, i, &num);
printf("%i^%i == ", i, i);
for(j = lim - 1; j > -1; --j)
printf("%09u", num[j]);
printf("\n");
free(num);
}
return 0;
}
/*
bigNum: Compute number base^exp and store it in num array
#base: Base number
#exp: Exponent number
#num: Pointer to array where it stores big number
Return: Array length of result number
*/
unsigned int bigNum(const unsigned short int base, const unsigned int exp, unsigned int **num)
{
unsigned int m, lim, mem;
unsigned int *v, *w, *k;
//Note: mem has the exactly amount memory to allocate (dinamic memory version)
mem = ( (unsigned int) (exp * log10( (float) base ) / 9 ) ) + 3;
v = (unsigned int *) malloc( mem * sizeof(unsigned int) );
w = (unsigned int *) malloc( mem * sizeof(unsigned int) );
for(m = BMASK; ( (m & exp) == 0 ) && m; m >>= 1 ) ;
v[0] = (m) ? 1 : 0;
for(lim = 1; m > 1; m >>= 1)
{
if( exp & m )
lim = scaleBigNum(base, lim, v);
lim = pow2BigNum(lim, v, w);
k = v;
v = w;
w = k;
}
if(exp & 0x1)
lim = scaleBigNum(base, lim, v);
free(w);
*num = v;
return lim;
}
/*
scaleBigNum: Make an (num[] <- scale*num[]) big number operation
#scale: Scalar that multiply big number
#lim: Length of source big number
#num: Source big number (array of unsigned int). Update it with new big number value
Return: Array length of operation result
Warning: This method can write in an incorrect position if we don't previous reallocate num (if it's necessary). bigNum method do it for us
*/
unsigned int scaleBigNum(const unsigned short scale, const unsigned int lim, unsigned int *num)
{
unsigned int i;
unsigned long long int n, t;
for(n = 0, t = 0, i = 0; i < lim; ++i)
{
t = (n / MODULE);
n = ( (unsigned long long int) scale * num[i] );
num[i] = (n % MODULE) + t; // (n % MODULE) + t always will be smaller than MODULE
}
num[i] = (n / MODULE);
return ( (num[i]) ? lim + 1 : lim );
}
/*
pow2BigNum: Make a (dst[] <- src[] * src[]) big number operation
#lim: Length of source big number
#src: Source big number (array of unsigned int)
#dst: Destination big number (array of unsigned int)
Return: Array length of operation result
Warning: This method can write in an incorrect position if we don't previous reallocate num (if it's necessary). bigNum method do it for us
*/
unsigned int pow2BigNum(const unsigned int lim, unsigned int *src, unsigned int *dst)
{
unsigned int i, j;
unsigned long long int n, t;
unsigned int k, c;
for(c = 0, dst[0] = 0, i = 0; i < lim; ++i)
{
for(j = i, n = 0; j < lim; ++j)
{
n = ( (unsigned long long int) src[i] * src[j] );
k = i + j;
if(i != j)
{
t = 2 * (n % MODULE);
n = 2 * (n / MODULE);
// (i + j)
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + (t % MODULE);
++k; // (i + j + 1)
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + ( (t / MODULE) + (n % MODULE) );
++k; // (i + j + 2)
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + (n / MODULE);
}
else
{
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + (n % MODULE);
++k; // (i + j)
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + (n / MODULE);
}
for(k = i + j; k < (lim + j); ++k)
{
dst[k + 1] += (dst[k] / MODULE);
dst[k] %= MODULE;
}
}
}
i = lim << 1;
return ((dst[i - 1]) ? i : i - 1);
}
/*
addBigNum: Make a (num2[] <- num1[] + num2[]) big number operation
#lim1: Length of source num1 big number
#num1: First source operand big number (array of unsigned int). Should be smaller than second
#lim2: Length of source num2 big number
#num2: Second source operand big number (array of unsigned int). Should be equal or greater than first
Return: Array length of operation result or 0 if num1[] > num2[] (dosen't do any op)
Warning: This method can write in an incorrect position if we don't previous reallocate num2
*/
unsigned int addBigNum(const unsigned int lim1, unsigned int *num1, const unsigned int lim2, unsigned int *num2)
{
unsigned long long int n;
unsigned int i;
if(lim1 > lim2)
return 0;
for(num2[lim2] = 0, n = 0, i = 0; i < lim1; ++i)
{
n = num2[i] + num1[i] + (n / MODULE);
num2[i] = n % MODULE;
}
for(n /= MODULE; n; ++i)
{
num2[i] += n;
n = (num2[i] / MODULE);
}
return (lim2 > i) ? lim2 : i;
}
To compile:
gcc -o bgn <name>.c -Wall -O3 -lm //Math library if you wants to use log func
To check result, use direct output as and input to bc. Easy shell script:
#!/bin/bash
select S in ` awk -F '==' '{print $1 " == " $2 }' | bc`;
do
0;
done;
echo "Test Finished!";
We have and array of unsigned int (4 bytes) where we store at each int of array a number of 9 digits ( % 1000000000UL ); hence num[0] we will have the first 9 digits, num[1] we will have digit 10 to 18, num[2]...
I use convencional memory to work but an improvement can do it with dinamic memory. Ok, but how length It could be the array? (or how many memory we need to allocate?). Using bc calculator (bc -l with mathlib) we can determine how many digits has a number:
l(a^N) / l(10) // Natural logarith to Logarithm base 10
If we know digits, we know amount integers we needed:
( l(a^N) / (9 * l(10)) ) + 1 // Truncate result
If you work with value such as (2^k)^N you can resolve it logarithm with this expression:
( k*N*l(2)/(9*l(10)) ) + 1 // Truncate result
to determine the exactly length of integer array. Example:
256^800 = 2^(8*800) ---> l(2^(8*800))/(9*l(10)) + 1 = 8*800*l(2)/(9*l(10)) + 1
The value 1000000000UL (10^9) constant is very important. A constant like 10000000000UL (10^10) dosen't work because can produce and indetected overflow (try what's happens with number 16^16 and 10^10 constant) and a constant more little such as 1000000000UL (10^8) are correct but we need to reserve more memory and do more steps. 10^9 is key constant for unsigned int of 32 bits and unsigned long long int of 64 bits.
The code has two parts, Multiply (easy) and Power by 2 (more hard). Multiply is just multiplication and scale and propagate the integer overflow. It take the principle of associative property in math to do exactly the inverse principle, so if k(A + B + C) we want kA + kB + kC where number will be k*A*10^18 + k*B*10^9 + kC. Obiously, kC operation can generate a number bigger than 999 999 999, but never more bigger than 0xFF FF FF FF FF FF FF FF. A number bigger than 64 bits can never occur in a multiplication because C is an unsigned integer of 32 bits and k is a unsigned short of 16 bits. In worts case, we will have this number:
k = 0x FF FF;
C = 0x 3B 9A C9 FF; // 999999999
n = k*C = 0x 3B 9A | 8E 64 36 01;
n % 1000000000 = 0x 3B 99 CA 01;
n / 1000000000 = 0x FF FE;
After Mul kB we need to add 0x FF FE from last multiplication of C ( B = kB + (C / module) ), and so on (we have 18 bits arithmetic offset, enough to guarantee correct values).
Power is more complex but is in essencial, the same problem (multiplication and add), so I give some tricks about code power:
Data types are important, very important
If you try to multiplication an unsigned integer with unsigned integer, you get another unsigned integer. Use explicit cast to get unsigned long long int and don't lose data.
Always use unsigned modifier, dont forget it!
Power by 2 can directly modify 2 index ahead of current index
gdb is your friend
I've developed another method that add big numbers. These last I don't prove so much but I think it works well. Don't be cruels with me if it has a bug.
...and that's all!
PD1: Developed in a
Intel(R) Pentium(R) 4 CPU 1.70GHz
Data length:
unsigned short: 2
unsigned int: 4
unsigned long int: 4
unsigned long long int: 8
Numbers such as 256^1024 it spend:
real 0m0.059s
user 0m0.033s
sys 0m0.000s
A bucle that's compute i^i where i goes to i = 1 ... 1024:
real 0m40.716s
user 0m14.952s
sys 0m0.067s
For numbers such as 65355^65355, spent time is insane.
PD2: My response is so late but I hope my code it will be usefull.
PD3: Sorry, explain me in english is one of my worst handicaps!
Last update: I just have had an idea that with same algorithm but other implementation, improve response and reduce amount memory to use (we can use the completely bits of unsigned int). The secret: n^2 = n * n = n * (n - 1 + 1) = n * (n - 1) + n.
(I will not do this new code, but if someone are interested, may be after exams... )
I don't know if you still need a solution, but I wrote an article about this problem. It shows a very simple algorithm which can be used to convert an arbitrary long number with base X to a corresponding number of base Y. The algorithm is written in Python, but it is really only a few lines long and doesn't use any Python magic. I needed such an algorithm for a C implementation, too, but decided to describe it using Python for two reasons. First, Python is very readable by anyone who understands algorithms written in a pseudo programming language and, second, I am not allowed to post the C version, because it I did it for my company. Just have a look and you will see how easy this problem can be solved in general. An implementation in C should be straight forward...
Here is a function that does what you want:
#include <math.h>
#include <stddef.h> // for size_t
double getval(unsigned char *arr, size_t len)
{
double ret = 0;
size_t cur;
for(cur = 0; cur < len; cur++)
ret += arr[cur] * pow(256, cur);
return ret;
}
That looks perfectly readable to me. Just pass the unsigned char * array you want to convert and the size. Note that it won't be perfect - for arbitrary precision, I suggest looking into the GNU MP BigNum library, as has been suggested already.
As a bonus, I don't like your storing your numbers in little-endian order, so here's a version if you want to store base-256 numbers in big-endian order:
#include <stddef.h> // for size_t
double getval_big_endian(unsigned char *arr, size_t len)
{
double ret = 0;
size_t cur;
for(cur = 0; cur < len; cur++)
{
ret *= 256;
ret += arr[cur];
}
return ret;
}
Just things to consider.
It may be too late or too irrelevant to make this suggestion, but could you store each byte as two base 10 digits (or one base 100) instead of one base 256? If you haven't implemented division yet, then that implies all you have is addition, subtraction, and maybe multiplication; those shouldn't be too hard to convert. Once you've done that, printing it would be trivial.
As I was not satisfied with the other answers provided, I decided to write an alternative solution myself:
#include <stdlib.h>
#define BASE_256 256
char *largenum2str(unsigned char *num, unsigned int len_num)
{
int temp;
char *str, *b_256 = NULL, *cur_num = NULL, *prod = NULL, *prod_term = NULL;
unsigned int i, j, carry = 0, len_str = 1, len_b_256, len_cur_num, len_prod, len_prod_term;
//Get 256 as an array of base-10 chars we'll use later as our second operand of the product
for ((len_b_256 = 0, temp = BASE_256); temp > 0; len_b_256++)
{
b_256 = realloc(b_256, sizeof(char) * (len_b_256 + 1));
b_256[len_b_256] = temp % 10;
temp = temp / 10;
}
//Our first operand (prod) is the last element of our num array, which we'll convert to a base-10 array
for ((len_prod = 0, temp = num[len_num - 1]); temp > 0; len_prod++)
{
prod = realloc(prod, sizeof(*prod) * (len_prod + 1));
prod[len_prod] = temp % 10;
temp = temp / 10;
}
while (len_num > 1) //We'll stay in this loop as long as we still have elements in num to read
{
len_num--; //Decrease the length of num to keep track of the current element
//Convert this element to a base-10 unsigned char array
for ((len_cur_num = 0, temp = num[len_num - 1]); temp > 0; len_cur_num++)
{
cur_num = (char *)realloc(cur_num, sizeof(char) * (len_cur_num + 1));
cur_num[len_cur_num] = temp % 10;
temp = temp / 10;
}
//Multiply prod by 256 and save that as prod_term
len_prod_term = 0;
prod_term = NULL;
for (i = 0; i < len_b_256; i++)
{ //Repeat this loop 3 times, one for each element in {6,5,2} (256 as a reversed base-10 unsigned char array)
carry = 0; //Set the carry to 0
prod_term = realloc(prod_term, sizeof(*prod_term) * (len_prod + i)); //Allocate memory to save prod_term
for (j = i; j < (len_prod_term); j++) //If we have digits from the last partial product of the multiplication, add it here
{
prod_term[j] = prod_term[j] + prod[j - i] * b_256[i] + carry;
if (prod_term[j] > 9)
{
carry = prod_term[j] / 10;
prod_term[j] = prod_term[j] % 10;
}
else
{
carry = 0;
}
}
while (j < (len_prod + i)) //No remaining elements of the former prod_term, so take only into account the results of multiplying mult * b_256
{
prod_term[j] = prod[j - i] * b_256[i] + carry;
if (prod_term[j] > 9)
{
carry = prod_term[j] / 10;
prod_term[j] = prod_term[j] % 10;
}
else
{
carry = 0;
}
j++;
}
if (carry) //A carry may be present in the last term. If so, allocate memory to save it and increase the length of prod_term
{
len_prod_term = j + 1;
prod_term = realloc(prod_term, sizeof(*prod_term) * (len_prod_term));
prod_term[j] = carry;
}
else
{
len_prod_term = j;
}
}
free(prod); //We don't need prod anymore, prod will now be prod_term
prod = prod_term;
len_prod = len_prod_term;
//Add prod (formerly prod_term) to our current number of the num array, expressed in a b-10 array
carry = 0;
for (i = 0; i < len_cur_num; i++)
{
prod[i] = prod[i] + cur_num[i] + carry;
if (prod[i] > 9)
{
carry = prod[i] / 10;
prod[i] -= 10;
}
else
{
carry = 0;
}
}
while (carry && (i < len_prod))
{
prod[i] = prod[i] + carry;
if (prod[i] > 9)
{
carry = prod[i] / 10;
prod[i] -= 10;
}
else
{
carry = 0;
}
i++;
}
if (carry)
{
len_prod++;
prod = realloc(prod, sizeof(*prod) * len_prod);
prod[len_prod - 1] = carry;
carry = 0;
}
}
str = malloc(sizeof(char) * (len_prod + 1)); //Allocate memory for the return string
for (i = 0; i < len_prod; i++) //Convert the numeric result to its representation as characters
{
str[len_prod - 1 - i] = prod[i] + '0';
}
str[i] = '\0'; //Terminate our string
free(b_256); //Free memory
free(prod);
free(cur_num);
return str;
}
The idea behind it all derives from simple math. For any base-256 number, its base-10 representation can be calculated as:
num[i]*256^i + num[i-1]*256^(i-1) + (···) + num[2]*256^2 + num[1]*256^1 + num[0]*256^0
which expands to:
(((((num[i])*256 + num[i-1])*256 + (···))*256 + num[2])*256 + num[1])*256 + num[0]
So all we have to do is to multiply, step-by step, each element of the number array by 256 and add to it the next element, and so on... That way we can get the base-10 number.