C/C++程序集

#include <stdio.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <arpa/inet.h>

int main()
{
        int s, afd;
        int len;
        struct sockaddr_in sin, csin;
        char buf[32] = "";

        memset(&sin, 0, sizeof(sin));

        sin.sin_family = AF_INET;
        sin.sin_port = htons(5678);
        sin.sin_addr.s_addr = INADDR_ANY;
        //sin.sin_addr.s_addr = INADDR_ANY;

        s = socket(PF_INET, SOCK_STREAM, 0);
        if (bind(s, (struct sockaddr *)&sin, sizeof(sin)) < 0) {
                printf("bind error\n");
        }
        if (listen(s, 32) < 0 )
                printf("listen error\n");

        while (1) {
                afd = accept(s, (struct sockaddr *)&csin, &len);
                if (afd < 0) {
                        printf("accept error\n");
                        continue;
                }
                if (read(afd, buf, sizeof(buf)) > 0)
                        printf("%s\n", buf);
                close(afd);
        }
}

执行‘telnet localhsot 5678’测试。 
tcp 一对一
udp 一对多 多对一 多对多

质数验证程序：

-bash-3.2$ more prime.c
#include <stdio.h>
#include <time.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
 
#define  E9        1072000000 
#define  E10       10000000000
#define  E11       100000000000
#define  E12       1000000000000
 
char table[E9];
 
int num = 0;
//int prime[E8];
 
void cal_table()
{
        int j, k;
        time_t t1, t2;
 
        t1 = time(NULL);
 
        memset(table, 1, E9);
        for (j = 2 ; j < E9; j++) {
                if ( table[j] ) {
                        num++;
                        for ( k = j + j; k < E9; k += j )
                        {
                                table[k] = 0;
                        }
                }
        }
        t2 = time(NULL);
 
        printf("Totaly %d primes until E9, cost %d time_t.\n", num, t2 - t1);
}
int main()
{
    int i;
    cal_table(); 
    return 0;
}
 
-bash-3.2$ gcc prime.c
-bash-3.2$ ./a.out
Totaly 54316419 primes until E9, cost 179 time_t.
-bash-3.2$

Little endian把低字节存放在内存的低位；而Big endian将低字节存放在内存的高位；Intel的CPU是little endian，Sun Sparc的CPU是big endian。

假设从地址0x00000000开始的一个字(WORD)中存有数据0x1234abcd。

Little endian机器，从低到高内存的存放顺序是：0x00000000-0xcd,0x00000001-0xab,0x00000002-0x34,0x00000003-0x12 

Big endian机器，从低到高内存的存放顺序是：0x00000000-0x12,0x00000001-0x34,0x00000002-0xab,0x00000003-0xcd 

#include <stdio.h> 
int main() 
{ 
    int a = 0x12345678; 
    char *p = (char * )&a; 
    printf("%x %x %x %x\n", *p, *(p+1), *(p+2), *(p+3)); 
    return 0; 
} 

-bash-3.2$ ./a.out 
78 56 34 12 
-bash-3.2$ uname -a 
Linux fedora.unix-center.net 2.6.27.10-grsec2.1.12 #2 SMP Fri May 8 07:04:03 CST 2009 i686 i686 i386 GNU/Linux 

-bash-3.00$ ./a.out 
12 34 56 78 
-bash-3.00$ uname -a 
SunOS t1000 5.10 Generic_118833-33 sun4v sparc SUNW,Sun-Fire-T1000 Solaris 

以下程序在Linux上用GCC编译运行，结果是0.000000，其实这种代码编译器应该报错，最起码是警告，可是.......

#include <stdio.h>
double sum(); 
int main() { 
     int x = 10; 
     int y = 10; 
     double z = sum(x,y); 
     printf("%lf\n", z); 
     return 0; 
} 
double sum(double x, double y) { 
     return (x+y); 
}

反向排列字符串

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

char a[128] = "I am fine thank you asdasd aaaaaa bbbbbb cccccc cccccccccccc"; 

void reserve (char * in, char *out) 
{ 
     char *p = in + strlen(in) - 1; 
     while (1) { 
         while (1) { 
             if (*p == ' ' || p == in) 
                 break; 
             p--; 
         } 
        if (p == in) { 
            strcat(out, p); 
            return; 
        } 
        strcat(out, p+1); 
        strcat(out, " "); 
        while (*(p - 1) == ' ') 
            p--; 
        *p = '\0'; 
    } 
} 

int main() 
{ 
    char *b; 
    b = malloc(128); 
    reserve(a, b); 
    printf("%s\n", b); 
    free(b); 
    return 0; 
} 

打印1-20的全排列。 

#include<stdio.h> 
void print_it(int n, int arr[]); 
void arrange_all(int size, int arr[], int pos); 
int main() 
{ 
    int a[20] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}; 
    arrange_all(20, a, 0); 
    return 0; 
} 
void arrange_all(int size, int arr[], int pos) 
{ 
    int i, tmp; 
    if (pos + 1== size)  { 
        print_it(size, arr); 
        return; 
    } 
    for (i = pos; i < size; i++)  { 
        tmp = arr[pos]; 
        arr[pos] = arr[i]; 
        arr[i] = tmp; 
        arrange_all(size, arr, pos + 1); 
        tmp = arr[pos]; 
        arr[pos] = arr[i]; 
        arr[i] = tmp; 
    } 
} 
void print_it(int n, int arr[]) 
{ 
    int i; 
    static int cnt = 0; 
    printf("%21d: ", ++cnt); 
    for (i = 0; i < n; i++) 
         printf("%4d ", arr[i]); 
    printf(""); 
} 

#include <stdio.h> 

int func(int n) 
{ 
    int a[30]; 
    int i, j, t; 
    int count=0; 

    for(i=0;i<n;i++) a[i]=i+1; 
    for(;;) { 
        printf("%4d: ", ++count); 
        for(i=0;i<n;i++) printf("%d ", a[i]); 
        printf("\n"); 
        for(i=n-2;i>=0;i--) if(a[i]<a[i+1]) break; 
        if(i<0) break; 
        for(j=n-1;;j--) if(a[j]>a[i]) break; 
        t=a[j], a[j]=a[i], a[i]=t; 
        i++, j=n-1; 
       while(i<j) { 
            t=a[j], a[j]=a[i], a[i]=t; 
            i++, j--; 
        } 
    } 
    return(0); 
} 

main() 
{ 
    func(20); 
}