I want to concatenate two integers using only bit operations since i need efficiency as much as possible.There are various answers available but they are not fast enough what I want is implementation that uses only bit operations like left shift,or and etc. Please guide me how to do it.
example
int x=32;
int y=12;
int result=3212;
I am having and FPGA implentation of AES.I need this on my system to reduce time consumption of some task
The most efficient way to do it, is likely something similar to this:
uint32_t uintcat (uint32_t ms, uint32_t ls)
{
uint32_t mult=1;
do
{
mult *= 10;
} while(mult <= ls);
return ms * mult + ls;
}
Then let the compiler worry about optimization. There's likely not a lot it can improve since this is base 10, which doesn't get along well with the various instructions of the computer, like shifting.
EDIT : BENCHMARK TEST
Intel i7-3770 2 3,4 GHz
OS: Windows 7/64
Mingw, GCC version 4.6.2
gcc -O3 -std=c99 -pedantic-errors -Wall
10 million random values, from 0 to 3276732767.
Result (approximates):
Algorithm 1: 60287 micro seconds
Algorithm 2: 65185 micro seconds
Benchmark code used:
#include <stdint.h>
#include <stdio.h>
#include <windows.h>
#include <time.h>
uint32_t uintcat (uint32_t ms, uint32_t ls)
{
uint32_t mult=1;
do
{
mult *= 10;
} while(mult <= ls);
return ms * mult + ls;
}
uint32_t myConcat (uint32_t a, uint32_t b) {
switch( (b >= 10000000) ? 7 :
(b >= 1000000) ? 6 :
(b >= 100000) ? 5 :
(b >= 10000) ? 4 :
(b >= 1000) ? 3 :
(b >= 100) ? 2 :
(b >= 10) ? 1 : 0 ) {
case 1: return a*100+b; break;
case 2: return a*1000+b; break;
case 3: return a*10000+b; break;
case 4: return a*100000+b; break;
case 5: return a*1000000+b; break;
case 6: return a*10000000+b; break;
case 7: return a*100000000+b; break;
default: return a*10+b; break;
}
}
static LARGE_INTEGER freq;
static void print_benchmark_results (LARGE_INTEGER* start, LARGE_INTEGER* end)
{
LARGE_INTEGER elapsed;
elapsed.QuadPart = end->QuadPart - start->QuadPart;
elapsed.QuadPart *= 1000000;
elapsed.QuadPart /= freq.QuadPart;
printf("%lu micro seconds", elapsed.QuadPart);
}
int main()
{
const uint32_t TEST_N = 10000000;
uint32_t* data1 = malloc (sizeof(uint32_t) * TEST_N);
uint32_t* data2 = malloc (sizeof(uint32_t) * TEST_N);
volatile uint32_t* result_algo1 = malloc (sizeof(uint32_t) * TEST_N);
volatile uint32_t* result_algo2 = malloc (sizeof(uint32_t) * TEST_N);
srand (time(NULL));
// Mingw rand() apparently gives numbers up to 32767
// worst case should therefore be 3,276,732,767
// fill up random data in arrays
for(uint32_t i=0; i<TEST_N; i++)
{
data1[i] = rand();
data2[i] = rand();
}
QueryPerformanceFrequency(&freq);
LARGE_INTEGER start, end;
// run algorithm 1
QueryPerformanceCounter(&start);
for(uint32_t i=0; i<TEST_N; i++)
{
result_algo1[i] = uintcat(data1[i], data2[i]);
}
QueryPerformanceCounter(&end);
// print results
printf("Algorithm 1: ");
print_benchmark_results(&start, &end);
printf("\n");
// run algorithm 2
QueryPerformanceCounter(&start);
for(uint32_t i=0; i<TEST_N; i++)
{
result_algo2[i] = myConcat(data1[i], data2[i]);
}
QueryPerformanceCounter(&end);
// print results
printf("Algorithm 2: ");
print_benchmark_results(&start, &end);
printf("\n\n");
// sanity check both algorithms against each other
for(uint32_t i=0; i<TEST_N; i++)
{
if(result_algo1[i] != result_algo2[i])
{
printf("Results mismatch for %lu %lu. Expected: %lu%lu, algo1: %lu, algo2: %lu\n",
data1[i],
data2[i],
data1[i],
data2[i],
result_algo1[i],
result_algo2[i]);
}
}
// clean up
free((void*)data1);
free((void*)data2);
free((void*)result_algo1);
free((void*)result_algo2);
}