添加长值显示不同的o / p

Question

我在添加长值时遇到问题 示例

 typedef unsigned short  UINT16;

   UINT16* flash_dest_ptr;  // this is equal to in hexa 0XFF910000

   UINT16 data_length ; // hex =  0x000002AA &  dec = 682

  //now when I add  

  UINT16 *memory_loc_ver = flash_dest_ptr + data_length ;

  dbug_printf( DBUG_ERROR | DBUG_NAVD, " ADD hex =0x%08X\n\r",memory_loc_ver   );

实际O / p = 0xFF910554

// shouldn't o/p be FF9102AA ?

Answer 1

它的指针算术，所以

UINT16 *memory_loc_ver = flash_dest_ptr + data_length ;

将flash_dest_ptr字节提升data_length * sizeof (UINT16)。

通常，sizeof (UINT16)为2，

2 * 0x2AA = 0x554

Answer 2

当您将整数添加到指针值时，实际上是将指针移动到将data_length UINT16移出内存所需的字节数，而不是data_length字节。