C基准语义是不是在调用者的code明确 [英] c reference semantic isn't explicit in caller's code

问题描述

有人能解释一下吗？

  

老的C程序员有时并不喜欢引用因为它们提供引用语义，是不是在调用者的code明确。

解决方案

当您使用C ++函数如下：

 无效make42（INT＆安培; X）{X = 42; }
 

，因此称之为：

  INT Y = 7;
make42（Y）; //Ÿ现在是42
 

再有就是的在 make42（Y）没有的指示调用是可以改变的。完全一样的函数调用将作出是否的不是的一通-by-reference参数：

 无效trymake42（INT X）{X = 42; }INT Y = 7;
trymake42（Y）; // y为7仍
 

在C code，它会是这样的：

 无效make42为（int * PX）{
    * PX = 42;
}
INT Y = 7;
make42（安培; Y）;
 

这就很清楚，在调用函数时更改是是一个现实的可能性。

那么，这些老线C codeRS会关心的是这样一行：

  someFunction（X）;
 

在那里他们就没有从理念的只是的这是否 X 将改变与否。

我不得不说，这些旧线codeRS之一，它远不如一个问题比报价似乎暗示。在没有看到样机，我甚至不能告诉的参数类型应该是什么，或者有多少，或它们的顺序。

所以，不得不参考原型说要弄清楚，如果参数是通过按值传递或按引用是不是一个真正的大问题。

C＃需要用退出和 REF 参数修饰符了不同的策略，但仍然使得它在呼叫明确该变量是通过按值传递或按引用。

Can anybody explain this?

Old line C programmers sometimes don't like references since they provide reference semantics that isn't explicit in the caller's code.

解决方案

When you use a C++ function like:

void make42 (int &x) { x = 42; }

and call it thus:

int y = 7;
make42 (y); // y is now 42

then there is no indication in the make42 (y) call that y can be changed. Exactly the same function call would be made if it wasn't a pass-by-reference parameter:

void trymake42 (int x) { x = 42; }

int y = 7;
trymake42 (y); // y is still 7

In C code, it would be something like:

void make42 (int *px) {
    *px = 42;
}
int y = 7;
make42 (&y);

making it clear that a change to y is a real possibility when calling the function.

So what these "old line" C coders would be concerned about is a line like:

someFunction (x);

where they would have no idea from just that whether x would change or not.

I have to say that, as one of these old line coders, it's far less of a problem than the quote seems to imply. Without seeing the prototype, I can't even tell what the parameter types should be, or how many there are, or their order.

So, having to refer to said prototype to figure out if a parameter is pass-by-value or pass-by-reference is not really an big deal.

C# takes a different tack with the out and ref parameter modifiers but that still makes it explicit in the call whether the variable is pass-by-value or pass-by-reference.

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