u8指针强制转换成u32

typedef unsigned long  u32;
typedef unsigned short u16;
typedef unsigned char  u8;

int main(void)
{
    u8 i;
    u8 tab[4] = {0x12, 0x34, 0x56, 0x78};
    u8 *p = tab;
    u32 *q = (u32*)p; 

    for (i=0; i<4; i++)
    {
        printf("%#x  ", tab+i);
        printf("%#x\n", tab[i]);
    }

    printf("\n");

    for (i=0; i<4; i++)
    {
        printf("%#x  ", p);
        printf("%#x\n", *p);
        p++;
    }
    
    printf("\n");
    printf("%#x\n", *q);
    
    return 0;
}


/*************************************************
在VC++6.0中的输出结果是：
0x18ff40  0x12
0x18ff41  0x34
0x18ff42  0x56
0x18ff43  0x78

0x18ff40  0x12
0x18ff41  0x34
0x18ff42  0x56
0x18ff43  0x78

0x78563412
Press any key to continue
*************************************************/