恼人的常见问题 [英] Annoying const problem
问题描述
我有一个总是占用16个字节数据的函数并且没有修改它/ b $ b它：
void func（字节const（* data） ）[16]）;
但是，如果我尝试使用非const数据调用它，编译器将无法执行转换：
static const byte bar1 [16] = {0};
int foo（）
{
byte bar2 [16] = {0};
func（& bar1）; / * OK * /
func（& bar2）; / *错误* /
}
除了定义const和
之外，还有其他解决方法吗？ func的nonconst版本，或者将指针传递给第一个元素
数据，从而丢失编译时长度检查？
Old Wolf写道：
>
我有一个总是占用16个字节数据的函数没有
修改它：
void func（字节常量（*数据）[16]）;
但是，如果我尝试使用非const数据调用它，编译器将无法执行转换：
static const byte bar1 [16] = {0};
int foo（）
{
byte bar2 [16] = {0} ;
func（& bar1）; / * OK * /
func（& bar2）; / *错误* /
}
除了定义const和
之外，还有其他解决方法吗？非对象版本的func，或者将指针传递给第一个元素
的数据，从而失去编译时长度检查？
可能用C编译器编译就足够了。看起来
就像你在使用C ++。
-
Chuck F（cinefalconer at maineline dot net）
可用于咨询/临时嵌入式和系统。
< http://cbfalconer.home.att.net>
在文章< 11 ********************** @ n67g2000cwd.googlegroups .com> ;,
Old Wolf< ; ol ***** @ inspire.net.nzwrote：
> void func（byte const（* data）[16]）;
[推测字节是typedefed。]
>但是，如果我尝试调用它非常数据，编译器无法执行转换：
你必须进行明确的演员：
func（（字节const（*）[16]）& bar2）;
当然，这会丢失类型检查，但仅限于
你施放它的情况。
- 理查德
-
"考虑应该在一些字母表中需要多达32个字符
- X3.4,1963。
CBFalconer写道：
Old Wolf写道：
< blockquote class =post_quotes>
void func（byte const（* data）[16]）;
static const byte bar1 [16] = {0};
int foo（）
{
byte bar2 [16] = {0};
func（& bar1）; / * OK * /
func（& bar2）; / *错误* /
}
可能用C编译器进行编译就足够了。看起来
就像你在使用C ++一样。
其实我不是。谢谢你的猜测。什么C编译器
您使用它编译此代码而不发出诊断？
typedef unsigned char byte;
void func （byte const（* data）[16]）{}
int main（）
{
byte bar [16] = {0 };
func（& bar）;
}
I have a function that always takes 16 bytes of data and doesn''t modify
it:
void func( byte const (*data)[16] );
However, if I try to call it with non-const data, the compiler is
unable to perform the conversion:
static const byte bar1[16] = { 0 };
int foo()
{
byte bar2[16] = { 0 };
func( &bar1 ); /* OK */
func( &bar2 ); /* Error */
}
Is there any work-around to this, other than defining a const and a
nonconst version of func, or passing a pointer to the first element
of data and thereby losing the compile-time length check?
Old Wolf wrote:>
I have a function that always takes 16 bytes of data and doesn''t
modify it:
void func( byte const (*data)[16] );
However, if I try to call it with non-const data, the compiler is
unable to perform the conversion:
static const byte bar1[16] = { 0 };
int foo()
{
byte bar2[16] = { 0 };
func( &bar1 ); /* OK */
func( &bar2 ); /* Error */
}
Is there any work-around to this, other than defining a const and a
nonconst version of func, or passing a pointer to the first element
of data and thereby losing the compile-time length check?Probably it will be enough to compile with a C compiler. Looks
like you are using C++.
--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>
In article <11**********************@n67g2000cwd.googlegroups .com>,
Old Wolf <ol*****@inspire.net.nzwrote:
>void func( byte const (*data)[16] );[Presumably byte is typedefed.]
>However, if I try to call it with non-const data, the compiler is
unable to perform the conversion:You''ll have to put in an explicit cast:
func( (byte const (*)[16]) &bar2 );
Of course, this loses the type checking, but only for the cases where
you cast it.
-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
CBFalconer wrote:Old Wolf wrote:void func( byte const (*data)[16] );
static const byte bar1[16] = { 0 };
int foo()
{
byte bar2[16] = { 0 };
func( &bar1 ); /* OK */
func( &bar2 ); /* Error */
}
Probably it will be enough to compile with a C compiler. Looks
like you are using C++.Actually I''m not. Thanks for guessing though. What C compiler are
you using that compiles this code without issuing a diagnostic?
typedef unsigned char byte;
void func( byte const (*data)[16] ) {}
int main()
{
byte bar[16] = { 0 };
func( &bar );
}
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