[英]print 64bit c++ full memory address
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: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:
int a = 3 ;
cout sizeof(&a) << " bytes" << endl ;
cout << &a << endl ;
The output is:输出是:
8 bytes
8 字节
0x7fff007bcce0
0x7fff007bcce0
Obviously, the address of a variable is 8 byte (64 bit system).显然,变量的地址是 8 字节(64 位系统)。 But when I print it, I get only 12 hexadecimal digits instead of 16.
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。
Why this?为什么这个? I think that is due to the fact that the 4 "lost" digits were leading zeroes, that were not printed.
我认为这是因为 4 个“丢失”的数字是前导零,没有打印出来。 But this is only my thought, and I wish to have a definitive and technically correct answer.
但这只是我的想法,我希望有一个明确且技术上正确的答案。
How could I print the entire address?我怎么能打印整个地址? 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?
是否有内置解决方案,或者我应该手动使用“sizeof”以获得真实长度,然后在地址中添加正确数量的零?
Forgive me, I googled for a day for an answer to my stupid question, but I wasn't able to find an answer.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。 I'm just a newbie.
我只是一个新手。 (On stackoverflow I did not find any question/answer about what I needed to know, but maybe I'm wrong.)
(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。)
Someone asks a pretty similar question here: c++ pointer on 64 bit machine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
Hope this helps :)希望这可以帮助 :)
To print the full 64bit address with leading zeros you can use:要打印带有前导零的完整 64 位地址,您可以使用:
std::cout
<< "0x"
<< std::hex
<< std::noshowbase
<< std::setw(16)
<< std::setfill('0')
<< n
<< std::endl ;
Got it from: How can I pad an int with leading zeros when using cout << operator?来自: 如何在使用 cout << 运算符时用前导零填充 int?
I am currently writing a book on C++ and windows 32-bit programming for peeps such as you, but unfortunately I am not yet done with it :(我目前正在为像你这样的人写一本关于 C++ 和 windows 32 位编程的书,但不幸的是我还没有完成:(
The following code demonstrates how you would display a 64-bit unsigned number using cout method:以下代码演示了如何使用 cout 方法显示 64 位无符号数:
// Define a 64-bit number. You may need to include <stdint.h> header file depending on your C++ compiler.
uint64_t UI64 = 281474976709632ULL; // Must include ULL suffix and this is C99 C++ compiler specific.
// unsigned __int64 UI64 = 281474976709632ULL; // Must include ULL suffix and this is Microsoft C++ compiler specific.
// Set decimal output.
cout << dec;
// Display message to user.
cout << "64-bit unsigned integer value in decimal is: " << UI64 << endl;
cout << "\n64-bit unsigned integer value in hexadecimal is: ";
// Set the uppercase flag to display hex value in capital letters.
cout << uppercase;
// Set hexadecimal output.
cout << hex;
// Set the width output to be 16 digits.
cout.width(16);
// Set the fill output to be zeros.
cout.fill('0');
// Set right justification for output.
right(cout);
// Display the 64-bit number.
cout << UI64 << endl;
You may need to (type) cast the address into a 64-bit unsigned value.您可能需要(键入)将地址转换为 64 位无符号值。 In this case, you can do the following:
在这种情况下,您可以执行以下操作:
// (type) cast pointer adddress into an unsigned 64-bit integer.
uint64_t UADD64 = (uint64_t)&UI64; // C99 C++ compiler specific.
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