Question

在试图通过以下法典将记忆中的数据读成一卷时,我得出了这些重要成果:

void read_memory (const unsigned, const unsigned, unsigned* const);

/* ... */

enum {DATA_0, DATA_1, DATA_2} data;
read_memory(base_addr, offset, &data);  //data = 0x0900
data >>= 8;  //data = 0x7e000000

我在此工作时采用了一种临时的未签署类型。但我很想知道,以前的方法为何不可行。

First of all, I m aware that the standard does not require a specific width for enum types, as long as all members can be represented. As a matter of fact, 6.7.2.2 states that:

Each enumerated type shall be compatible with char, a signed integer type, or an unsigned integer type.

But since the raw data read from memory fits in a char, I think it should not be an issue. Moreover, if I understand correctly, "compatible" means that you can use it as if it were of such a type. In particular, objects of enumerated type can be operands of bitwise shift operators. I m also aware that signedness may be a problem as we don t know if enum are signed or not. But, as far as I can tell, 0x0900 does not appear to be signed.

因此,问题何在?

Answer 1

在抽烟外测试该守则之后,它发现该守则的行为是正确的,用 g布显示的信息是不准确的。

In fact, analyzing the disassembled machine code revealed that it did not match C source line numbers. As a result, the epilogue of the function was thought (by gdb) to be part of the data >>= 8; instruction. This issue remains strange to me since I compiled and linked all source and object files with -g -O0.

也许汇编者中的一杯......

Edit:实际上,它与gdb没有任何关系,这是汇编者所排放的丑化标志是错误的。

Answer 2

你们(我认为,有坏名)读到多少数据? 如果与<>数据编码>的数值相同,则你有问题。仅仅因为所列举的数值小到<条码>。 <><>has >到<> > > 仅选一个char <>/code>; 或许操作的指令(<>int-型号>的编号更快,而且(或)存在一致性问题。

Answer 3

签署“Enum”型号,因此,当你把标书(1)放在数字轨道上。

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