Convert octal representation of a number to decimal - c++

Question

Assume one has an vector or array of N elements (N can be very large) containing the octal representation of a non negative integer. How do I get the decimal representation of the number from this array? The code has to be really fast.
EDIT: array A of N elements contains octal representation of a non-negative integer K, i.e. each element of A belongs to the interval [0; 7] (both ends included)
Example: A[0] = 2; A[1] = 6; A[2] = 3
Now a naive calculation would be 2*8pow0 + 6*8pow1 + 3*8pow2 = 2+ 48+ 192 = 242
I tried this but it does not seem to work for large inputs > 6K
//vector<int> A is the input
using namespace std;
vector<int>::iterator it = A.begin();
unsigned int k = 0;
unsigned int x = 0;
while(it < A.end()){
x = x | (*it<<3*k);
k++;
it++;
}
I am also having problems converting a hexadecimal string to its decimal representation? Is this the correct way to do this in C++:
//Assume S to be your input string containing a hex representation
//like F23
std::stringstream ss;
ss << std::hex << S;
ss >> x;

Answer 1

Arbitray precision octal to decimal conversion is rather annoying because there is no way to localize the computation. In other words a change in the most significant digit of the octal number will change even the least significant digit in the decimal representation.
That said I think I would convert the octal number to a say base-1000000000 number and then I'd print that (this is instead a trivial problem, each base-1000000000 digit just maps trivially to 9 base-10 digits).
The conversion to base-1000000000 is simple because you only need to support incrementing and multiplying by two (just consider the input as binary with three bits for each octal digit).
EDIT
I tried to implement it in C++ and this is the resulting code
#include <stdio.h>
#include <vector>
int main(int argc, const char *argv[]) {
// Base 100,000,000 accumulator
// Initialized with one digit = 0
std::vector<unsigned> big(1);
const unsigned DIGIT = 100000000;
for (int c=getchar(); c >= '0' && c <= '7'; c=getchar()) {
// Multiply accumulator by 8 and add in incoming digit
int carry = c - '0';
for (int i=0,n=big.size(); i<n; i++) {
unsigned x = big[i] * 8 + carry;
carry = x / DIGIT;
big[i] = x - carry * DIGIT;
}
if (carry) big.push_back(carry);
}
// Output result in decimal
printf("%i", big.back());
for (int i=int(big.size())-2; i>=0; i--) {
printf("%08i", big[i]);
}
putchar('\n');
return 0;
}
On my PC the time to convert an 80,000 digit octal number to decimal (resulting in 72246 digits) is about 1.2 seconds. Doing the same using python eval/str the time is about 3 seconds. The number used was "01234567" * 10000.
The code above uses 100,000,000 as base so that it can process one digit (3 bits) at a time with 32-bit arithmetic not overflowing with the intermediate results. I tried also using 64 bit integers or the 53 bit integer part of a double but the code was running always slower than in this case (one reason is probably the division in the inner loop that can be converted to a multiplication in the 32 bit case).
This is still a simple O(n^2) implementation that would take ages to convert a 10,000,000-digits octal number.

Answer 2

This is what I came up with:
template<typename Iter>
int read_octal(Iter begin, Iter end)
{
int x = 0;
int f = 1;
for (; begin != end; ++begin)
{
x += *begin * f;
f *= 8;
}
return x;
}
int main()
{
int test[] = {2, 6, 3};
int result = read_octal(test + 0, test + 3);
std::cout << result << '\n';
}
I tried this but it does not seem to work for large inputs > 6K
What exactly do you mean by 6k? An int usually has 32 bits, and an octal digit has 3 bits. Thus, you cannot have more than 10 elements in your range, otherwise x will overflow.
I am also having problems converting a hexadecimal string to its decimal representation?
Well, you could always write a function to parse a string in hex format yourself:
int parse_hex(const char* p)
{
int x = 0;
for (; *p; ++p)
{
x = x * 16 + digit_value(*p);
}
return x;
}
With the most portable version of digit_value being:
int digit_value(char c)
{
switch (c)
{
case '0': return 0;
case '1': return 1;
case '2': return 2;
case '3': return 3;
case '4': return 4;
case '5': return 5;
case '6': return 6;
case '7': return 7;
case '8': return 8;
case '9': return 9;
case 'A':
case 'a': return 10;
case 'B':
case 'b': return 11;
case 'C':
case 'c': return 12;
case 'D':
case 'd': return 13;
case 'E':
case 'e': return 14;
case 'F':
case 'f': return 15;
}
}