blocks|key|1205674|text|您可以将其转换为intptr_t类型。它是一个int类型，保证足够大以包含一个指针。使用#include+<cstdint>来定义它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1205675|entityMap^0|8|8|N|3|18|I|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|$]]

You can cast it to an <code>intptr_t</code> type. It's an <code>int</code> type guaranteed to be big enough to contain a pointer. Use <code>#include &lt;cstdint&gt;</code> to define it.

blocks|key|1206017|text|再说一次，上面的所有答案都严重错失了要点。OP想要将指针值转换为int值，而大多数答案都试图将arg指针的内容错误地转换为int值。而且，这些方法中的大多数甚至不能在gcc4上工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1206018|正确的答案是，如果人们不介意失去数据精度，|1206019|int+x+=+*((int*)(&arg));|code-block|syntax|javascript|1206020|这在GCC4上有效。|1206021|最好的方法是，如果可以的话，不要进行这样的强制转换，相反，如果必须为指针和int共享相同的内存地址(例如，为了节省RAM)，请使用union，并确保只有当您知道内存地址最后被设置为int时，才将其视为int。|1206022|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Again, all of the answers above missed the point badly. The OP wanted to convert a pointer value to a int value, instead, most the answers, one way or the other, tried to wrongly convert the content of arg points to to a int value. And, most of these will not even work on gcc4.

The correct answer is, if one does not mind losing data precision,

<pre><code>int x = *((int*)(&amp;arg));
</code></pre>

This works on GCC4.

The best way is, if one can, do not do such casting, instead, if the same memory address has to be shared for pointer and int (e.g. for saving RAM), use union, and make sure, if the mem address is treated as an int only if you know it was last set as an int.

blocks|key|1205621|text|而不是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1205622|int+x+=+(int)arg;|code-block|syntax|javascript|1205623|使用：|1205624|int+x+=+(long)arg;|1205625|在大多数平台上，指针和长整型的大小相同，但整型和指针在64位平台上的大小通常不同。如果将(void*)转换为(long)不会丢失精度，则通过将(long)赋值给(int)，它会正确地截断数字以使其适合。|offset|length|style|CODE|1205626|entityMap^0|0|0|0|0|19|5|1J|4|20|4|29|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]|$M|13|N|14|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

Instead of:

<pre><code>int x = (int)arg;
</code></pre>

use:

<pre><code>int x = (long)arg;
</code></pre>

On most platforms pointers and longs are the same size, but ints and pointers often are not the same size on 64bit platforms. If you convert (<code>void*</code>) to (<code>long</code>) no precision is lost, then by assigning the (<code>long</code>) to an (<code>int</code>), it properly truncates the number to fit.

blocks|key|1420985|text|在一般情况下，没有合适的方法将其转换为int。C99标准库提供了intptr_t和uintptr_t+typedefs，只要需要执行这样的转换，就应该使用它们。如果您的标准库(即使不是C99)恰好提供了这些类型，请使用它们。如果没有，请检查您的平台上的指针大小，相应地定义这些typedefs并使用它们。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1420986|entityMap^0|J|3|W|8|15|9|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|$]]

There's no proper way to cast this to <code>int</code> in general case. C99 standard library provides <code>intptr_t</code> and <code>uintptr_t</code> typedefs, which are supposed to be used whenever the need to perform such a cast comes about. If your standard library (even if it is not C99) happens to provide these types - use them. If not, check the pointer size on your platform, define these typedefs accordingly yourself and use them.

blocks|key|1420907|text|将指针转换为空*并返回是reinterpret_cast<>的有效用法。所以你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1420908|pthread_create(&thread,+NULL,+myFcn,+new+int(5));+//+implicit+cast+to+void*+from+int*|code-block|syntax|javascript|1420909|然后在myFcn中：|1420910|void*+myFcn(void*+arg)
{
++++int*++data+=+reinterpret_cast<int*>(arg);
++++int+++x++++=+*data;
++++delete+data;|1420911|注意:正如sbi所指出的那样，这需要对OP调用进行更改以创建线程。|1420912|我想强调的是，当你从一个平台移动到另一个平台时，从int到指针的转换可能会出现问题。但是，将指针转换为void*然后再转换回来是非常受支持的(无论在哪里)。|1420913|因此，动态生成指针并使用它可能不太容易出错。使用后记得删除指针，这样我们就不会泄漏。|1420914|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Casting a pointer to void* and back is valid use of reinterpret_cast&lt;>. So you could do this:

<pre><code>pthread_create(&amp;thread, NULL, myFcn, new int(5)); // implicit cast to void* from int*
</code></pre>

Then in myFcn:

<pre><code>void* myFcn(void* arg)
{
 int* data = reinterpret_cast&lt;int*&gt;(arg);
 int x = *data;
 delete data;
</code></pre>

Note: As sbi points out this would require a change on the OP call to create the thread.

What I am trying to emphasis that conversion from int to pointer and back again can be frough with problems as you move from platform to platform. BUT converting a pointer to void* and back again is well supported (everywhere).

Thus as a result it may be less error prone to generate a pointer dynamcially and use that.
Remembering to delete the pointer after use so that we don't leak.

blocks|key|1206053|text|应该强制转换为size_t，而不是使用long强制转换。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1206054|int+val=+(int)((size_t)arg);|code-block|syntax|javascript|1206055|entityMap^0|7|6|J|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Instead of using a <code>long</code> cast, you should cast to <code>size_t</code>.

<pre><code>int val= (int)((size_t)arg);
</code></pre>

blocks|key|1421206|text|我也遇到了这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1421207|ids[i]+=+(int)+arg;+//+error+occur+here+=>+I+change+this+to+below.

ids[i]+=+(uintptr_t)+arg;|code-block|syntax|javascript|1421208|然后我就可以继续编译了。也许你也可以试试这个。|1421209|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|$]]

I meet this problem too.

<pre><code>ids[i] = (int) arg; // error occur here =&gt; I change this to below.

ids[i] = (uintptr_t) arg;
</code></pre>

Then I can continue compiling. Maybe you can try this too.

blocks|key|1205564|text|正确的方法是将其转换为另一个pointer+type。将void*转换为int是一种不可移植的方式，可能会工作，也可能不会！如果需要保留返回的地址，只需将其保留为void*即可。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1205565|entityMap^0|E|C|S|5|10|3|29|5|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|$]]

The proper way is to cast it to another <code>pointer type</code>. Converting a <code>void*</code> to an <code>int</code> is non-portable way that may work or may not! If you need to keep the returned address, just keep it as <code>void*</code>.

blocks|key|1421011|text|最安全的方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1421012|static_cast<int>(reinterpret_cast<long>(void+*+your_variable));|code-block|syntax|javascript|1421013|long保证任何机器上的指针大小。Windows的32位长度也只有64位。因此，您需要将其更改为long+long，而不是64位windows中的long。所以reinterpret_cast已经将它转换为long类型，然后static_cast安全地将long转换为int，如果您准备好了，就截断数据。|offset|length|style|CODE|1421014|entityMap^0|0|0|0|4|1C|9|21|4|28|G|2V|4|34|B|3J|4|3Q|3|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]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|N|$]]

Safest way : 

<pre><code>static_cast&lt;int&gt;(reinterpret_cast&lt;long&gt;(void * your_variable));
</code></pre>

<code>long</code> guarantees a pointer size on Linux on any machine. Windows has 32 bit long only on 64 bit as well. Therefore, you need to change it to <code>long long</code> instead of <code>long</code> in windows for 64 bits.
So <code>reinterpret_cast</code> has casted it to <code>long</code> type and then <code>static_cast</code> safely casts <code>long</code> to <code>int</code>, if you are ready do truncte the data.

blocks|key|1206387|text|在32位整数中存储64位指针没有“正确”的方法。问题不在于强制转换，而在于目标类型丢失了指针的一半。存储在int中的剩余32位不足以重新构造指向线程函数的指针。大多数答案只是试图从参数中提取32个无用的比特。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1206388|正如Ferruccio所说，必须用intptr_t替换int才能使程序有意义。|1206389|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/4086/ferruccio^0|1H|3|0|H|8|2|9|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|H|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

There is no "correct" way to store a 64-bit pointer in an 32-bit integer. The problem is not with casting, but with the target type loosing half of the pointer. The 32 remaining bits stored inside <code>int</code> are insufficient to reconstruct a pointer to the thread function. Most answers just try to extract 32 useless bits out of the argument.

As <a href="https://stackoverflow.com/users/4086/ferruccio">Ferruccio</a> said, int must be replaced with <code>intptr_t</code> to make the program meaningful.

blocks|key|1205816|text|不要将int作为void*传递，而是将int*传递给int，这样您就可以将void*强制转换为int*，并将取消引用的指针复制到int。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1205817|int+x+=+*static_cast<int*>(arg);|code-block|syntax|javascript|1205818|entityMap^0|8|5|J|4|Q|3|11|5|1B|4|1S|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|L|$]]

Don't pass your int as a <code>void*</code>, pass a <code>int*</code> to your <code>int</code>, so you can cast the <code>void*</code> to an <code>int*</code> and copy the dereferenced pointer to your <code>int</code>.

<pre><code>int x = *static_cast&lt;int*&gt;(arg);
</code></pre>

blocks|key|1205907|text|您可能需要的是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1205908|int+x+=+reinterpret_cast<int>(arg);|code-block|syntax|javascript|1205909|这允许您将void+*重新解释为int。|offset|length|style|CODE|1205910|entityMap^0|0|0|5|6|G|3|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]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

What you may want is

<pre><code>int x = reinterpret_cast&lt;int&gt;(arg);
</code></pre>

This allows you to reinterpret the <code>void *</code> as an <code>int</code>.

blocks|key|1420993|text|//new_fd+is+a+int
pthread_create(&threads[threads_in_use]+,+NULL,+accept_request,+(void*)((long)new_fd));

//inside+the+method+I+then+have
int+client;
client+=+(long)new_fd;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1420994|希望这能有所帮助|unstyled|1420995|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>//new_fd is a int
pthread_create(&amp;threads[threads_in_use] , NULL, accept_request, (void*)((long)new_fd));

//inside the method I then have
int client;
client = (long)new_fd;
</code></pre>

Hope this helps

blocks|key|1206208|text|在我的例子中，我使用了一个需要传递给OpenGL函数的32位值作为空*，表示缓冲区中的偏移量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1206209|您不能只是将32位变量转换为指针，因为在64位计算机上，该指针的长度是它的两倍。你的指针有一半会变成垃圾。你需要传递一个实际的指针。|1206210|这将从32位偏移量获得一个指针：|1206211|int32+nOffset+++=+762;+++++++//+random+offset
char+*+pOffset++=+NULL;++++++//+pointer+to+hold+offset
pOffset+++++++++%2B=+nOffset;++//+this+will+now+hold+the+value+of+762+correctly
glVertexAttribPointer(binding,+nStep,+glType,+glTrueFalse,+VertSize(),+pOffset);|code-block|syntax|javascript|1206212|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

In my case, I was using a 32-bit value that needed to be passed to an OpenGL function as a void * representing an offset into a buffer.

You cannot just cast the 32-bit variable to a pointer, because that pointer on a 64-bit machine is twice as long. Half your pointer will be garbage. You need to pass an actual pointer.

This will get you a pointer from a 32 bit offset:

<pre><code>int32 nOffset = 762; // random offset
char * pOffset = NULL; // pointer to hold offset
pOffset += nOffset; // this will now hold the value of 762 correctly
glVertexAttribPointer(binding, nStep, glType, glTrueFalse, VertSize(), pOffset);
</code></pre>

blocks|key|1206274|text|函数指针与void*+(以及任何其他非函数指针)不兼容|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1206275|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|$]]

A function pointer is incompatible to void* (and any other non function pointer)

blocks|key|1420750|text|它之所以这样做，是因为您将64位指针转换为32位整数，从而丢失了信息。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1420751|你可以使用64位整数来代替，但是我通常使用具有正确原型的函数，并且我将函数类型转换为:例如。|1420752|void+thread_func(int+arg){
...
}|code-block|syntax|javascript|1420753|我创建了这样的线程：|1420754|pthread_create(&tid,+NULL,+(void*(*)(void*))thread_func,+(void*)arg);|1420755|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Well it does this because you are converting a 64 bits pointer to an 32 bits integer so you loose information. 

You can use a 64 bits integer instead howerver I usually use a function with the right prototype and I cast the function type :
eg.

<pre><code>void thread_func(int arg){
...
}
</code></pre>

and I create the thread like this :

<pre><code>pthread_create(&amp;tid, NULL, (void*(*)(void*))thread_func, (void*)arg);
</code></pre>

I have a function with prototype <code>void* myFcn(void* arg)</code> which is used as the starting point for a pthread. I need to convert the argument to an int for later use:

<pre><code>int x = (int)arg;
</code></pre>

The compiler (GCC version 4.2.4) returns the error:

<pre><code>file.cpp:233: error: cast from 'void*' to 'int' loses precision
</code></pre>

What is the proper way to cast this?

error: cast from 'void*' to 'int' loses precision

编译器

Windows

我有一个带有prototype void* myFcn(void* arg)的函数，它被用作pthread的起点。我需要将参数转换为int类型，以便以后使用：int x = (int)arg;编译器(GCC 4.2.4版)返回错误：file.cpp:233: error: cast from 'void*' to 'int' loses precision什么是正确的投射方式？

问错误:从'void*‘到'int’的转换失去精度
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回答 16

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问错误:从'void*‘到'int’的转换失去精度EN

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