blocks|key|307110|text|没有明确的答案。这取决于平台。它是由实现定义的。它可以是2、4或其他值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|307111|int背后的想法是，它应该与给定平台上的自然“字”大小相匹配:+16位平台上为16位，32位平台上为32位，64位平台上为64位。但是，出于向后兼容的目的，一些编译器更喜欢坚持使用32位int，即使是在64位平台上也是如此。|offset|length|style|CODE|307112|不过，2字节int的时代已经一去不复返了(16位平台？)除非您使用的是具有16位字长的嵌入式平台。你的课本可能已经很旧了。|307113|entityMap^0|0|0|3|2L|3|0|6|3|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]|$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|K|$]]

There's no specific answer. It depends on the platform. It is implementation-defined. It can be 2, 4 or something else.

The idea behind <code>int</code> was that it was supposed to match the natural "word" size on the given platform: 16 bit on 16-bit platforms, 32 bit on 32-bit platforms, 64 bit on 64-bit platforms, you get the idea. However, for backward compatibility purposes some compilers prefer to stick to 32-bit <code>int</code> even on 64-bit platforms.

The time of 2-byte <code>int</code> is long gone though (16-bit platforms?) unless you are using some embedded platform with 16-bit word size. Your textbooks are probably very old.

blocks|key|524071|text|这个问题的答案取决于您使用的平台。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|524072|但与平台无关，您可以可靠地假设以下类型：|524073|+[8-bit]+signed+char:+-127+to+127
+[8-bit]+unsigned+char:+0+to+255
+[16-bit]signed+short:+-32767+to+32767
+[16-bit]unsigned+short:+0+to+65535
+[32-bit]signed+long:+-2147483647+to+2147483647
+[32-bit]unsigned+long:+0+to+4294967295
+[64-bit]signed+long+long:+-9223372036854775807+to+9223372036854775807
+[64-bit]unsigned+long+long:+0+to+18446744073709551615|code-block|syntax|javascript|524074|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The answer to this question depends on which platform you are using. 
But irrespective of platform, you can reliably assume the following types:

<pre><code> [8-bit] signed char: -127 to 127
 [8-bit] unsigned char: 0 to 255
 [16-bit]signed short: -32767 to 32767
 [16-bit]unsigned short: 0 to 65535
 [32-bit]signed long: -2147483647 to 2147483647
 [32-bit]unsigned long: 0 to 4294967295
 [64-bit]signed long long: -9223372036854775807 to 9223372036854775807
 [64-bit]unsigned long long: 0 to 18446744073709551615
</code></pre>

blocks|key|307213|text|这取决于实现，但通常取决于x86和其他流行的体系结构，如ARM+int需要4个字节。您总是可以在编译时使用sizeof(int)或您想要检查的任何其他类型进行检查。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|307214|如果要确保使用特定大小的类型，请使用<stdint.h>中的类型|307215|entityMap^0|W|3|1H|B|0|I|A|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|$]]

This depends on implementation, but usually on x86 and other popular architectures like ARM <code>int</code>s take 4 bytes. You can always check at compile time using <code>sizeof(int)</code> or whatever other type you want to check.

If you want to make sure you use a type of a specific size, use the types in <code>&lt;stdint.h&gt;</code>

blocks|key|307251|text|#include+<stdio.h>

int+main(void)+{
++++printf("size+of+int:+%25d",+(int)sizeof(int));
++++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|307252|这将返回4，但它可能与机器相关。|unstyled|307253|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>#include &lt;stdio.h&gt;

int main(void) {
 printf("size of int: %d", (int)sizeof(int));
 return 0;
}
</code></pre>

This returns 4, but it's probably machine dependant.

blocks|key|524142|text|This+is+a+good+source+for+answering+this+question.|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|524143|但这个问题在某种程度上是一个真理，答案是“是的，两者都是。”|524144|这取决于您的体系结构。如果你打算在16位或更少的机器上工作，它不能是4字节(=32位)。如果你在32位或更好的机器上工作，它的长度是32位。|524145|为了解决这个问题，让你的程序准备好输出一些可读的东西，并使用"sizeof“函数。它返回声明的数据类型的字节大小。但是在数组中使用这个是很有意义的。|524146|如果你声明为int+t[12];，它将返回12*4字节。要获得该数组的长度，只需使用sizeof(t)/sizeof(t[0])。如果你要构建一个函数，它应该计算发送数组的大小，记住如果|style|CODE|524147|typedef+int+array[12];
int+function(array+t){
++++int+size_of_t+=+sizeof(t)/sizeof(t[0]);
++++return+size_of_t;
}
void+main(){
++++array+t+=+{1,1,1};++//remember:+t=+[1,1,1,0,...,0]
++++int+a+=+function(t);++++//remember:+sending+t+is+just+a+pointer+and+equal+to+int*+t
+++print(a);+++//+output+will+be+1,+since+t+will+be+interpreted+as+an+int+itselve.+
}|code-block|syntax|javascript|524148|所以这甚至不会返回任何不同的东西。如果您定义了一个数组，然后尝试获取长度，请使用sizeof。如果您将数组发送到函数，请记住，send值只是第一个元素上的指针。但是在第一种情况下，你总是知道你的数组的大小。第二种情况可以通过定义两个函数来解决，但会丢失一些性能。定义函数(数组t)和定义函数(数组t，int+size_of_t)。调用"function(t)“通过一些复制工作测量长度，并将结果发送到function2，在那里您可以对可变数组大小执行任何操作。|524149|entityMap|0|LINK|mutability|MUTABLE|url|http://www.rapidtables.com/prog/cref/data_types.htm^0|0|1E|0|0|0|0|0|6|A|16|M|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|6|7|16|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|17|8|@]|9|@]|C|$]]|$1|H|3|I|5|6|7|18|8|@]|9|@]|C|$]]|$1|J|3|K|5|6|7|19|8|@$A|1A|B|1B|L|M]|$A|1C|B|1D|L|M]]|9|@]|C|$]]|$1|N|3|O|5|P|7|1E|8|@]|9|@]|C|$Q|R]]|$1|S|3|T|5|6|7|1F|8|@]|9|@]|C|$]]|$1|U|3|-4|5|6|7|1G|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]]]

<a href="http://www.rapidtables.com/prog/cref/data_types.htm" rel="nofollow noreferrer">This is a good source for answering this question.</a>

But this question is a kind of a always truth answere "Yes. Both."

It depends on your architecture. If you're going to work on a 16-bit machine or less, it can't be 4 byte (=32 bit). If you're working on a 32-bit or better machine, its length is 32-bit. 

To figure out, get you program ready to output something readable and use the "sizeof" function. That returns the size in bytes of your declared datatype. But be carfull using this with arrays. 

If you're declaring <code>int t[12];</code> it will return 12*4 byte. To get the length of this array, just use <code>sizeof(t)/sizeof(t[0])</code>. 
If you are going to build up a function, that should calculate the size of a send array, remember that if

<pre><code>typedef int array[12];
int function(array t){
 int size_of_t = sizeof(t)/sizeof(t[0]);
 return size_of_t;
}
void main(){
 array t = {1,1,1}; //remember: t= [1,1,1,0,...,0]
 int a = function(t); //remember: sending t is just a pointer and equal to int* t
 print(a); // output will be 1, since t will be interpreted as an int itselve. 
}
</code></pre>

So this won't even return something different. If you define an array and try to get the length afterwards, use sizeof. If you send an array to a function, remember the send value is just a pointer on the first element. But in case one, you always knows, what size your array has. Case two can be figured out by defining two functions and miss some performance. Define function(array t) and define function2(array t, int size_of_t). Call "function(t)" measure the length by some copy-work and send the result to function2, where you can do whatever you want on variable array-sizes.

Does an Integer variable in C occupy 2 bytes or 4 bytes? What are the factors that it depends on?

Most of the textbooks say integer variables occupy 2 bytes.
But when I run a program printing the successive addresses of an array of integers it shows the difference of 4.

Is the size of C "int" 2 bytes or 4 bytes?

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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C中的Integer变量占用2个字节还是4个字节？它依赖的因素是什么？大多数教科书都说整数变量占用2个字节。但是，当我运行一个程序，打印一个整数数组的连续地址时，它显示了4的差值。

问C "int“的大小是2字节还是4字节？
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问C "int“的大小是2字节还是4字节？
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