如何告诉static_assert constexpr函数参数是const？

问题描述

我有一个constexpr函数看起来像这样：

  constexpr int foo（int bar）
 { 
 static_assert（bar> arbitrary_number，请使用较低的数字）; 
 
 return something_const;但是，使用GCC 4.6.3编译这段代码时总是告诉我
。
 
 
 
 
错误：'bar'不能出现在常量表达式中
 
 
 
我尝试过类似
  constexpr int foo（constexpr const int bar）
 {
 static_assert（bar> arbitrary_number，Use a lower number please 
 
 return something_const; 
} 
  
但是constexpr不能用于函数参数。
 
 
 
有一些简单的方法告诉编译器bar总是一个编译时常数？
解决方案
如果 bar 始终是编译时常量，则应将函数写为：
  template< int bar> 
 constexpr int foo（）
 {
 static_assert（bar> arbitrary_number，请使用较低的数字）; 
 return something_const; 
} 
  
因为如果你不这样做，已经写入，则在这种情况下，可以使用非const 参数调用函数它只是当你传递非const参数，那么函数将丢失它的constexpr   - 性。
 
 
 
请注意，在上面的代码 arbitrary_number 也应该是常量表达式，否则将无法编译。
 
I have a constexpr function that looks something like this:
constexpr int foo(int bar)
{
   static_assert(bar>arbitrary_number, "Use a lower number please");

   return something_const;
}
However, compiling this with GCC 4.6.3 keeps telling me 
error: 'bar' cannot appear in a constant-expression
I tried something like
constexpr int foo(constexpr const int bar)
{
   static_assert(bar>arbitrary_number, "Use a lower number please");

   return something_const;
}
but constexpr can't be used for function arguments.
Is there some simple way to tell the compiler that bar is always a compile time constant?
 解决方案 
If bar is always compile-time constant, then you should write your function as:
template<int bar>
constexpr int foo()
{
   static_assert(bar>arbitrary_number, "Use a lower number please");
   return something_const;
}
Because if you don't do so, and instead write what you've already written, then in that case, the function can be called with non-const argument as well; it is just that when you pass non-const argument, then the function will loss it's constexpr-ness.
Note that in the above code arbitrary_number should be constant expression as well, or else it will not compile.
                        
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