64位整数

本文介绍了64位整数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

您好，我相信这有一个快速简便的解决方案，但我找不到 吧。 我需要存储19' '9'是一个整数，这意味着我需要一个无符号的64 位 整数。 在Visual Studio中的代码如下： __int64 test = 9999999999999999999; 我使用g ++版本3.3.5（suse 9.3）并尝试过两种方法： unsigned long long test = 999999999999999999; 和： uint64_t test = 9999999999999999999; 两者都使用命令g ++ test.cpp进行编译错误 test.cpp：5：错误：整数常量对于long而言太大了类型 我打算包含一些或不通过g ++的选项，或者 有另一种方法可以做64位整数存储？ 感谢您的帮助

解决方案

Jay写道：

您好，我确信这有一个快速简便的解决方案，但我找不到它。 我需要以整数存储19''9'，这意味着我需要一个无符号的64位整数。在Visual Studio中，代码为： __int64 test = 9999999999999999999; 双下划线是赠品，这不是C ++，而是一个 Microsoft 扩展名。 我使用g ++版本3.3.5（suse 9.3）并尝试过两种方法： unsigned long long test = 999999999999999999; 和： uint64_t test = 9999999999999999999; 使用命令g ++ test.cpp进行编译错误测试错误：5：错误：整数常量对于long而言太大了type

合理的，按照标准规则它可能确实拒绝代码。如果 你需要GCC扩展到C ++，你应该在GCC小组中询问。 HTH， Michiel Salters

Jay写道：

您好，我确信这有一个快速简便的解决方案，但我找不到它。 我需要存储19''9'的整数，这意味着我需要一个无符号的64位整数。

[snip] 或者你可以使用任意精度的数字类。各种内置整数类型的大小取决于平台，因此你需要查看你的操作系统/编译器文档。 干杯！ --M

所以没有使用扩展名就没有办法在C ++中这样做吗？ 如果这是错误的这个问题的小组，你碰巧知道 哪个群体更合适？ 谢谢

Hello, I am sure this has a quick and easy solution but I can''t find it. I need to store 19 ''9''s in an integer which means i need an unsigned 64 bit integer. in Visual Studio the code would be: __int64 test = 9999999999999999999; I am using g++ version 3.3.5(suse 9.3) and have tried both: unsigned long long test = 999999999999999999; and: uint64_t test = 9999999999999999999; both have this compile error using the command "g++ test.cpp" test.cpp:5: error: integer constant is too large for "long" type Am I forgeting to include something or not passing an option to g++, or is there another way to do 64 bit integer storage? Thank you for your help

解决方案

Jay wrote:

Hello, I am sure this has a quick and easy solution but I can''t find it. I need to store 19 ''9''s in an integer which means i need an unsigned 64 bit integer. in Visual Studio the code would be: __int64 test = 9999999999999999999; The double underscore is a giveaway that this isn''t C++, but a Microsoft extension. I am using g++ version 3.3.5(suse 9.3) and have tried both: unsigned long long test = 999999999999999999; and: uint64_t test = 9999999999999999999; both have this compile error using the command "g++ test.cpp" test.cpp:5: error: integer constant is too large for "long" type

Reasonable, by the standard rules it may indeed reject the code. If you need an GCC extension to C++, you should ask in a GCC group. HTH, Michiel Salters

Jay wrote:

Hello, I am sure this has a quick and easy solution but I can''t find it. I need to store 19 ''9''s in an integer which means i need an unsigned 64 bit integer.

[snip] Or you could use a number class with arbitrary precision. The sizes of the various built-in integral types are platform-dependent, so you''ll want to check with your OS/compiler documentation. Cheers! --M

So is there no way to do this in C++ without using an extension? If this is the wrong group for this question, do you happen to know which group would be more appropriate? Thank you