Is it possible to implement bitwise operators using integer arithmetic?

You can operate bit-by-bit (as Mark Byers suggested), by extracting every bit which will be slow.

Or you could accelerate process and use 2d lookup tables that store results, say, for two 4-bit operands and operate on those. You'll need less extractions than if you were operating on bits.

You can also do everything using addition, subtraction and >= operation. Every bitwise operation can be unrolled into something like this using macros:

/*I didn't actually compile/test it, it is just illustration for the idea*/
uint16 and(uint16 a, uint16 b){
    uint16 result = 0;
    #define AND_MACRO(c) \
        if (a >= c){ \ 
            if (b >= c){\
                result += c;\
                b -= c;\
            }\
            a -= c;\
        }\
        else if (b >= c)\
            b -= c;

    AND_MACRO(0x8000)
    AND_MACRO(0x4000)
    AND_MACRO(0x2000)
    AND_MACRO(0x1000)
    AND_MACRO(0x0800)
    AND_MACRO(0x0400)
    AND_MACRO(0x0200)
    AND_MACRO(0x0100)
    AND_MACRO(0x0080)
    AND_MACRO(0x0040)
    AND_MACRO(0x0020)
    AND_MACRO(0x0010)
    AND_MACRO(0x0008)
    AND_MACRO(0x0004)
    AND_MACRO(0x0002)
    AND_MACRO(0x0001)
    #undef AND_MACRO
    return result;
}

You'll need 3 variables to implement this.

Every bitwise operation will revolve around macros similar to AND_MACRO - you compare remaining values of a and b to the "mask" (which is "c" parameter). then add mask to the result in the if branch that is suitable for your operation. And you subtract mask from values, if bit is set.

Depending on your platform, it may be faster than extracting every bit using % and / , and then putting it back using multiplication.

See for yourself whichever is better for you.

First solutions for shifting (shift is the shift distance, must not be negative, a is the operand to be shifted and contains also the result when done). The power table is used by all three shift operations.

// table used for shift operations
powtab = { 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, -32768 };

// logical shift left
if (shift > 15) {
     a = 0; // if shifting more than 15 bits to the left, value is always zero
} else {
     a *= powtab[shift];
}

// logical shift right (unsigned)
if (shift > 15) {
    a = 0; // more than 15, becomes zero
} else if (shift > 0) {
    if (a < 0) {
        // deal with the sign bit (15)
        a += -32768;
        a /= powtab[shift];
        a += powtab[15 - shift];
    } else {
        a /= powtab[shift];
    }
}

// arithmetic shift right (signed)
if (shift >= 15) {
    if (a < 0) {
        a = -1;
    } else {
        a = 0;
    }
} else if (shift > 0) {
    if (a < 0) {
        // deal with the sign bit
        a += -32768;
        a /= powtab[shift];
        a -= powtab[15 - shift];
    } else {
        // same as unsigned shift
        a /= powtab[shift];
    }
}

For AND, OR and XOR i could not come up with a simple solution, so i'll do it with looping over each single bit. There might be a better trick to do this. Pseudocode assumes a and b are input operands, c is the result value, x is the loop counter (each loop must run exactly 16 times):

// XOR (^)
c = 0;
for (x = 0; x <= 15; ++x) {
    c += c;
    if (a < 0) {
        if (b >= 0) {
            c += 1;
        }
    } else if (b < 0) {
        c += 1;
    }
    a += a;
    b += b;
}

// AND (&)
c = 0;
for (x = 0; x <= 15; ++x) {
    c += c;
    if (a < 0) {
        if (b < 0) {
            c += 1;
        }
    }
    a += a;
    b += b;
}

// OR (|)
c = 0;
for (x = 0; x <= 15; ++x) {
    c += c;
    if (a < 0) {
        c += 1;
    } else if (b < 0) {
        c += 1;
    }
    a += a;
    b += b;
}

Thats assuming that all variables are 16 bits and all operations behave as signed (so a<0 actually is true when bit 15 is set).

EDIT: i actually tested all possible operand values (-32768 to 32767) for shifts ranging from 0 to 31 for correctness and it works correctly (assuming integer divides). For the AND/OR/XOR code an exhaustive test takes too long on my machine, but since the code for these is pretty simple there should be no edge cases anyway.

In this environment it might be best if you could set up to actually use arithmatic operators to peel out components of integers.

E.G.

if (a & 16)  becomes if ((a % 32) > 15)
a &= 16 becomes if ((a % 32) < 15) a += 16

The transforms for these operators are obvious enough if you restrict RHS to a constant power of 2.

Peeling off two or four bits is also easy to do.

An incomplete answer on an old question, here concentrating on AND, OR, XOR. Once a solution is found for one of these bitwise operations, the other two can be derived. There are several ways, one is shown in the following test program (compiled on gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5)).

In December 2018 I discovered an error in the solution. The XOR commented below only works because intermediate results in a+b-2*AND(a,b) are promoted to int, which is larger than 16 bits for all modern compilers.

#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>

//#define XOR(a,b) (a + b - 2*AND(a,b)) // Error. Intermediate overflow
#define XOR(a,b) (a - AND(a,b) +  b - AND(a,b) )
#define IOR(a,b) XOR(XOR(a,b),AND(a,b)) // Credit to Jan Gray, Gray Research LLC, for IOR
static const uint16_t andlookup[256] = {
#define C4(a,b) ((a)&(b)), ((a)&(b+1)), ((a)&(b+2)), ((a)&(b+3))
#define L(a) C4(a,0), C4(a,4), C4(a,8), C4(a,12)
#define L4(a) L(a), L(a+1), L(a+2), L(a+3)
    L4(0), L4(4), L4(8), L4(12)
#undef C4
#undef L
#undef L4
};

uint16_t AND(uint16_t a, uint16_t b) {
    uint16_t r=0, i;

    for ( i = 0; i < 16; i += 4 ) {
            r = r/16 + andlookup[(a%16)*16+(b%16)]*4096;
            a /= 16;
            b /= 16;
    }
    return r;
}

int main( void ) {
    uint16_t a = 0, b = 0;

    do {
            do {
                    if ( AND(a,b) != (a&b) ) return printf( "AND error\n" );
                    if ( IOR(a,b) != (a|b) ) return printf( "IOR error\n" );
                    if ( XOR(a,b) != (a^b) ) return printf( "XOR error\n" );
            } while ( ++b != 0 );
            if ( (a & 0xff) == 0 )
                    fprintf( stderr, "." );
    } while ( ++a != 0 );
    return 0;
}