元组std :: get（）不适用于变量定义的常量

问题描述

我一直遇到以下问题：

std::tuple<Ts...> some_tuple = std::tuple<Ts...>();
for (int i = 0; i < 2; i++) {
  const int j = i;
  std::get<j>(temp_row) = some_value;
}

不编译：（在xcode中）它说没有匹配的函数调用'get'。

Does not compile: (in xcode) it says " no matching function call for 'get' ".

但是，以下工作正常：

std::tuple<Ts...> some_tuple = std::tuple<Ts...>();
for (int i = 0; i < 2; i++) {
  const int j = 1;
  std::get<j>(temp_row) = some_value;
}

是否与将常量的值定义为

Does it have something to do with defining the value of a constant to be the value of a variable?

谢谢！

编辑：区别在于 const int j = i; 与 const int j = 1;

推荐答案

这里的问题是C ++具有两种不同的常量。有编译时间常数和运行时间常数。编译时常数是在编译时已知的常数，并且是可以在模板或数组大小中使用的唯一有效常数。运行时常数是一个不能更改的值，但该值直到运行时才知道。这些常量不能用作模板或数组大小的值。因此，在

The problem here is C++ has two different kinds of constants. There are compile time constants and there are run time constants. A compile time constant is a constant that is know at compile time and is the only valid constant that can be used in a template or as an array size. A run time constant is a value that cannot change but the value is something that is not known until run time. These constants cannot be used as a value for a template or an array size. So in

std::tuple<Ts...> some_tuple = std::tuple<Ts...>();
for (int i = 0; i < 2; i++) {
  const int j = i;
  std::get<j>(temp_row) = some_value;
}

i 是一个使得 j 为运行时常数的运行时值，您无法使用它实例化模板。但是在

i is a run time value which makes j a run time constant and you cannot instantiate a template with it. However in

std::tuple<Ts...> some_tuple = std::tuple<Ts...>();
for (int i = 0; i < 2; i++) {
  const int j = 1;
  std::get<j>(temp_row) = some_value;
}

const int j = 1; 是一个编译时间常数，可用于实例化模板，因为模板的值在编译时是已知的。

const int j = 1; is a compile time constant and can be used to instantiate the template as its value is known at compile time.

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