C/C++程序集

作者: 司马懿 | 来源:发表于2013-04-08 11:11 被阅读483次
    #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); 
    }
    

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