blocks|key|795615|text|不是的。C#规范严格定义了int是恰好32位的System.Int32的别名。改变这一点将是一个重大的突破性变化。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|795616|entityMap^0|D|3|N|C|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

No. The C# specification rigidly defines that <code>int</code> is an alias for <code>System.Int32</code> with exactly 32 bits. Changing this would be a major breaking change.

blocks|key|3086200|text|C#中的int关键字被定义为System.Int32类型的别名，它(根据名称判断)是一个32位整数。对于规范：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3086201|3086202|+CLI+specification第8.2.2节(内置值和引用类型)中有一个表，其中包含以下内容：|blockquote|3086203|3086204|3086205|3086206|System.Int32带符号的32位整数|unordered-list-item|3086207|3086208|C#+specification第8.2.1节(预定义类型)有一个类似的表：|3086209|3086210|int+-32位有符号整型|3086211|3086212|这保证了CLR中的System.Int32和C#中的int都是32位的。|3086213|entityMap|0|LINK|mutability|MUTABLE|url|http://www.ecma-international.org/publications/files/ECMA-ST/Ecma-335.pdf|1|http://www.ecma-international.org/publications/files/ECMA-ST/Ecma-334.pdf^0|4|3|E|C|0|0|1|H|0|0|0|0|0|0|C|0|0|0|G|1|0|0|0|3|0|0|9|C|Q|3|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|J|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|1L|8|@$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|P|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1P|8|@]|D|@$9|1Q|A|1R|1|1S]]|E|$]]|$1|S|3|-4|5|6|7|1T|8|@]|D|@]|E|$]]|$1|T|3|U|5|O|7|1U|8|@$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1X|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|1Y|8|@$9|1Z|A|20|B|C]|$9|21|A|22|B|C]]|D|@]|E|$]]|$1|Y|3|-4|5|6|7|23|8|@]|D|@]|E|$]]]|Z|$10|$5|11|12|13|E|$14|15]]|16|$5|11|12|13|E|$14|17]]]]

The <code>int</code> keyword in C# is defined as an alias for the <code>System.Int32</code> type and this is (judging by the name) meant to be a 32-bit integer. To the specification:
<blockquote>
<a href="http://www.ecma-international.org/publications/files/ECMA-ST/Ecma-335.pdf" rel="noreferrer">CLI specification</a> section 8.2.2 (Built-in value and reference types) has a table with the following:
<ul>
<li><code>System.Int32</code> - Signed 32-bit integer</li>
</ul>
<a href="http://www.ecma-international.org/publications/files/ECMA-ST/Ecma-334.pdf" rel="noreferrer">C# specification</a> section 8.2.1 (Predefined types) has a similar table:
<ul>
<li><code>int</code> - 32-bit signed integral type</li>
</ul>
</blockquote>
This guarantees that both <code>System.Int32</code> in CLR and <code>int</code> in C# will always be 32-bit.

blocks|key|795847|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|795848|+(TestInt)的大小会变成8吗？|blockquote|795849|795850|795851|不，sizeof(testInt)是错误的。testInt是一个局部变量。sizeof运算符需要一个类型作为其参数。这永远不会是8，因为它始终是一个错误。|795852|795853|+VS2010将c#托管整数编译为4字节，即使在64位计算机上也是如此。|795854|795855|795856|对，是这样。我注意到C#规范的18.5.8节将sizeof(int)定义为编译时常量4。也就是说，当您说sizeof(int)时，编译器只是将其替换为4；这就像您在源代码中说的是"4“一样。|offset|length|style|CODE|795857|有没有人知道C#中的标准“|795858|”会不会变成64位？|795859|795860|795861|绝不可能。C#规范的第4.1.4节规定，"int“是"System.Int32”的同义词。|795862|如果你想要的是一个“指针大小的整数”，那么使用IntPtr。IntPtr在不同的架构上改变其大小。|795863|entityMap^0|0|0|0|0|0|0|0|0|0|N|B|1G|B|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|14|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|15|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|16|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|17|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|18|7|@]|8|@]|9|$]]|$1|H|3|-4|4|5|6|19|7|@]|8|@]|9|$]]|$1|I|3|J|4|C|6|1A|7|@]|8|@]|9|$]]|$1|K|3|-4|4|5|6|1B|7|@]|8|@]|9|$]]|$1|L|3|-4|4|5|6|1C|7|@]|8|@]|9|$]]|$1|M|3|N|4|5|6|1D|7|@$O|1E|P|1F|Q|R]|$O|1G|P|1H|Q|R]]|8|@]|9|$]]|$1|S|3|T|4|5|6|1I|7|@]|8|@]|9|$]]|$1|U|3|V|4|C|6|1J|7|@]|8|@]|9|$]]|$1|W|3|-4|4|5|6|1K|7|@]|8|@]|9|$]]|$1|X|3|-4|4|5|6|1L|7|@]|8|@]|9|$]]|$1|Y|3|Z|4|5|6|1M|7|@]|8|@]|9|$]]|$1|10|3|11|4|5|6|1N|7|@]|8|@]|9|$]]|$1|12|3|-4|4|5|6|1O|7|@]|8|@]|9|$]]]|13|$]]

<blockquote>
 Will sizeof(testInt) ever be 8? 
</blockquote>

No, sizeof(testInt) is an error. testInt is a local variable. The sizeof operator requires a type as its argument. This will never be 8 because it will always be an error.

<blockquote>
 VS2010 compiles a c# managed integer as 4 bytes, even on a 64 bit machine. 
</blockquote>

Correct. I note that section 18.5.8 of the C# specification defines <code>sizeof(int)</code> as being the compile-time constant 4. That is, when you say <code>sizeof(int)</code> the compiler simply replaces that with 4; it is just as if you'd said "4" in the source code.

<blockquote>
 Does anyone know if/when the time will come that a standard "int" in C# will be 64 bits?
</blockquote>

Never. Section 4.1.4 of the C# specification states that "int" is a synonym for "System.Int32". 

If what you want is a "pointer-sized integer" then use IntPtr. An IntPtr changes its size on different architectures.

blocks|key|5341538|text|在所有平台上，int始终是Int32的同义词。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5341539|微软不太可能在未来改变这一点，因为它会破坏许多假设int是32位的现有代码。|5341540|entityMap^0|7|3|D|5|0|P|3|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

<code>int</code> is always synonymous with <code>Int32</code> on all platforms.

It's very unlikely that Microsoft will change that in the future, as it would break lots of existing code that assumes <code>int</code> is 32-bits.

blocks|key|3086409|text|我认为你可能会感到困惑的是，int是Int32的别名，所以它总是4个字节，但是IntPtr被假定为匹配CPU体系结构的字大小，所以在32位系统上是4个字节，在64位系统上是8个字节。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3086410|entityMap^0|E|3|I|5|13|6|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

I think what you may be confused by is that <code>int</code> is an alias for <code>Int32</code> so it will always be 4 bytes, but <code>IntPtr</code> is suppose to match the word size of the CPU architecture so it will be 4 bytes on a 32-bit system and 8 bytes on a 64-bit system.

blocks|key|3932152|text|根据C#规范ECMA-334+"11.1.4简单类型“一节，保留字int将别名为System.Int32。因为这是在规范中的，所以不太可能改变。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3932153|entityMap|0|LINK|mutability|MUTABLE|url|http://www.ecma-international.org/publications/files/ECMA-ST/Ecma-334.pdf^0|X|3|14|C|6|8|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

According to the C# specification <a href="http://www.ecma-international.org/publications/files/ECMA-ST/Ecma-334.pdf" rel="nofollow noreferrer">ECMA-334</A>, section "11.1.4 Simple Types", the reserved word <code>int</code> will be aliased to <code>System.Int32</code>. Since this is in the specification it is very unlikely to change.

blocks|key|5341479|text|无论你使用的是32位版本的CLR还是64位版本的CLR，在C#中，int总是意味着System.Int32，而long总是意味着System.Int64。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5341480|entityMap^0|X|3|15|C|1J|4|1S|C|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

No matter whether you're using the 32-bit version or 64-bit version of the CLR, in C# an <code>int</code> will always mean <code>System.Int32</code> and <code>long</code> will always mean <code>System.Int64</code>.

blocks|key|3932488|text|以下命令将在C#中执行always+be+true操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3932489|sbyte带符号8位，1字节|style|BOLD|3932490|字节无符号8位，1字节|3932491|短带符号16位，2字节|3932492|ushort无符号16位，2字节|3932493|整数带符号32位，4字节|3932494|uint无符号32位，4字节|3932495|长带符号64位，8字节|3932496|ulong无符号64位，8字节|3932497|整型文字只是一个数字序列(例如314159)，没有任何这些显式类型。C#将其分配给适合它的序列中的第一个类型(int、uint、long、ulong)。这似乎在上面的至少一个回应中有点混乱。|CODE|3932498|出现在一串数字前的一元减号运算符(减号)不会将选择减少为(+Weirdly+，long)。文字始终是正的；减号实际上是一个运算符。所以想必-314159和-((int)314159)完全是一回事。显然，有一种特殊的情况可以让-2147483648直接进入int；否则它就是-((uint)2147483648)。我猜想这会造成一些不愉快的事情。|3932499|不知何故，似乎可以放心地预测，C#+(和朋友)永远不会为>=128位整数的“模糊名称”类型而烦恼。一旦处理器支持这么广泛的数学运算，我们将获得对任意大整数的良好支持，以及对UInt128、UInt256等的超精确支持，而且几乎不会使用任何数学运算。64位地址空间真的很大。如果它们太小，可能是因为某种深奥的原因，比如ASLR或更高效的MapReduce或其他原因。|3932500|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/library/exx3b86w.aspx|1|https://msdn.microsoft.com/en-us/library/aa664674(v=vs.71).aspx^0|B|E|0|0|0|5|0|0|2|0|0|1|0|0|6|0|0|2|0|0|4|0|0|1|0|0|5|0|F|6|1J|3|1N|4|1S|4|1X|5|0|U|7|13|4|3J|3|1X|7|25|E|34|B|3S|J|U|7|1|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|D|3|E|5|6|7|1G|8|@$A|1H|B|1I|F|G]]|9|@]|C|$]]|$1|H|3|I|5|6|7|1J|8|@$A|1K|B|1L|F|G]]|9|@]|C|$]]|$1|J|3|K|5|6|7|1M|8|@$A|1N|B|1O|F|G]]|9|@]|C|$]]|$1|L|3|M|5|6|7|1P|8|@$A|1Q|B|1R|F|G]]|9|@]|C|$]]|$1|N|3|O|5|6|7|1S|8|@$A|1T|B|1U|F|G]]|9|@]|C|$]]|$1|P|3|Q|5|6|7|1V|8|@$A|1W|B|1X|F|G]]|9|@]|C|$]]|$1|R|3|S|5|6|7|1Y|8|@$A|1Z|B|20|F|G]]|9|@]|C|$]]|$1|T|3|U|5|6|7|21|8|@$A|22|B|23|F|G]]|9|@]|C|$]]|$1|V|3|W|5|6|7|24|8|@$A|25|B|26|F|X]|$A|27|B|28|F|G]|$A|29|B|2A|F|G]|$A|2B|B|2C|F|G]|$A|2D|B|2E|F|G]]|9|@]|C|$]]|$1|Y|3|Z|5|6|7|2F|8|@$A|2G|B|2H|F|G]|$A|2I|B|2J|F|G]|$A|2K|B|2L|F|G]|$A|2M|B|2N|F|X]|$A|2O|B|2P|F|X]|$A|2Q|B|2R|F|X]|$A|2S|B|2T|F|X]]|9|@$A|2U|B|2V|1|2W]]|C|$]]|$1|10|3|11|5|6|7|2X|8|@]|9|@]|C|$]]|$1|12|3|-4|5|6|7|2Y|8|@]|9|@]|C|$]]]|13|$14|$5|15|16|17|C|$18|19]]|1A|$5|15|16|17|C|$18|1B]]]]

The following will <a href="https://msdn.microsoft.com/en-us/library/exx3b86w.aspx" rel="nofollow">always be true</a> in C#:

sbyte signed 8 bits, 1 byte

byte unsigned 8 bits, 1 byte

short signed 16 bits, 2 bytes

ushort unsigned 16 bits, 2 bytes

int signed 32 bits, 4 bytes

uint unsigned 32 bits, 4 bytes

long signed 64 bits, 8 bytes

ulong unsigned 64 bits, 8 bytes

An integer literal is just a sequence of digits (eg <code>314159</code>) without any of these explicit types. C# assigns it the first type in the sequence (int, uint, long, ulong) in which it fits. This seems to have been slightly muddled in at least one of the responses above.

<a href="https://msdn.microsoft.com/en-us/library/aa664674(v=vs.71).aspx" rel="nofollow">Weirdly</a> the unary minus operator (minus sign) showing up before a string of digits does not reduce the choice to (int, long). The literal is always positive; the minus sign really is an operator. So presumably <code>-314159</code> is exactly the same thing as <code>-((int)314159)</code>. Except apparently there's a special case to get <code>-2147483648</code> straight into an int; otherwise it'd be <code>-((uint)2147483648)</code>. Which I presume does something unpleasant.

Somehow it seems safe to predict that C# (and friends) will never bother with "squishy name" types for >=128 bit integers. We'll get nice support for arbitrarily large integers and super-precise support for UInt128, UInt256, etc. as soon as processors support doing math that wide, and hardly ever use any of it. 64-bit address spaces are really big. If they're ever too small it'll be for some esoteric reason like ASLR or a more efficient MapReduce or something.

blocks|key|3086036|text|是的，正如乔恩所说，与“C/C%2B%2B世界”不同，Java和C#并不依赖于它们所运行的系统。它们严格定义了byte/short/int/long和单/双精度浮点数的长度，在每个系统上都是一样的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3086037|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Yes, as Jon said, and unlike the 'C/C++ world', Java and C# aren't dependent on the system they're running on. They have strictly defined lengths for byte/short/int/long and single/double precision floats, equal on every system.

blocks|key|795890|text|不带后缀的int可以是32位也可以是64位，这取决于它所表示的值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|795891|如MSDN中所定义：|795892|795893|当整型文字没有后缀时，它的类型是可以表示其值的第一个类型:+int、uint、long、ulong。|blockquote|795894|795895|795896|这是地址：https://msdn.microsoft.com/en-us/library/5kzh1b5w.aspx|offset|length|795897|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/library/5kzh1b5w.aspx^0|0|0|0|0|0|0|5|1I|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|Y|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|11|8|@]|9|@$L|12|M|13|1|14]]|A|$]]|$1|N|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|O|$P|$5|Q|R|S|A|$T|U]]]]

int without suffix can be either 32bit or 64bit, it depends on the value it represents.

as defined in MSDN:

<blockquote>
 When an integer literal has no suffix, its type is the first of these types in which its value can be represented: int, uint, long, ulong.
</blockquote>

Here is the address:
<a href="https://msdn.microsoft.com/en-us/library/5kzh1b5w.aspx" rel="nofollow">https://msdn.microsoft.com/en-us/library/5kzh1b5w.aspx</a>

In my C# source code I may have declared integers as:

<pre><code>int i = 5;
</code></pre>

or

<pre><code>Int32 i = 5;
</code></pre>

In the currently prevalent 32-bit world they are equivalent. However, as we move into a 64-bit world, am I correct in saying that the following will become the same?

<pre><code>int i = 5;
Int64 i = 5;
</code></pre>

Is an int a 64-bit integer in 64-bit C#?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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功能1上新10个字符

功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符。

功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符。

功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符

功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符

功能4上新

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在我的C#源代码中，我可能将整数声明为：int i = 5;或Int32 i = 5;在当前流行的32位世界中，它们是等价的。然而，当我们进入64位世界时，我说下面的内容将变得相同，这样说对吗？int i = 5;Int64 i = 5;

问整数是64位C#中的64位整数吗？
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回答 10

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回答 10

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问整数是64位C#中的64位整数吗？
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