面试2

//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  1 #include 
  2 
  3 struct data {
  4     long l;
  5     int i;
  6     int *pi;
  7     char c;
  8 
  9     short a[5];
 10 };
 11 
 12 int main(void)
 13 {
 14     struct data *dat;
 15     
 16     dat = 0x100000;
 17 
 18     printf("sizeof(struct data) = %d\n", sizeof(struct data));
 19     printf("dat = %p\n", dat);
 20     printf("dat + 0x200 = %p\n", dat + 0x200);
 21     printf("(unsigned long)dat + 0x200 = %p\n", (unsigned long)dat + 0x200);
 22     printf("(char *)dat + 0x200 = %p\n", (char *)dat + 0x200);
 23 
 24     return 0;
 25 }
 /* 输出
sizeof(struct data) = 24
dat = 0x100000
dat + 0x200 = 0x103000
(unsigned long)dat + 0x200 = 0x100200
(char *)dat + 0x200 = 0x100200
*/



//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#include 

int main(void)
{
    char *array0[10][10];
    char (*array1[10][10]);
    char (*array2)[10][10];
    char (*array3[10])[10];
    char (*array4)[][100];
    char (*array5)[10][100];

    printf("sizeof(array0) = %d\n", sizeof(array0));
    printf("sizeof(array1) = %d\n", sizeof(array1));
    printf("sizeof(array2) = %d\n", sizeof(array2));
    printf("sizeof(array3) = %d\n", sizeof(array3));
    printf("sizeof(array4) = %d\n", sizeof(array4));
    printf("sizeof(array5) = %d\n", sizeof(array5));

    return 0;
}
/* 输出
sizeof(array0) = 400
sizeof(array1) = 400
sizeof(array2) = 4
sizeof(array3) = 40
sizeof(array4) = 4
sizeof(array5) = 4
*/


//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  1 #include 
  2 #include 
  3 
  4 int main(void)
  5 {
  6     char a[20];
  7     strcpy(&a[0], "ces");
  8     strcpy(&a[1], "adfdfa");
  9     strcpy(&a[2], "hgfa");
 10 
 11     printf("%s\n", a);
 12 
 13     return 0;
 14 }
/* 输出
  cahgfa
 */



//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  1 #include 
  2 #include 
  3 
  4 int main(void)
  5 {
  6     char a[] = "hello\0world";
  7 
  8     char b[20];
  9 
 10     memcpy(b, a, 12);
 11 
 12     printf("strlen(a) = %d\n",strlen(a));
 13     printf("sizeof(b) = %d\n", sizeof(b));
 14 
 15     return 0;
 16 }
/* 输出
strlen(a) = 5
sizeof(b) = 20
*/


编程题:

//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1. 冒泡->从小到大排序
#include 
#include 

void sort(char *p, int len)
{
    int i, j;
    char tmp;

    for(i = 0; i < len-1; ++i)
    {
        for(j = 0; j < len-i-1; ++j)
        {
            if(p[j] > p[j+1])
            {
                 tmp = p[j];
                 p[j] = p[j+1];
                 p[j+1] = tmp;
            }
        }
    }
}

int main(void)
{
    int i;
    char a[] = {4, 2, 3, 8 ,7 ,6, 9, 0};
    int len = sizeof(a);

    sort(a, len);
    for(i = 0; i < len; ++i)
    {
        printf("%d ", a[i]);
    }
    printf("\n");

    return 0;
}
/* 输出
0 2 3 4 6 7 8 9 
*/


//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/* 2.
 * 输出字符串中第一个未重复的字符, 如有字符串"dggadgcdp",
 * 结果应该输出'a'.
 */