元组 std::get() 不适用于变量定义的常量
问题说明
我遇到了以下问题:
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变量的值有关?
Does it have something to do with defining the value of a constant to be the value of a variable?
谢谢！
区别在于 const int j = i; 与 const int j = 1;
正确答案
这里的问题是 C 有两种不同的常量.有编译时常量和运行时常量.编译时常量是在编译时已知的常量，并且是唯一可以在模板中使用或用作数组大小的有效常量.运行时常数是一个不能改变的值，但该值是在运行时才知道的.这些常量不能用作模板或数组大小的值.所以在
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 是一个运行时值，它使 j 成为运行时常量，您不能用它实例化模板.然而在
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; 是编译时常量，可用于实例化模板，因为其值在编译时已知.
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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