打印 64 位 C++ 完整内存地址 [英] print 64bit c++ full memory address

问题描述

当我在 64 位机器上为 C++ 程序编写代码时，我注意到打印变量的地址(例如)只返回 12 个十六进制字符，而不是 16 个.这是一个示例代码:

While I was writing code on a 64 bit machine for a c++ program,I noticed that printing the address of a variable (for example) returns just 12 hexadecimal characters, instead of 16. Here's an example code:

int a = 3 ;
cout sizeof(&a) << " bytes" << endl ;
cout << &a << endl ;

输出为:

8 个字节

0x7fff007bcce0

0x7fff007bcce0

显然，变量的地址是 8 字节(64 位系统).但是当我打印它时，我只得到 12 个十六进制数字而不是 16 个.

Obviously, the address of a variable is 8 byte (64 bit system). But when I print it, I get only 12 hexadecimal digits instead of 16.

• 为什么会这样?我认为这是由于 4 个丢失"的事实.数字是未打印的前导零.但这只是我的思想，我希望有一个明确的和技术上正确的回答.

• Why this? I think that is due to the fact that the 4 "lost" digits were leading zeroes, that were not printed. But this is only my thought, and I wish to have a definitive and technically correct answer.

如何打印完整地址?有没有内置的解决方案，或者我应该手动使用sizeof"吗?为了得到真正的长度和然后在地址中添加正确数量的零?

How could I print the entire address? Is there a built-in solution, or should I manually use "sizeof" in order to get the real lenght and then add to the address the right number of zeroes?

原谅我，我在谷歌上搜索了一天来回答我愚蠢的问题，但我找不到答案.我只是一个新手.(在 stackoverflow 上，我没有找到关于我需要知道的内容的任何问题/答案，但也许我错了.)

Forgive me, I googled for a day for an answer to my stupid question, but I wasn't able to find an answer. I'm just a newbie. (On stackoverflow I did not find any question/answer about what I needed to know, but maybe I'm wrong.)

推荐答案

有人在这里问了一个非常相似的问题:64 位机器上的 C++ 指针

Someone asks a pretty similar question here: c++ pointer on 64 bit machine

希望这有帮助:)

要打印带前导零的完整 64 位地址，您可以使用:

To print the full 64bit address with leading zeros you can use:

std::cout
<< "0x"
<< std::hex
<< std::noshowbase
<< std::setw(16)
<< std::setfill('0')
<< n
<< std::endl ;

来自:如何在使用 cout << 时用前导零填充 int运算符?

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