C Programming FAQs: Frequently Asked Questions 笔记

// 使用qsort 和 strcmpint pstrcmp(const void *p1, const void *p2) { return strcmp(*(char * const *)p1, *(char * const *)p2);}int main(int argc, char **argv) { char *strings[NSTRING]; int nstrings; qsort(strings, nstrings, sizeof(char *), pstrcmp); }

 

// 同时输出到两个地方#include <stdio.h>#include <stdarg.h>void f2printf(FILE *fp1, FILE *fp2, char *fmt, ...) { va_list argp; va_start(argp, fmt); vfprintf(fp1, fmt, argp); va_end(argp); va_start(argp, fmt); vfprintf(fp2, fmt, argp); va_end(argp);}

 

// convert integers to binary or hexadecimalchar *baseconv(unsigned int num, int base) {static char retbuf[33];char *p; if(base < 2 || base > 16) { return NULL; }p = &retbuf[sizeof(retbuf)-1];*p = '/0';do {*--p = "0123456789abcdef"[num % base];num /= base;} while(num != 0);return p;}// determine whether a machine's byte order is big-endian or little-endianint isLE() { int x = 1; return (*(char *)&x == 1) ? 1 : 0;}// count the number of bits which are set in an integerint bitcount(unsigned int u) { const static int bitcounts[] = {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4};int n = 0; for(; u != 0; u >>= 4) {n += bitcounts[u & 0x0f]; }return n;}// 调换字节void swap(char *ptr, int nwords) { char *p = ptr; while (nwords--) { char tmp = *p; *p = *(p+1); *(p+1) = tmp; p += 2; }}

 

// match for * ?int match(char *pat, char *str) {switch(*pat) { case '/0': return !*str; case '*': return match(pat+1, str) || *str && match(pat, str+1); case '?': return *str && match(pat+1, str+1); default: return *pat == *str && match(pat+1, str+1);}}

 

// match for * . ^ $ from beautiful code int matchstar(int c, char *regexp, char *text);// matchhere: search for regexp at beginning of textint matchhere(char *regexp, char *text) { if (regexp[0] == '/0') { return 1; } if (regexp[1] == '*') { return matchstar(regexp[0], regexp+2, text); } if (regexp[0] == '$' && regexp[1] == '/0') { return *text == '/0'; } if (*text!='/0' && (regexp[0]=='.' || regexp[0]==*text)) { return matchhere(regexp+1, text+1); } return 0;}// matchstar: search for c*regexp at beginning of textint matchstar(int c, char *regexp, char *text) { do {// a * matches zero or more instances if (matchhere(regexp, text)) { return 1; } } while (*text != '/0' && (*text++ == c || c == '.')); return 0;}// match: search for regexp anywhere in textint match(char *regexp, char *text) { if (regexp[0] == '^') { return matchhere(regexp+1, text); } do {// must look even if string is empty */ if (matchhere(regexp, text)) { return 1; } } while (*text++ != '/0'); return 0;}

 

// 0 = Sundayint dayofweek(int y, int m, int d) { static int t[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4}; y -= m < 3; return (y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;}

char *commaprint(unsigned long n) {static int comma = '/0';static char retbuf[30]; char *p = &retbuf[(4*sizeof(long)*CHAR_BIT+2)/3/3+1];int i = 0;if(comma == '/0') {struct lconv *lcp = localeconv();if(lcp != NULL) {comma = (lcp->thousands_sep != NULL && *lcp->thousands_sep != '/0') ? *lcp->thousands_sep : ',';}}*p = '/0';do { if(i%3 == 0 && i != 0) {*--p = comma; }*--p = '0' + n % 10;n /= 10;i++;} while(n != 0);return p;}

 

// Here is a portable C implementation of the ``minimal standard'' generator proposed by Park and Miller: #define a 16807#define m 2147483647#define q (m / a)#define r (m % a)static long int seed = 1;long int PMrand() {long int hi = seed / q;long int lo = seed % q;long int test = a * lo - r * hi; seed = (test > 0) ? test : test + m;return seed;}#include <stdlib.h>(int)((double)rand() / ((double)RAND_MAX +1) * N)rand() / (RAND_MAX / N + 1)unsigned int x = (RAND_MAX + 1U) / N;unsigned int y = x * N;unsigned int r;do { r = rand();} while (r >= y);return r / x;#include <time.h>srand((unsigned int )time((time_t *)NULL));// [M,N]范围的随机整数M + rand() / (RAND_MAX /(N - M + 1)+1)

 

// 怎样产生正态分布或高斯分布的随机数// Exploit the Central Limit Theorem (``law of large numbers'') and add up several uniformly-distributed random numbers: // 中心极限定理（大数定律）#include <stdlib.h>#include <math.h>#define NSUM 25double gaussrand() { double x = 0; for(int i = 0; i < NSUM; i++) { x += (double)rand() / RAND_MAX; }x -= NSUM / 2.0;x /= sqrt(NSUM / 12.0); return x;}// Use a method described by Abramowitz and Stegun: #include <stdlib.h>#include <math.h>#define PI 3.141592654double gaussrand() {static double U, V;static int phase = 0;double Z;if(phase == 0) {U = (rand() + 1.) / (RAND_MAX + 2.);V = rand() / (RAND_MAX + 1.);Z = sqrt(-2 * log(U)) * sin(2 * PI * V); } else {Z = sqrt(-2 * log(U)) * cos(2 * PI * V); }phase = 1 - phase;return Z;}// Use a method discussed in Knuth and due originally to Marsaglia: #include <stdlib.h>#include <math.h>double gaussrand() {static double V1, V2, S;static int phase = 0;double X;if(phase == 0) {do {double U1 = (double)rand() / RAND_MAX;double U2 = (double)rand() / RAND_MAX;V1 = 2 * U1 - 1;V2 = 2 * U2 - 1;S = V1 * V1 + V2 * V2; } while(S >= 1 || S == 0);X = V1 * sqrt(-2 * log(S) / S); } else {X = V2 * sqrt(-2 * log(S) / S); }phase = 1 - phase;return X;}

 

// Here is a function which concatenates an arbitrary number of strings into malloc'ed memory: #include <stdlib.h>/* for malloc, NULL, size_t */#include <stdarg.h>/* for va_ stuff */#include <string.h>/* for strcat et al. */char *vstrcat(const char *first, ...) {va_list argp; if(first == NULL) { return NULL; }size_t len = strlen(first);va_start(argp, first); while((char *p = va_arg(argp, char *)) != NULL) {len += strlen(p); }va_end(argp);char *retbuf = malloc(len + 1);/* +1 for trailing /0 */ if(retbuf == NULL) { return NULL; }(void)strcpy(retbuf, first);va_start(argp, first);/* restart; 2nd scan */ while((char *p = va_arg(argp, char *)) != NULL) {(void)strcat(retbuf, p); }va_end(argp);return retbuf;}// example :char *str = vstrcat("Hello, ", "world!", (char *)NULL);

 

// 输入一个百分比进度器for(i = 0; i < lotsa; i++) {printf("/r%3d%%", (int)(100L * i / lotsa));fflush(stdout);do_timeconsuming_work();}printf("/ndone./n");// 旋转棒printf("working: ");for(i = 0; i < lotsa; i++) {printf("%c/b", "|/-//"[i%4]);fflush(stdout);do_timeconsuming_work();}printf("done./n");

 

#include <limits.h>/* for CHAR_BIT */#define BITMASK (b) (1 << ((b) % CHAR_BIT))#define BITSLOT (b) ((b) / CHAR_BIT)#define BITSET (a, b) ((a)[BITSLOT(b)] |= BITMASK(b))#define BITCLEAR (a, b) ((a)[BITSLOT(b)] &= ~BITMASK(b))#define BITTEST (a, b) ((a)[BITSLOT(b)] & BITMASK(b))#define BITNSLOTS(nb) ((nb + CHAR_BIT - 1) / CHAR_BIT)//#define XOR(a,b) (!!(a) ^ !!(b))#define XOR(a,b) (!!(a)!=!!(b))// 声明一个47位的"数组"char bitarray[BITNSLOTS(47)];// 置第23位BITSET(bitarray, 23);// 测试第35位if(BITTEST(bitarray, 35)) ...// 计算两个位数组的并，再将结果放入另一个数组for(i = 0; i < BITNSLOTS(47); i++) { array3[i] = array1[i] | array2[i];}// 要计算交，使用 & 而不是 |/****功能 | 示例 | 位运算 ----------------------+--------------------- ----+-------------------- 去掉最后一位 | (101101->10110) | x >> 1 在最后加一个0 | (101101->1011010) | x < < 1 在最后加一个1 | (101101->1011011) | x < < 1+1 把最后一位变成1 | (101100->101101) | x | 1 把最后一位变成0 | (101101->101100) | x | 1-1 最后一位取反 | (101101->101100) | x ^ 1 把右数第k位变成1 | (101001->101101,k=3) | x | (1 < < (k-1)) 把右数第k位变成0 | (101101->101001,k=3) | x & ~ (1 < < (k-1)) 右数第k位取反 | (101001->101101,k=3) | x ^ (1 < < (k-1)) 取末三位 | (1101101->101) | x & 7 取末k位 | (1101101->1101,k=5) | x & ((1 < < k)-1) 取右数第k位 | (1101101->1,k=4) | x >> (k-1) & 1 把末k位变成1 | (101001->101111,k=4) | x | (1 < < k-1) 末k位取反 | (101001->100110,k=4) | x ^ (1 < < k-1) 把右边连续的1变成0 | (100101111->100100000) | x & (x+1) 把右起第一个0变成1 | (100101111->100111111) | x | (x+1) 把右边连续的0变成1 | (11011000->11011111) | x | (x-1) 取右边连续的1 | (100101111->1111) | (x ^ (x+1)) >> 1 去掉右起第一个1的左边 | (100101000->1000) | x & (x ^ (x-1)) 判断奇数 | | (x&1)==1 判断偶数 | | (x&1)==0 ***/// 用筛选法（Sieve of Eratosthenes）来计算素数的快速实现#include <stdio.h>#include <string.h>#define MAX 10000int main() {char bitarray[BITNSLOTS(MAX)];int i, j;memset(bitarray, 0, BITNSLOTS(MAX));for(i = 2; i < MAX; i++) {if(!BITTEST(bitarray, i)) {printf("%d/n", i); for(j = i + i; j < MAX; j += i) { BITSET(bitarray, j); }}}return 0;}