为什么（类型*）指针不等于*（类型**）指针？ [英] Why (type*)pointer isn't equal to *(type**)pointer?
问题描述
为什么（类型*）指针不等于*（类型**）指针，
在代码片段中，它显示：
（int *）==（int **），
（int *）！=（*（int **））。
打字改变地址？或者不要打字吗
什么？
1 int main（无效）
2 {
3 char ch;
4 int * p;
5 int * p2;
6 int * p3;
8 ch =''c'';
9 p =（int *）（& ch）;
10 p2 = *（int **）（& ch）;
11 p3 =（int **）（& ch）;
12
13 printf（"％c"，* p）;
14 printf（"％c"，* p2）; / *错误参考* /
15 printf（"％c"，* p3）;
16
17返回0 ;
18}
19
~
~
谢谢
lovecreatesbeauty
lovecreatesbeauty写道：为什么（类型*）指针不等于*（类型**）指针，
在代码片段中，它显示：
（int *）==（int **），
（int *）！=（*（int **））。
类型转换会改变地址吗？或者没有打字做什么？
1 int main（无效）
2 {
3 char ch;
4 int * p;
5 int * p2;
6 int * p3;
7
8 ch =''c'';
9 p =（int *）（ & ch）;
10 p2 = *（int **）（& ch）;
您正在转向指向int的指针并取消引用
结果，这将是''c''和垃圾。
Ian
< blockquote> lovecreatesbeauty写道：
为什么（类型*）指针不等于*（类型**）指针，
好，主要原因是第一个没有取消引用
指针，但第二个确实如此。你不可能期望两个
表达式具有相同的值，如果一个取消引用指针
而另一个没有。
- Logan
是的，我可以抽象int的4个字节中的第一个字节。这种
类型的播放是正常的，经常发生，我想。
Ian写道：
你正在转向指向int的指针并取消引用
结果，这将是''c''和垃圾。
Ian
Why (type*)pointer isn''t equal to *(type**)pointer,
In the code snippet, it shows that:
(int *) == (int **) ,
(int *) != (*(int **)) .
Does type-casting change the address? or doesn''t type-casting do
anything?
1 int main(void)
2 {
3 char ch;
4 int *p;
5 int *p2;
6 int *p3;
7
8 ch = ''c'';
9 p = (int *)(&ch);
10 p2 = *(int **)(&ch);
11 p3 = (int **)(&ch);
12
13 printf("%c", *p);
14 printf("%c", *p2); /* error in de-reference */
15 printf("%c", *p3);
16
17 return 0;
18 }
19
~
~
Thank you
lovecreatesbeauty
lovecreatesbeauty wrote:Why (type*)pointer isn''t equal to *(type**)pointer,
In the code snippet, it shows that:
(int *) == (int **) ,
(int *) != (*(int **)) .
Does type-casting change the address? or doesn''t type-casting do
anything?
1 int main(void)
2 {
3 char ch;
4 int *p;
5 int *p2;
6 int *p3;
7
8 ch = ''c'';
9 p = (int *)(&ch);
10 p2 = *(int **)(&ch);
Your are casting to a pointer to pointer to int and dereferencing the
result, which will be ''c'' and garbage.
Ian
lovecreatesbeauty wrote:Why (type*)pointer isn''t equal to *(type**)pointer,
Well, the main reason is that the first does not dereference the
pointer, but the second does. You can''t possibly expect two
expressions to have the same value if one dereferences a pointer
and the other one does not.
- Logan
Yes, I can abstract the first byte among the 4 bytes of the int. This
kind of type-casting is normal and occurs frequently, I think.
Ian wrote:
Your are casting to a pointer to pointer to int and dereferencing the
result, which will be ''c'' and garbage.
Ian
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