blocks|key|1026912|text|您可以使用以下宏，摘自这里|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1026913|#define+sizeof_var(+var+)+((size_t)(&(var)%2B1)-(size_t)(&(var)))+|code-block|syntax|javascript|1026914|其思想是使用指针算法((&(var)%2B1))确定变量的偏移量，然后减去变量的原始地址，得到变量的大小。例如，如果int16_t+i变量位于0x0002，则将从0x0006中减去0x0002，从而获得0x4或4个字节。|style|CODE|1026915|但是，我认为不使用sizeof并不是一个合理的理由，但我相信您必须有一个理由。|1026916|entityMap|0|LINK|mutability|MUTABLE|url|http://www.geocities.com/vijoeyz/faq/c/sizeof_impl.html^0|B|2|0|0|0|B|A|1K|9|1X|6|27|6|2G|6|2R|3|0|9|6|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|11|8|@$A|12|B|13|K|L]|$A|14|B|15|K|L]|$A|16|B|17|K|L]|$A|18|B|19|K|L]|$A|1A|B|1B|K|L]|$A|1C|B|1D|K|L]]|9|@]|C|$]]|$1|M|3|N|5|6|7|1E|8|@$A|1F|B|1G|K|L]]|9|@]|C|$]]|$1|O|3|-4|5|6|7|1H|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

You can use the following macro, taken from <a href="http://www.geocities.com/vijoeyz/faq/c/sizeof_impl.html" rel="noreferrer">here</a>:

<pre><code>#define sizeof_var( var ) ((size_t)(&amp;(var)+1)-(size_t)(&amp;(var))) 
</code></pre>

The idea is to use pointer arithmetic (<code>(&amp;(var)+1)</code>) to determine the offset of the variable, and then subtract the original address of the variable, yielding its size. For example, if you have an <code>int16_t i</code> variable located at <code>0x0002</code>, you would be subtracting <code>0x0002</code> from <code>0x0006</code>, thereby obtaining <code>0x4</code> or 4 bytes.

However, I don't really see a valid reason not to use <code>sizeof</code>, but I'm sure you must have one.

blocks|key|991062|text|我已经很久没有编写任何C代码了，而且我从来不擅长它，但这看起来是正确的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|991063|int+i+=+1;
size_t+size+=+(char*)(&i%2B1)-(char*)(&i);
printf("%25zi\n",+size);|code-block|syntax|javascript|991064|我相信有人可以告诉我为什么这是错误的理由，但它给我打印了一个合理的价值。|991065|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

It's been ages since I wrote any C code and I was never good at it, but this looks about right:

<pre><code>int i = 1;
size_t size = (char*)(&amp;i+1)-(char*)(&amp;i);
printf("%zi\n", size);
</code></pre>

I'm sure someone can tell me plenty of reasons why this is wrong, but it prints a reasonable value for me.

blocks|key|1025790|text|尝尝这个,|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1025791|#define+sizeof_type(+type+)++((size_t)((type*)1000+%2B+1+)-(size_t)((type*)1000))|code-block|syntax|javascript|1025792|对于以下用户定义的数据类型，|1025793|struct+x
{
++++char+c;
++++int+i;
};

sizeof_type(x)++++++++++=+8
(size_t)((x*)1000+%2B+1+)+=+1008
(size_t)((x*)1000)++++++=+1000|1025794|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try this,

<pre><code>#define sizeof_type( type ) ((size_t)((type*)1000 + 1 )-(size_t)((type*)1000))
</code></pre>

For the following user-defined datatype,

<pre><code>struct x
{
 char c;
 int i;
};

sizeof_type(x) = 8
(size_t)((x*)1000 + 1 ) = 1008
(size_t)((x*)1000) = 1000
</code></pre>

blocks|key|2517003|text|int+*a,+*b,c,d;/*+one+must+remove+the+declaration+of+c+from+here*/
int+c=10;
a=&c;
b=a;
a%2B%2B;
d=(int)a-(int)b;
printf("size+is+%25d",d);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2517004|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int *a, *b,c,d;/* one must remove the declaration of c from here*/
int c=10;
a=&amp;c;
b=a;
a++;
d=(int)a-(int)b;
printf("size is %d",d);
</code></pre>

blocks|key|1026958|text|#include<stdio.h>

#include<conio.h>

struct+size1


++{
int+a;
char+b;
float+c;
};

void+main()
{
struct+size1+*sptr=0;++//declared+one+pointer+to+struct+and+initialise+it+to+zero//
sptr%2B%2B;+++++++++++++++++
printf("size:%25d\n",*sptr);
getch();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1026959|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

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

#include&lt;conio.h&gt;

struct size1


 {
int a;
char b;
float c;
};

void main()
{
struct size1 *sptr=0; //declared one pointer to struct and initialise it to zero//
sptr++; 
printf("size:%d\n",*sptr);
getch();
}
</code></pre>

blocks|key|991133|text|这将给出变量的大小。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|991134|#define+mySizeof(type)+((uint)((type+*)0%2B1))|code-block|syntax|javascript|991135|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This should give you the size of your variable

<pre><code>#define mySizeof(type) ((uint)((type *)0+1))
</code></pre>

blocks|key|991137|text|这很管用..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|991138|int+main()+{
++++int+a;+//try+changing+this+to+char/double/float+etc+each+time//
++++char+*p1,+*p2;
++++p1+=+&a;
++++p2+=+(&a)+%2B+1;
++++printf("size+of+variable+is:%25d\n",+p2+-+p1);
}|code-block|syntax|javascript|991139|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This works..

<pre><code>int main() {
 int a; //try changing this to char/double/float etc each time//
 char *p1, *p2;
 p1 = &amp;a;
 p2 = (&amp;a) + 1;
 printf("size of variable is:%d\n", p2 - p1);
}
</code></pre>

blocks|key|1025805|text|在不使用sizeof运算符的情况下查找变量大小的程序|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1025806|#include<stdio.h>
int+main()
++{
++int+*p,*q;
++int+no;
++p=&no;
+++printf("Address+at+p=%25u\n",p);
++q=((&no)%2B1);
++printf("Address+at+q=%25u\n",q);
++printf("Size+of+int+'no':+%25d+Bytes\n",(int)q-(int)p);

++char+*cp,*cq;
++++char+ch;
++cp=&ch;
++printf("\nAddress+at+cp=%25u\n",cp);
++cq=cp%2B1;
++printf("Address+at+cq=%25u\n",cq);
++printf("Size+of+Char=%25u+Byte\n",(int)cq-(int)cp);

++float+*fp,*fq;
++float+f;
++fp=&f;
+printf("\nAddress+at+fp=%25u\n",fp);
++fq=fp%2B1;
++printf("Address+at+fq=%25u\n",fq);
++printf("Size+of+Float=%25u+Bytes\n",(int)fq-(int)fp);

++return+0;
}|code-block|syntax|javascript|1025807|entityMap^0|4|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Program to find Size of the variable without using <code>sizeof</code> operator

<pre><code>#include&lt;stdio.h&gt;
int main()
 {
 int *p,*q;
 int no;
 p=&amp;no;
 printf("Address at p=%u\n",p);
 q=((&amp;no)+1);
 printf("Address at q=%u\n",q);
 printf("Size of int 'no': %d Bytes\n",(int)q-(int)p);

 char *cp,*cq;
 char ch;
 cp=&amp;ch;
 printf("\nAddress at cp=%u\n",cp);
 cq=cp+1;
 printf("Address at cq=%u\n",cq);
 printf("Size of Char=%u Byte\n",(int)cq-(int)cp);

 float *fp,*fq;
 float f;
 fp=&amp;f;
 printf("\nAddress at fp=%u\n",fp);
 fq=fp+1;
 printf("Address at fq=%u\n",fq);
 printf("Size of Float=%u Bytes\n",(int)fq-(int)fp);

 return 0;
}
</code></pre>

blocks|key|1025839|text|下面的声明将给出通用解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1025840|printf("%25li\n",+(void+*)(&i+%2B+1)+-+(void+*)(&i));|code-block|syntax|javascript|1025841|i是一个变量名，可以是任何数据类型(char、short、int、float、double、struct)。|offset|length|style|CODE|1025842|entityMap^0|0|0|0|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Below statement will give generic solution:

<pre><code>printf("%li\n", (void *)(&amp;i + 1) - (void *)(&amp;i));
</code></pre>

<code>i</code> is a variable name, which can be any data type (char, short, int, float, double, struct).

blocks|key|991190|text|我希望下面的代码能够解决您在c%2B%2B中的问题，而不需要使用sizeof()操作符|type|unstyled|depth|inlineStyleRanges|entityRanges|data|991191|对于任何变量，如(int，char，float，double，char，short等).|991192|这里我取整数，|991193|int+a;|code-block|syntax|javascript|991194|然后将其显示为字节可寻址输出，|991195|cout<<(char+*)(&a+%2B+1)+-+(char+*)(&a);|991196|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I hope that below code would solve your problem in c++ without using sizeof() operator

for any variables like (int, char, float, double, char, short and many more...)

here I take integer,

<pre><code>int a;
</code></pre>

then 
show it as byte addressable output,

<pre><code>cout&lt;&lt;(char *)(&amp;a + 1) - (char *)(&amp;a);
</code></pre>

blocks|key|991215|text|这应该能行。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|991216|#define+xsizeof(x)+(char+*)(&x%2B1)+-+(char+*)&x|code-block|syntax|javascript|991217|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This should work.
<pre><code>#define xsizeof(x) (char *)(&amp;x+1) - (char *)&amp;x
</code></pre>

blocks|key|2517127|text|#define+GET_SIZE(myvar)+++((char)(+((char*)(&myvar%2B1)+)-+((char*)&myvar)+))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2517128|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#define GET_SIZE(myvar) ((char)( ((char*)(&amp;myvar+1) )- ((char*)&amp;myvar) ))
</code></pre>

Let us assume I have declared the variable 'i' of certain datatype (might be int, char, float or double) ... 

NOTE: Simply consider that 'i' is declared and dont bother if it is an int or char or float or double datatype. Since I want a generic solution I am simply mentioning that variable 'i' can be of any one of the datatypes namely int, char, float or double.

Now can I find the size of the variable 'i' without sizeof operator?

How to find the size of a variable without using sizeof

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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让我们假设我已经声明了某些数据类型的变量“i”(可能是int、char、float或double) .注意：只是考虑'i‘是声明的，如果它是int、char、float或double数据类型，就不用麻烦了。因为我想要一个通用的解决方案，我只是简单地提到变量'i‘可以是任意一种数据类型，即int、char、float或d...

问如何找到变量的大小而不使用
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问如何找到变量的大小而不使用EN