在C和C寄存器变量的地址++ [英] address of register variable in C and C++

问题描述

我知道寄存器变量的概念和它的使用案例，但也有基于什么我试图在我的脑海几个问题。

1. 我无法访问在C寄存器变量的地址，但我能做到这一点的C ++！为什么？是否有访问寄存器变量的地址？

的任何问题


2. 假设，如果我用C声明一个字符串变量++为寄存器，然后在那里将变量存储？什么是声明存储类非数字数据类型，如C'串'++是注册??

点

更新：
我认为C ++允许我们取一个寄存器变量的地址，因为我没有得到我的程序中的任何错误是如下：

 ＃包括LT＆;＆iostream的GT;
＃包括LT＆;＆time.h中GT;使用命名空间std;clock_t表示求，连接;诠释主（）{    INT J，K = 0;    求=时钟（）;
    对于（寄存器INT I = 0; I＆LT;千万，我++）{
        / *如果（K == 0）{
            COUT＆LT;＆LT;＆安培; I＆LT;＆LT; ENDL; //如果code是取消注释，那么C ++拒绝的建议，使我为寄存器
            ķ++;
        } * /
    }
    EN =时钟（）;    COUT＆LT;＆LT; EN-求＆LT;＆LT; ENDL;    COUT＆LT;＆LT;＆放大器; J＆LT;＆LT;＆ENDL LT;＆LT;＆安培; k;    返回0;
}
 

我所观察到的是，如果我把这些变量'我'为寄存器，不要尝试使用的地址打印'和;我'，那么C ++接受这项建议，并存储在寄存器'我'，这可能是从运行的时间循环，这将始终是围绕4-12毫秒，如果'我'是寄存器infered。但是，如果我尝试打印变量i的话，虽然我没有得到任何错误，但C ++拒绝该建议，这可以从循环的执行时间，这始终是超过25，如果我没有登记被infered地址！ ！

所以，基本上我不能存储类取一个变量的地址在C寄存器以及C ++！为什么？

解决方案

C和C ++是不同的语言。

• 在C，你不能把一个变量与地址寄存器存储。参看C11 6.7.1 / 6：

  
    

  的标识符与存储类说明的对象的声明注册
    建议访问对象尽可能快。到这样的程度
    建议是有效的是实现定义的。

  
    
    

  脚注：实现可能只是将任何注册声明为汽车声明。 [...]

  
  




• 在C ++中，注册是去precated，有没有效果（也许除了作为一个编译器提示）无意义的关键字，和变量声明为注册仍然只是具有自动存储。特别是，C ++的没有的一个注册存储类。 （它只是存储类别的说明的，从C继承）参看C ++ 11 7.1.1 / 3：

  
    

  A 注册符是一个提示，这么声明的变量将被大量使用的实施。 [注：该提示可以忽略，而且在大多数实现中，如果变量的地址取它会被忽略。这种使用德precated [...]

  
  

即使在C没有什么实际保证有关如何注册存储的实现（实现可随意对待注册为汽车），但无论语言规则。

I know the concept of register variable and it's use cases but there are few questions in my mind based on what I have tried.

1. I cannot access the address of a register variable in C though I can do it C++! Why? Is there any issue in accessing the addressing of a register variable?

2. Suppose if I declare a string variable in C++ as register, then where will that variable be stored? What is the point in declaring the storage class of non-numeric data types such as 'string' in C++ to be register??

UPDATE: I thought that C++ allows us to fetch the address of a register variable, as I was not getting any error in my program which is as follows:

#include<iostream>
#include<time.h>

using namespace std;

clock_t beg, en;

int main(){

    int j, k=0;

    beg=clock();
    for(register int i=0;i<10000000;i++){
        /*if(k==0){
            cout<<&i<<endl;    // if this code is uncommented, then C++ rejects the recommendation to make 'i' as register
            k++;
        }*/
    }
    en=clock();

    cout<<en-beg<<endl;

    cout<<&j<<endl<<&k;

    return 0;
}

What I have observed is, if I make the variable 'i' as register and don't try to print the address using '&i' then C++ accepts the recommendation and stores 'i' in register, this can be infered from running time of for loop which will always be around 4-12 ms if 'i' is in register. But if I try to print address of variable 'i' then though I don't get any error but C++ rejects the recommendation and this can be infered from the time of execution of loop which is always more than 25 if i is not register!!

So, basically I cannot fetch address of a variable with storage class as register in both C as well as C++!! WHY?

解决方案

C and C++ are different languages.

• In C, you cannot take the address of a variable with register storage. Cf. C11 6.7.1/6:

  A declaration of an identifier for an object with storage-class specifier register suggests that access to the object be as fast as possible. The extent to which such suggestions are effective is implementation-defined.

  Footnote: The implementation may treat any register declaration simply as an auto declaration. [...]

• In C++, register is a deprecated, meaningless keyword that has no effect (except perhaps serve as a compiler hint), and variables declared as register still just have automatic storage. In particular, C++ doesn't have a "register" storage class. (It just has the storage class specifier, inherited from C.) Cf. C++11, 7.1.1/3:

  A register specifier is a hint to the implementation that the variable so declared will be heavily used. [ Note: The hint can be ignored and in most implementations it will be ignored if the address of the variable is taken. This use is deprecated [...]

Even in C nothing is actually guaranteed about how register storage is implemented (implementations are free to treat register as auto), but the language rules apply regardless.

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