Int to char [4] [英] Int to char[4]
问题描述
您好，我想知道如何将一个由四个元素组成的char数组
分配给int中使用的四个字节。截至目前我的代码是：
cNameSize =（char）（（void）NameSize）;
cFileSize =（char）（（void）FileSize）;
其中NameSize和FileSize是整数，cNameSize和
cFileSize是4个元素数组。这不行。
Hello, just wondering how I would assign a char array of four elements
to the four bytes used in an int. As of right now my code is:
cNameSize = (char)((void)NameSize);
cFileSize = (char)((void)FileSize);
Where NameSize and FileSize are the integers, and cNameSize and
cFileSize are 4 element arrays. This doesn''t work.
推荐答案
Th ************ @ gmail.com 写道：
您好，只是想知道如何将一个包含四个元素的char数组分配给int中使用的四个字节。截至目前我的代码是：
cNameSize =（char）（（void）NameSize）;
cFileSize =（char）（（void）FileSize）;
其中NameSize和FileSize是整数，cNameSize和
cFileSize是4个元素数组。这不起作用。
Hello, just wondering how I would assign a char array of four elements
to the four bytes used in an int. As of right now my code is:
cNameSize = (char)((void)NameSize);
cFileSize = (char)((void)FileSize);
Where NameSize and FileSize are the integers, and cNameSize and
cFileSize are 4 element arrays. This doesn''t work.
以上两者均无效。你的第一个问题是你不能完全分配数组。最简单的方法就是使用memcpy：
int NameSize;
char cNameSize [4];
memcpy（cNameSize，& NameSize，4）;
Neither of the above are valid. Your first issue is you can''t
assign arrays at all. The easiest way is just to use memcpy:
int NameSize;
char cNameSize[4];
memcpy(cNameSize, &NameSize, 4);
在文章< 11 ******************* **@j27g2000cwj.googlegroups。 com>，
Th************@gmail.com 说...
In article <11*********************@j27g2000cwj.googlegroups. com>,
Th************@gmail.com says...
您好，我想知道如何将四个元素的char数组分配给四个字节
用于int。截至目前我的代码是：
cNameSize =（char）（（void）NameSize）;
cFileSize =（char）（（void）FileSize）;
其中NameSize和FileSize是整数，cNameSize和
cFileSize是4个元素数组。这不行。
Hello, just wondering how I would assign a char array of four elements
to the four bytes used in an int. As of right now my code is:
cNameSize = (char)((void)NameSize);
cFileSize = (char)((void)FileSize);
Where NameSize and FileSize are the integers, and cNameSize and
cFileSize are 4 element arrays. This doesn''t work.
你不可能做任何完全便携的事情。关于它的好处是：
union type_pun {
int i;
char c [sizeof（int）] ;
};
正式来说，它给出了未定义的行为，但是使用典型的编译器你可以将
写入一个int ，然后通过c访问它的各个字节。
-
后来，
杰里。
宇宙是自己想象的虚构。
You can''t do anything entirely portably. About as good as it gets is:
union type_pun {
int i;
char c[sizeof(int)];
};
Officially, it gives undefined behavior, but with a typical compiler you
can write an int into i, and then access its individual bytes via c.
--
Later,
Jerry.
The universe is a figment of its own imagination.
Jerry Coffin写道：
Jerry Coffin wrote:
在文章< 11 ********************* @ j27g2000cwj.googlegroups中。 com>，
Th************@gmail.com 说...
In article <11*********************@j27g2000cwj.googlegroups. com>,
Th************@gmail.com says...
>>您好，我想知道如何将四个元素的char数组分配给四个字节用于int。截至目前我的代码是：
cNameSize =（char）（（void）NameSize）;
cFileSize =（char）（（void）FileSize）;
其中NameSize和FileSize是整数，cNameSize和
cFileSize是4个元素数组。这不行。
>>Hello, just wondering how I would assign a char array of four elements
to the four bytes used in an int. As of right now my code is:
cNameSize = (char)((void)NameSize);
cFileSize = (char)((void)FileSize);
Where NameSize and FileSize are the integers, and cNameSize and
cFileSize are 4 element arrays. This doesn''t work.
你不可能做任何完全便携的事情。关于它的好处是：
union type_pun {
int i;
char c [sizeof（int）] ;
};
正式来说，它给出了未定义的行为，但是使用典型的编译器你可以将
写入一个int ，然后通过c访问它的各个字节。
You can''t do anything entirely portably. About as good as it gets is:
union type_pun {
int i;
char c[sizeof(int)];
};
Officially, it gives undefined behavior, but with a typical compiler you
can write an int into i, and then access its individual bytes via c.
假设你不关心字节顺序。
-
Ian Collins。
Assuming you aren''t concerned with the byte order.
--
Ian Collins.
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