如何在C ++中找到字符的内存地址？ [英] How Do I Find The Memory Address Of A Character In C++?

问题描述

我想找到一个角色的记忆地址，但我注意到了一些事情，我无法理解。



char name [] = {hello};



//查找内存地址这就是我所做的名字数组



cout<<名称[]的内存地址是：<<& name []< < endl;



//这个工作正常，现在我想在

名称数组中找到一个字符的内存地址



//说我想找到char h ||的内存地址char e

//所以这就是我找到的地址



cout<<名称的内存地址[0 ]是：<< &（name [0]）<< endl;

//现在这里是问题，我得到错误的结果，我得到的是你好而不是获取内存地址



//我在网上找到了这个技巧，找到了名字[0]的内存地址

cout<<的内存地址name [0]是：<<< static_cast< void>（&（name [0]）<< endl;



我不明白为什么&（name [0]）dint工作的正常方法是什么，static_cast< void>（&（name [0]）的含义是什么？它是做什么的？

请任何人用简单的语言解释一下。谢谢

I want to find the memory address of a character but i have noticed somethings and i am unable to understand it.

char name[] = {"hello"};

// to find the memory address of name array this is what i did

cout<<"The memory address of name[] is:"<<&name[]<<endl;

//this works fine and now i want to find the memory address of a character in the
name array

// say i want to find the memory address of char h || char e
//so this is what i did for finding the address

cout<<"The memory address of name[0] is:"<< &(name[0])<<endl;
//now here is the problem, i get wrong results, what i get is "hello" instead of getting the memory address

// i looked online and found this trick to find the memory address of name[0]
cout<<"the memory address of name[0] is:"<<static_cast<void>(&(name[0])<<endl;

I dont understand why the normal approach of &(name[0]) dint work and what is the meaning of static_cast<void>(&(name[0]) and what does it do?
Please can anyone explain me in simple terms. Thanks

推荐答案

你所拥有的是一个字符数组。 name [0] 将返回第一个字符，& name [0] 将返回指向该字符的指针（内存地址）（ char * ），如你所料。

发生了什么，是 cout 正在解释 char * 作为字符串和打印输出字符而不是地址。



使用 static_cast 找到的技巧已经关闭，但它缺少一个关键部分 - 要转换的类型，目的是将 char * 转换为 void * 以便 cout 不会将其视为字符串。

What you have is an array of chars. name[0] will return the first char, and &name[0] will return a pointer (the memory address) to that char (a char*), as you expect.
What's happening though, is that cout is interpreting that char* as a string and printing out the characters instead of the address.

The trick you found using static_cast is close, but it's missing a crucial part - the type to cast to, the aim is to cast the char* to a void* so that cout won't treat it as a string.

char name[] = "Hello";
cout<<"the memory address of name[0] is:"<<static_cast<void*>(&name[0])<<endl;

有关不同类型演员表的更多信息（因为 static_cast 不是唯一一个），请参阅 http://www.cplusplus.com/doc/tutorial/typecasting/ [ ^ ]

For more info on different types of casts (as static_cast is not the only one) see http://www.cplusplus.com/doc/tutorial/typecasting/[^]

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