这个电话的类型是什么？ [英] what is the type for this call?
问题描述
如果我有：
typedef struct str32 {
uint32_t word1;
uint32_t word2;
} word_array;
word_array * my_array;
这是什么数据类型：
myarray- > word1
为什么它是uint32_t类型并且给出了word1的值？不是吗？b $ b它还是一个没有被解除引用的指针吗？
我以为我应该做*（my_array-> word1）才能到达
uint32_t类型，怎么回事？
If i have a :
typedef struct str32{
uint32_t word1;
uint32_t word2;
} word_array;
word_array *my_array;
what would be the data type of this:
myarray->word1
How come it is a uint32_t type and is giving the value of word1? isn''t
it still a pointer which is not dereferenced yet?
I was thinking I should have to do *(my_array->word1) to get to
uint32_t type, what goes?
推荐答案
Neo写道：
Neo wrote:
如果我有：
typedef struct str32 {
uint32_t word1;
uint32_t word2;
} word_array;
If i have a :
typedef struct str32{
uint32_t word1;
uint32_t word2;
} word_array;
（你为什么要将结构称为str32而类型为word_array？
这两个名字似乎都不合适关系很脆弱。）
(Why are you calling the struct a "str32" and the type "word_array"?
Neither name seems appropriate and their relationship is tenuous.)
word_array * my_array;
这是什么数据类型：
myarray-> word1
word_array *my_array;
what would be the data type of this:
myarray->word1
unint32_t。
unint32_t.
为什么它是uint32_t类型并给出word1的值？
How come it is a uint32_t type and is giving the value of word1?
你将`word1`声明为unint32_t，这就是'它是什么。
You declared `word1` as a unint32_t, so that''s what it is.
难道它还不是一个没有被解除引用的指针吗？
isn''t it still a pointer which is not dereferenced yet?
No.
No.
我以为我应该做*（my_array-> word1）得到
uint32_t类型，怎么回事？
I was thinking I should have to do *(my_array->word1) to get to
uint32_t type, what goes?
`A-> B`是`（*（A.B））`;箭头已经为你做了解除引用。
-
" 2008就是这一切都改变了，你必须做好准备。 " Unsaid / Torchwood /
Hewlett-Packard Limited注册办事处：Cain Road，Bracknell，
注册号：690597 England Berks RG12 1HN
`A->B` is `(*(A.B))`; the arrow has done the dereference for you.
--
"2008 is when it all changes, and you''ve got to be ready." Unsaid /Torchwood/
Hewlett-Packard Limited registered office: Cain Road, Bracknell,
registered no: 690597 England Berks RG12 1HN
Neo写道：
Neo wrote:
如果我有：
typedef struct str32 {
uint32_t word1;
uint32_t word2;
} word_array;
word_array * my_array;
这将是什么数据类型：
myarray-> word1
为什么它是uint32_t类型并且正在给予word1的值？不是吗？b $ b它还是一个没有被解除引用的指针吗？
我以为我应该做*（my_array-> word1）才能到达
uint32_t类型，怎么回事？
If i have a :
typedef struct str32{
uint32_t word1;
uint32_t word2;
} word_array;
word_array *my_array;
what would be the data type of this:
myarray->word1
How come it is a uint32_t type and is giving the value of word1? isn''t
it still a pointer which is not dereferenced yet?
I was thinking I should have to do *(my_array->word1) to get to
uint32_t type, what goes?
据我所知，你使用（* my_array）.word1或my_array-> word1。
括号需要优先。
To my understanding either u use (*my_array).word1 or my_array->word1.
Parenthesis are need for precedence.
Chris Dollin写道：
Chris Dollin wrote:
`A-> B`是`（*（AB））` ;箭头已经为你取消引用。
`A->B` is `(*(A.B))`; the arrow has done the dereference for you.
B不是指针，所以我认为你的符号是错误的。
你应该使用（* A）.B而不是
B is not a pointer, so I think your notation is wrong.
you should use (*A).B instead
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