Decoding integer value

Question

I have this program that encodes integer values:

#include "stdafx.h"
#define _SECURE_SCL_DEPRECATE 0
#include <iostream>
#include <list>
#include <vector>
#include <algorithm>

using namespace std;

template<class T>
vector<unsigned char>  nToB(T );
unsigned long ByteToint(vector<unsigned char> v)
{
    unsigned long int a = 0;
    int s = v.size();
    for (int i = 0; i<s ; i++)
    {
        a |= (v[s - 1 - i] << (8 * (s - i - 1)));
    }
    return a;
}

static unsigned  long  int  Encode7Bits(unsigned long int);

int main()
{   
    cout << Encode7Bits(420);
    getchar();
    return 0;
}

static unsigned  long  int  Encode7Bits( unsigned long int x)
{
    vector<unsigned char> Result;

    do
    {
        unsigned long  int tmp = x & 0x7f;
        x = x >> 7;
        if (x > 0)
            tmp |= 0x80;        
        Result.push_back((unsigned char )tmp);
    } while (x > 0);

    return ByteToint(Result);
}

If the argument to this function is 420 it will return 932.

My question is whether it is possible to do the reverse operation, a decoding function that given 932, returns 420.

This is a trapdoor function. Forget about inverting it.

Detonar
– Detonar

2017-11-24 08:49:04 +00:00
Commented Nov 24, 2017 at 8:49 — Detonar
– Detonar, Commented Nov 24, 2017 at 8:49

Bathsheba · Accepted Answer · 2017-11-24 08:53:45Z

1

No it isn't.

|= is non-invertible in the sense that if you write c = a | b, then given c, and either a or b, you can't recover the other variable.

The bitwise operators << and >> are obviously lossy since they introduce 0 bits.

You'll have better luck with XOR: if you write c = a ^ b, then c ^ b will be a.

edited Nov 24, 2017 at 8:53

answered Nov 24, 2017 at 8:47

Bathsheba

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2 Comments

Detonar Over a year ago

Same goes for x = x >> 7. Not invertable.

user555045 Over a year ago

I feel like this misses the point though. The right shift and the OR are not what make this non-invertible, that part is just taking blocks of 7 bits and encoding them with a "continue"-bit, a pretty standard variable-length integer encoding. The problem is that this expands the output for high enough values, but the code forces the output back down to its original size.

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