为什么不允许'stexpr'参数?

2022-02-27 00:00:00 language-lawyer c++ compile-time c++14

为了区分编译器已知值,从而能够在编译时检测错误,使用‘constexpr’参数会很有用。示例:

int do_something(constexpr int x)
{
  static_assert(x > 0, "x must be > 0");
  return x + 5;
}

int do_something(int x)
{
  if(x > 0) { cout << "x must be > 0" << endl; exit(-1); }
  return x + 5;
}

int var;

do_something(9); //instance 'do_something(constexpr int x)' and check arg validity at compile time

do_something(0); //produces compiler error

do_something(var); //instance 'do_something(int x)'

目前这是无效代码。有人能给我解释一下为什么这不能实施吗?

编辑:

使用模板时,用户必须始终将文字作为模板参数传递,而不是作为函数参数传递,这非常不舒服:

template<int x>
int do_something()
{
  static_assert(x > 0, "x must be > 0");
  return x + 5;
}

int do_something(int x)
{
  if(x > 0) { cout << "x must be > 0" << endl; exit(-1); }
  return x + 5;
}

int var;

do_something(9); //instance 'do_something(int x)' and doesn't check validity at compile time

do_something(0); //same as above, if check was performed - compiler error should occur

do_something<9>(); //instance template 'do_something<int>()'

do_something<0>(); //produces compiler error

do_something(var); //instance 'do_something(int x)'

解决方案

如果我了解您尝试正确执行的操作,则您请求的功能已经可用。这不是最雅致的,但我认为这已经足够好了。

您希望在编译时和运行时使用相同的语法调用函数,如果可能,让它在编译时求值,否则它应该在运行时求值。无论何时调用函数,您都需要对该函数计算断言。

我相信这会满足您的要求:

constexpr int do_something(int x)
{
    if(x <= 0)
    {
        std::cout << "x must be > 0" << std::endl; exit(-1);
    }
    return x + 5;
}

constexpr int compiletime_good = do_something(5);
constexpr int compiletime_bad = do_something(0);    // Fails at compile-time

int runtime_good = do_something(5);
int runtime_bad = do_something(0);    // Fails at runtime

constexpr int val_good = 5;
constexpr int val_bad = 0;

do_something(val_good);
do_something(val_bad);    // Fails at run-time

int valrun_good = 5;
int valrun_bad = 0;

do_something(valrun_good);
do_something(valrun_bad);    // Fails at run-time

这里的诀窍是在编译时以不需要静电_ASSERT的方式失败,在运行时也会失败。

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