C modulus with using unions and anonymous structs

Question

I am using anonymous structs in unions to get a%b quickly.

Do you know any other way to get a%b without using powers of 2 for b.

include list:

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

and declaration of unions:

//C99
//my first program to test rand() for any periodicity
union //for array indexing without modulus operator
{
unsigned int counter; //32 bit on my computer
struct
{
    unsigned int lsb:16; //16 bit 
    unsigned int msb:16; //16 bit   
};
struct 
{
    unsigned int bit[32];
};
} modulo;

union // for rand()%256
{
unsigned int randoom; //from rand() 
struct
{
unsigned int lsb:5;//equivalent to rand()%32 without any calculations
unsigned int msb:27;
};
}random_modulus;

and here the main function:

int main()
{
srand(time(0));

modulo.counter=0;//init of for-loop counter

// i am takin (counter%65536) for my array index which is modulus.lsb
unsigned int array_1[65536]; 
float array_mean=0,delta_square=0;
clock_t clock_refe;


//taking counter%65536 using lsb (2x faster)
clock_refe=clock();
for(;modulo.counter<1000000000;modulo.counter++)
{
// i need accumulated randoms for later use for some std. dev. thing.
random_modulus.randoom=rand();
array_1[modulo.lsb]+=random_modulus.lsb;
}

//getting average clock cycles
for(int i=0;i<65536;i++)
{
array_mean+=array_1[i];

}
array_mean/=65536.0f;

//getting square of deltas
float array_2[65536];

for(int i=0;i<65536;i++)
{
array_2[i]=(array_1[i]-array_mean)*(array_1[i]-array_mean); 
}


//horizontal histogram for resoluton of 20 elements
for(int i=0;i<65536;i+=(65536)/20)
{


for(int j=0;j<(array_2[i]*0.01);j++)
{

    printf("*");
}
printf("\n");
}

//calculations continue .....
return 0;

}

Pre-calculated values in some array maybe? But if i use the % calc part for only once, this is same. Can you give some book references about bitwise operations manuals?

Answer 1

Bitwise operations are the most portable method in this case. Here's some example functions; they are written for readability, but can be easily made faster:

int lsb(int input) {
    int mask = 0x00FF; // 0x00FF is 0000000011111111 in binary
    return input & mask;
}

int msb(int input) {
    int mask = 0xFF00; // 0xFF00 is 1111111100000000 in binary
    return (input & mask) >> 8;
}

In general, mask the bits you want with &, and then align them to the right with >>. K&R 2nd edition (The C Programming Language by Brian Kernighan & Dennis Ritchie) has information on just about every C topic, including bitmasks.

If you want a % b where b is not a power of 2, the native % operator is the fastest way (in modern compilers it's as fast as the bitwise operations, even when b is a power of two). Bitwise operations are useful in other contexts, though.