| x=0; | |
| //thread 1 | |
| { | |
| lock(); | |
| x++; | |
| unlock(); | |
| } | |
| //thread 2 | |
| { | |
| lock(); |
| #if LONG_MAX == 2147483647L | |
| #define DETECT(X) (((X) - 0x01010101) & ~(X) & 0x80808080) | |
| #else | |
| #if LONG_MAX == 9223372036854775807L | |
| #define DETECT(X) (((X) - 0x0101010101010101) & ~(X) & 0x8080808080808080) | |
| #else | |
| #error long int is not a 32bit or 64bit type. | |
| #endif | |
| #endif |
| #include <stdio.h> | |
| int main(void) | |
| { | |
| int i, result; | |
| for(i=2147483647;i>-2147483648;i--) | |
| { | |
| result = i&(i-1); | |
| if(result==0) | |
| printf("%d: %d\n",i,result); | |
| } |
| #include <stdio.h> | |
| //#define MAX 2147483647 | |
| //#define MIN -MAX+1 | |
| #define MAX 15 | |
| #define MIN -16 | |
| #define BYTETOBINARYPATTERN "%d%d%d%d%d%d%d%d" | |
| #define BYTETOBINARY(byte) \ | |
| (byte & 0x80 ? 1 : 0), \ | |
| (byte & 0x40 ? 1 : 0), \ | |
| (byte & 0x20 ? 1 : 0), \ |
| int square(volatile int *ptr) //make sure each call to this function the CPU move data form MEM to REG | |
| { | |
| int rtn = *ptr; //fetch the value pointer points to only once.. | |
| return rtn*rtn; | |
| } | |
| //避免原來例子兩次從記憶體搬運資料時*ptr遭thread其他修改 | |
| //或是compiler 最佳時跳過部分指令, 像是下面的例子如果沒有volatile, | |
| // compile with O2 得到的assembly 是 rep ret | |
| void callsquare(volatile int *ptr) | |
| { |
| #include <stdint.h> | |
| #include <stdio.h> | |
| #include <limits.h> | |
| //code from Jserv | |
| //int max(int32_t a, int32_t b) { return a * sign(a - b) + b * (1 ^ sign(a - b)); } | |
| int incline sign(int32_t a) { return 1 ^ ((a >> 31) & 0x1); } | |
| //end of code from Jserv | |
| int max(int32_t a, int32_t b) { |
##Binary Search. ###Question: We are going to find a number A in a sequence of numbers N_1, ..., N_2^n N_i != N_j for all i!=j, i,j >=1 <=n The only way we interact with the numbers is a guessing function F(), which takes argument of integer and returns 1 for A>the parameter, -1 for A< the parameter and 0 for correct guess.
Show that we can find A in less than n guess, call F() in less/equal to n times.
| #define u 0 /* don't care */ | |
| #include <stdint.h> | |
| #include <stdio.h> | |
| static uint8_t table[64] = { | |
| 32, 31, u, 16, u, 30, 3, u, | |
| 15, u, u, u, 29, 10, 2, u, | |
| u, u, 12, 14, 21, u, 19, u, | |
| u, 28, u, 25, u, 9, 1, u, | |
| 17, u, 4, u, u, u, 11, u, | |
| 13, 22, 20, u, 26, u, u, 18, |
| @prologue... | |
| @move x to r0 | |
| @r2=n=1 | |
| cmp r0, #0 | |
| beq .L6 | |
| LSR r1, r0, LSR #16 | |
| cmp r1 #0 | |
| beq .L1; | |
| LSR r1, r0, LSR #24 | |
| cmp r1 #0 |
| #include <stdio.h> | |
| #include <stdint.h> | |
| int clz(uint32_t v) { | |
| uint32_t index; | |
| static const uint8_t pos[32] = {32, 31, 4, 30, 3, 18, 8, 29, | |
| 2, 10, 12, 17, 7, 15, 28, 24, 1, 5, 19, 9, 11, 13, | |
| 16, 25, 6, 20, 14, 26, 21, 27, 22, 23}; | |
| v |= v >> 1;//start | |
| v |= v >> 2; | |
| v |= v >> 4; |
