Code:Testing version : Elysia ------------------------------ Testing on AMD Athlon II X2 250 (3 GHz) OS = Windows 7 Ultimate 64 bit -- Compiler = Microsoft Visual Studio 2010 The value of num is: -1234567890 The formatted value of num is: -1,234,567,890 Elapsed time: 208 ms to perform 10,000,000 cycles ------------------------------------------------------- handling 0 ---> 0 handling 12 ---> 12 handling 256 ---> 256 handling 1000 --> 1,000 handling 1000000 ---> 1,000,000 handling 1000000000 ---> 1,000,000,000 handling 2147483647 -> 2,147,483,647 handling -2147483648 -> -2,147,483,648 Press any key to continue . . .There you go. Quick and dirt. x64 version (beats x86 version by ~2x).Code:// ---------------------------------------------------------------------------------------------- // Prog_name: testpad.c // A program for testing functions formatting integer numbers with // thousand separators and sign. //----------------------------------------------------------------------------------------------- // Created with Pelles C for Windows 6.00.4 // Standard ISO C99 //----------------------------------------------------------------------------------------------- // Date: 09 july 2010 // Testpad by frktons //----------------------------------------------------------------------------------------------- #include <stdio.h> #include <time.h> #include <string.h> //----------------------------------------------------------------------------------------------- // Function prototype for itoa(). Takes in the number to // format, and the pointer of the buffer to fill. Returns an int. //----------------------------------------------------------------------------------------------- void itoa(long num, char []); char digits_table[999999][8]; //----------------------------------------------------------------------------------------------- int main(void) { for (int i = 0; i < 999999; i++) { for (int j = 0; j < 8; j++) { digits_table[i][0] = ','; digits_table[i][1] = '0' + char(i / 100000); digits_table[i][2] = '0' + char(i / 10000 % 10); digits_table[i][3] = '0' + char(i / 1000 % 10); digits_table[i][4] = ','; digits_table[i][5] = '0' + char(i / 100 % 10); digits_table[i][6] = '0' + char(i / 10 % 10); digits_table[i][7] = '0' + char(i % 10); } } long num = -1234567890; // test number int cycles = 0; clock_t start = 0; // initial time clock_t stop = 0; // end time const char *ver = "Elysia"; // Program version char buffer[50] = {'\0'}; // string array for the formatted number // 10 digits max + 3 separators max + sign + NULL long times = 10000000; // for testing performance set this number for repetitions //----------------------------------------------------------------------------------------------- // HW-SW platform //----------------------------------------------------------------------------------------------- const char *OS = "Windows 7 Ultimate 64 bit"; const char *CPU = "AMD Athlon II X2 250 (3 GHz)"; const char *compiler = "Microsoft Visual Studio 2010"; //----------------------------------------------------------------------------------------------- printf("\n\t\t Testing version : %s \n",ver); printf("\t\t ------------------------------\n"); printf(" Testing on %s\n",CPU); printf(" OS = %s -- Compiler = %s\n",OS,compiler); printf("\n The value of num is: %d\n",num); start = clock(); for (cycles = 0; cycles < times; cycles++) { itoa(num,buffer); } // end for stop = clock(); printf("\n The formatted value of num is: %s\n",buffer); itoa((long)(stop-start)/(CLOCKS_PER_SEC/1000),buffer); printf("\n Elapsed time: %s ms",buffer); itoa(times,buffer); printf(" to perform %s cycles\n",buffer); //------------------------------------------------------------------------------ // Test some numbers //------------------------------------------------------------------------------ printf(" -------------------------------------------------------\n"); long test = -2147483647 - 1; itoa(0,buffer); printf("\n handling 0 ---> %s",buffer); itoa(12,buffer); printf("\n handling 12 ---> %s",buffer); itoa(256,buffer); printf("\n handling 256 ---> %s",buffer); itoa(1000,buffer); printf("\n handling 1000 --> %s",buffer); itoa(1000000,buffer); printf("\n handling 1000000 ---> %s",buffer); itoa(1000000000,buffer); printf("\n handling 1000000000 ---> %s",buffer); itoa(2147483647,buffer); printf("\n handling 2147483647 -> %s",buffer); itoa(test,buffer); printf("\n handling -2147483648 -> %s\n",buffer); return 0; } //--------------------------------------------------------------- // itoa() - Elysia's version //--------------------------------------------------------------- __forceinline void itoa(long num, char buffer[]) { int remain = 0; // integer variable to store the remainder bool neg = false; const int sizeof_buffer = 30; unsigned long newnum; memset(buffer, 0, 50); if (num == 0) { strcpy(buffer, "0"); return; } if (num < 0) { neg = true; buffer[0] = '-'; newnum = -num; // transform number to positive if negative } else newnum = num; int x, copy_from_offset = 0; for (x = sizeof_buffer - 1; newnum; x -= 8) { if (newnum < 100000) { copy_from_offset = 1; if (newnum < 10) copy_from_offset = 6; else if (newnum < 100) copy_from_offset = 5; else if (newnum < 1000) copy_from_offset = 4; else if (newnum < 10000) copy_from_offset = 2; } remain = newnum % 1000000; newnum /= 1000000; memcpy(&buffer[x - 8], digits_table[remain], 8); } int copy_to = 0; if (neg) copy_to++; memmove(&buffer[copy_to], &buffer[x + 1] + copy_from_offset, sizeof_buffer - x - 1 - copy_from_offset); buffer[sizeof_buffer - x - 1] = '\0'; }
Originally Posted by Adak
