正确的方式来访问数组成员为不同类型的？ [英] Proper way to access array members as a different type?

问题描述

我有一个大阵的uint16_t 。

它的大部分成员是 uint16_t ，但有些是 int16_t 有的 uint8_t有。

你会如何处理这个问题？

---

顺便说一句，我想：

1. 指针：

   用于2球，有一个 int16_t * 和其他 uint8_t有* ，既初始化数组的开始，访问数组成员是 int16_t 和 uint8_t有。

   （即最初的工作，但我遇到了问题时，后面的程序别的东西改变了指针的值，所以我不相信它。）




2. 类型定义为联合。

   在file.h：

    的typedef工会{
     uint16_t U16 [NO_OF_WORDS] //由于uint16_t
     int16_t S16 [NO_OF_WORDS] //由于int16_t
     uint8_t有U8 [2 * NO_OF_WORDS]。 //由于uint8_t有
   } ram_params_t;
   EXTERN ram_params_t ram_params [];
    

   在file.c中：

     ram_params_t ram_params [] = {0};
    

   （这确实轰炸了。）




3. 铸造。

   （我没有走得很远这一点。）

解决方案

你尝试＃2的问题是，你做一个数组的数组。

做这一点：

 的typedef工会{
  uint16_t U16; //由于uint16_t
  int16_t S16; //由于int16_t
  uint8_t有U8 [2]; //由于uint8_t有
} ram_params_t;
EXTERN ram_params_t ram_params [NO_OF_WORDS]ram_params_t ram_params [NO_OF_WORDS]uval16 = ram_params [I] .u16;
sval16 = ram_params [I] .s16;
uval8_1 = ram_params [I] .u8 [0];
uval8_2 = ram_params [Ⅰ] .u8 [1];
 

或者你做到这一点：

 的typedef工会{
  uint16_t U16 [NO_OF_WORDS] //由于uint16_t
  int16_t S16 [NO_OF_WORDS] //由于int16_t
  uint8_t有U8 [2 * NO_OF_WORDS]。 //由于uint8_t有
} ram_params_t;
EXTERN ram_params_t ram_params;ram_params_t ram_params;uval16 = ram_params.u16 [I]
sval16 = ram_params.s16 [I]
uval8_1 = ram_params.u8 [2 * i];
uval8_2 = ram_params.u8 [2 * I + 1];
 

我看不出什么毛病你尝试＃1无论是。我想，我可能会做，而不是使用一个联盟。

I have a large array of uint16_t.

Most of its members are uint16_t, but some are int16_t and some uint8_t.

How would you handle that?

---

By the way, I tried:

1. Pointers:

   Used 2 pointers, one int16_t* and the other uint8_t*, both initialized to the start of the array, to access member of the array that are int16_t and uint8_t.

   (That worked initially, but I ran into problems when later in the program something else changed the value of the pointers, so I don't trust it.)

2. Type definition with a union.

   In file.h:

   typedef union {
     uint16_t  u16[NO_OF_WORDS];     // As uint16_t
     int16_t   s16[NO_OF_WORDS];     // As int16_t
     uint8_t   u8[2 * NO_OF_WORDS];  // As uint8_t
   } ram_params_t;
   extern ram_params_t ram_params[];
   

   In file.c:

   ram_params_t ram_params[] = {0};
   

   (That really bombed.)

3. Casting.

   (I didn't get very far with that.)

解决方案

The problem with your attempt #2 was that you made an array of arrays.

Either do this:

typedef union {
  uint16_t  u16;     // As uint16_t
  int16_t   s16;     // As int16_t
  uint8_t   u8[2];   // As uint8_t
} ram_params_t;
extern ram_params_t ram_params[NO_OF_WORDS];

ram_params_t ram_params[NO_OF_WORDS];

uval16  = ram_params[i].u16;
sval16  = ram_params[i].s16;
uval8_1 = ram_params[i].u8[0];
uval8_2 = ram_params[i].u8[1];

Or you do this:

typedef union {
  uint16_t  u16[NO_OF_WORDS];     // As uint16_t
  int16_t   s16[NO_OF_WORDS];     // As int16_t
  uint8_t   u8[2 * NO_OF_WORDS];  // As uint8_t
} ram_params_t;
extern ram_params_t ram_params;

ram_params_t ram_params;

uval16  = ram_params.u16[i];
sval16  = ram_params.s16[i];
uval8_1 = ram_params.u8[2*i];
uval8_2 = ram_params.u8[2*i+1];

I don't see anything wrong with your attempt #1 either. I think that I would probably do that rather than using a union.

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