模板专门化函数的参数 [英] template to specialize a function's parameter

问题描述

我有一个函数 int f（int x，int y）需要调用自己很多次，其中一个参数固定，如

I have a function int f (int x, int y) which needs to call itself plenty of times, with one of its parameters fixed, as in

int f(int x, int y) { 
      ...
      int i = f(z,y);
      ...
}

有没有通过模板定义的方法一个函数 int g（int x）使得 g（z）：= f（z，y）上述调用将是 int i = g（z）？

Is there any way of defining via a template a function int g (int x) such that g(z) := f(z,y) so that the above call would have been int i = g(z)?

推荐答案

You can just define that without any templating,

auto f( int x, int y )
    -> int
{
    auto g = [=]( int z ) -> int { return f( z, y ); };
    // ...
    int i = g( z );
}

您可以省略 - >

You can omit the -> int result type specification for g if you want.

免责声明：代码没有被触及。

Disclaimer: code not touched by compiler's hands.

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