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Pointer arithmetic and arrays.

Started by MeshGearFox Oct 14, 2008 at 10:17 PM 8 replies 1.2k views

Original Post

Oct 14, 2008 at 10:17 PM

I'm well aware that this is odd and that this is probably something I SHOULDN'T do in a professional/more serious context. What I want to do is loop through an array without integer's to access specific array positions. This is a two or three dimensional array, consisting of chars. The arrays are all 16 elements in size. Instead of using actual integer values to access positions in these arrays, I instead have two (or three) arrays of booleans, four elements each, representing binary numbers from 0 to 15. Now, doing this in the more rational way, if I wanted to access something in an array, I'd just do something like cout << z[14][2] << endl; or whatever. However, instead, I'm trying to handle this using that binary number array thing.

    char array[16] = "abcdefghijklmno";
    char *i;
    bool byte[4] = {1,0,0,0};
    i = array;
    i++;
    cout << *i << endl;    
    i = array;
    cout << *i << endl;
    if(byte[0] == 1) {
        i++;
        i++;
        i++;
        i++;
        i++;
        i++;
        i++;
        i++;        
    }
    if(byte[1] == 1) {
        i++;
        i++;
        i++;
        i++;
    }
    if(byte[2] == 1) {
        i++;
        i++; 
    }
    if(byte[3] == 1) {
        i++; 
    }    
    cout << *i << endl;

This is sort of how I'm doing it right now (While loops would obviously be used in each of the if statements. I'm not really sure why I wrote that like I did, there). Anyway, is there any way to do something similar, for a 2D or 3D array? Or some other way I could iterate through the char array wihout directly accessing the various locations in the array using actual integers?

ApochPiQ

Oct 14, 2008 at 10:24 PM

Can you describe exactly why you need this behaviour? It seems to me there may be a cleaner solution to the root problem at hand.

Hodgman

Oct 14, 2008 at 10:58 PM

I'm not exactly sure what you're trying to do, but may I suggest a replacement for your if/++ section?

bool bits[4] = {1,0,0,0};bool* pBit = bits;int numBits = sizeof(bits)/sizeof(bool);while( numBits ){    --numBits;    if(*pBit == 1) {        i += 1<<numBits;         }    ++pBit;}

theOcelot

Oct 14, 2008 at 11:15 PM

Use more byte arrays for more dimensions? But why on earth are you simulating binary numbers when the normal practice is just to use the real thing?

DevFred

Oct 15, 2008 at 6:02 AM

Pointer arithmetic is more than just the ++ operator. This

Quote:

i = array;    if(byte[0] == 1) {        i++;        i++;        i++;        i++;        i++;        i++;        i++;        i++;    }    if(byte[1] == 1) {        i++;        i++;        i++;        i++;    }    if(byte[2] == 1) {        i++;        i++;    }    if(byte[3] == 1) {        i++;    }

can be replaced by a single line:

i = array + (byte[0] << 3 | byte[1] << 2 | byte[2] << 1 | byte[3]);

Note that if you made byte little endian instead of big endian, the code would look more symmetric:

i = array + (byte[3] << 3 | byte[2] << 2 | byte[1] << 1 | byte[0]);

Oh, and please don't name pointers i. It's very confusing.

Zahlman

Oct 15, 2008 at 11:55 PM

Quote:
Original post by MeshGearFox
Anyway, is there any way to do something similar, for a 2D or 3D array?

Here's a hint:

int test[4][4] = {  {0, 1, 2, 3},  {4, 5, 6, 7},  {8, 9, 10, 11},  {12, 13, 14, 15}};int* memory = reinterpret_cast<int*>(test);for (int i = 0; i < 16; ++i) {  std::cout << memory << '\n';}

MeshGearFox

Oct 16, 2008 at 2:01 AM

I think this is doing what I want it to do but I still don't understand that thing you were doing with the left-shifts.

#include <iostream>using namespace std;int main(){    struct byte {        bool eight;        bool four;        bool two;        bool one;    };        int array[16][16];        for(int y = 0; y < 16; y++) {        for(int x = 0; x < 16; x++) {            array[x][y] = x,y;        }    }                        byte xcoord = {0,0,0,0};     byte ycoord = {1,1,1,1};        int *i;    i = &array[0][0];        cout << *i << endl;    if(xcoord.eight) { i += 8; }    if(xcoord.four) { i += 4;}    if(xcoord.two) { i += 2; }    if(xcoord.one) { i += 1;}    if(ycoord.eight) { i += (8 * 16); }    if(ycoord.four) { i += (4 * 16);}    if(ycoord.two) { i += (2 * 16); }    if(ycoord.one) { i += 16; }    cout << *i << endl;        char bill;    cin >> bill;    return 0;}

DevFred

Oct 16, 2008 at 6:13 AM

Quote:

Original post by MeshGearFox

    for(int y = 0; y < 16; y++) {        for(int x = 0; x < 16; x++) {            array[x][y] = x,y; // <--- this probably doesn't do what you want        }    }

Do you know what the comma operator does? It evaluates the first operand, throws the result away, and then evaluates the second operand. Thus x,y is equivalent to y.

And this

Quote:

Original post by MeshGearFox

    int *i;    i = &array[0][0];        if(xcoord.eight) { i += 8; }    if(xcoord.four) { i += 4;}    if(xcoord.two) { i += 2; }    if(xcoord.one) { i += 1;}    if(ycoord.eight) { i += (8 * 16); }    if(ycoord.four) { i += (4 * 16);}    if(ycoord.two) { i += (2 * 16); }    if(ycoord.one) { i += 16; }

can be replaced by

    int *i = &array    [ycoord.eight << 3 | ycoord.four << 2 | ycoord.two << 1 | ycoord.one]    [xcoord.eight << 3 | xcoord.four << 2 | xcoord.two << 1 | xcoord.one];

Nyarlath

Oct 16, 2008 at 9:22 AM

Quote:
Original post by DevFred
Thus x,y is equivalent to y.

But the comma operator has less priority to respect to the assignment, so array[x][y] = x,y; is equivalent to array[x][y] = x; y; and array[x][y] = x;

DevFred

Oct 16, 2008 at 10:38 AM

You are right.

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