指针的字节数组

问题描述

我在遇到一些麻烦一个指针声明，我的同事的人想的，因为MISRA C的要求使用。米斯拉（安全关键指标）不会让我们单纯的程序员使用指针，反而会让我们的字节数组操作。他打算指针procur以字节数组（所以我们没有在栈上传递的实际阵列。）

I'm having some trouble with a pointer declaration that one of my co-workers wants to use because of Misra C requirements. Misra (Safety Critical guideline) won't let us mere Programmers use pointers, but will let us operate on arrays bytes. He intends to procur a pointer to an array of bytes (so we don't pass the actual array on the stack.)

// This is how I would normally do it
//
void Foo(uint8_t* pu8Buffer, uint16_t u16Len)
{
}

// This is how he has done it
//
void Foo(uint8_t (*pu8Buffer)[], uint16_t u16Len)
{
}

调用函数看起来类似;

The calling function looks something like;

void Bar(void)
{
    uint8_t  u8Payload[1024]
    uint16_t u16PayloadLen;

    // ...some code to fill said array...

    Foo(u8Payload, u16PayloadLen);
}

但是，当pu8Buffer在美孚（访问），该阵列是错误的。显然不是通过什么期待。该阵列是在调用函数正确的，但里面没有美孚（）

But, when pu8Buffer is accessed in Foo(), the array is wrong. Obviously not passing what it is expecting. The array is correct in the calling function, but not inside Foo()

我觉得他创造了一个指针数组的字节，而不是一个指针字节数组。

I think he has created an array of pointers to bytes, not a pointer to an array of bytes.

有人在乎澄清？美孚（安培; u8Payload，u16PayloadLen）;也不行。

Anyone care to clarify? Foo(&u8Payload, u16PayloadLen); doesn't work either.

推荐答案

在无效美孚（uint8_t有（* pu8Buffer）[]，uint16_t u16Len）， pu8Buffer 是一个指向 uint8_t有的（不完全）阵列。 pu8Buffer 有一个不完整的类型;这是一个指向数组的大小是未知的。它可能不会在需要规模前pressions使用。（如指针运算; pu8Buffer + 1 是不允许的）

In void Foo(uint8_t (*pu8Buffer)[], uint16_t u16Len), pu8Buffer is a pointer to an (incomplete) array of uint8_t. pu8Buffer has an incomplete type; it is a pointer to an array whose size is unknown. It may not be used in expressions where the size is required (such as pointer arithmetic; pu8Buffer+1 is not allowed).

然后 * pu8Buffer 是一个数组，其大小是未知的。因为它是一个数组，它会自动在大多数情况下转换为指针到它的第一个元素。因此， * pu8Buffer 成为一个指向第一个 uint8_t有的数组。该类型的转换 * pu8Buffer 齐全;这是一个指向 uint8_t有，因此它可能在地址运算中使用; *（* pu8Buffer + 1），（* pu8Buffer）[1] 和 1 [* pu8Buffer] 是对 uint8_t有一举超越 * pu8Buffer 。

Then *pu8Buffer is an array whose size is unknown. Since it is an array, it is automatically converted in most situations to a pointer to its first element. Thus, *pu8Buffer becomes a pointer to the first uint8_t of the array. The type of the converted *pu8Buffer is complete; it is a pointer to uint8_t, so it may be used in address arithmetic; *(*pu8Buffer + 1), (*pu8Buffer)[1], and 1[*pu8Buffer] are all valid expressions for the uint8_t one beyond *pu8Buffer.

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