我将用代码说明我的问题:
I'll illustrate my question with code:
#include <iostream>
void PrintInt(const unsigned char*& ptr)
{
int data = 0;
::memcpy(&data, ptr, sizeof(data));
// advance the pointer reference.
ptr += sizeof(data);
std::cout << std::hex << data << " " << std::endl;
}
int main(int, char**)
{
unsigned char buffer[] = { 0x11, 0x11, 0x11, 0x11, 0x22, 0x22, 0x22, 0x22, };
/* const */ unsigned char* ptr = buffer;
PrintInt(ptr); // error C2664: ...
PrintInt(ptr); // error C2664: ...
return 0;
}
当我运行这段代码(在 VS2008 中)时,我得到这个:错误 C2664:'PrintInt':无法将参数 1 从 'unsigned char *' 转换为 'const unsigned char *&'.如果我取消注释const"注释,它就可以正常工作.
When I run this code (in VS2008) I get this: error C2664: 'PrintInt' : cannot convert parameter 1 from 'unsigned char *' to 'const unsigned char *&'. If I uncomment the "const" comment it works fine.
但是指针不应该隐式转换为const指针然后引用吗?我期望这行得通有错吗?谢谢!
However shouldn't pointer implicitly convert into const pointer and then reference be taken? Am I wrong in expecting this to work? Thanks!
如果指针被转换为 const 指针,如您所建议的,那么转换的结果是一个临时值,一个 rvalue.您不能将非常量引用附加到右值 - 这在 C++ 中是非法的.
If the pointer gets converted to a const pointer, as you suggest, then the result of that conversion is a temporary value, an rvalue. You cannot attach a non-const reference to an rvalue - it is illegal in C++.
例如,由于类似的原因,此代码将无法编译
For example, this code will not compile for a similar reason
int i = 42;
double &r = i;
即使 int
类型可以转换为 double
类型,它仍然不意味着你可以附加一个 double &
引用到该转换的结果.
Even though type int
is convertible to type double
, it still doesn't mean that you can attach a double &
reference to the result of that conversion.
然而,const 引用(即引用到const 类型的引用)可以附加到右值,这意味着这段代码可以完美地编译
However, a const reference (i.e. a reference of reference-to-const type) can be attached to an rvalue, meaning that this code will compile perfectly fine
int i = 42;
const double &r = i;
在您的情况下,如果您将函数声明为
In your case if you declare your function as
void PrintInt(const unsigned char* const& ptr) // note the extra `const`
代码将被编译.
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