blocks|key|885687|text|在这种情况下，你要做一个演员。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|885688|你只是在说编译。嘿，我肯定这是一种类型(A)。|885689|编译器检查C%2B%2B样式转换。C风格的强制转换不会也可能在运行时失败|885690|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

In this case you are making a cast. 

You are just saying for the compile. Hey, I'm sure that this is a type (A).

C++ style casts are checked by the compiler. C style casts aren't and can fail at runtime

blocks|key|884085|text|在C中，它不是必需的，但在某些情况下，它有助于为隐式强制转换提供无声的编译器警告。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|884086|然而，在C%2B%2B中，需要显式转换才能将空指针强制转换为不同类型的指针。|884087|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

It is not necessary in C but it helps silent compiler warnings for implicit casts in some cases.

In C++ however you need explicit casts to cast a void pointer into a pointer of a different type.

blocks|key|2094476|text|它被称为cast。这意味着它在(A)类型中转换b。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2094477|char+c+=+'a';+//+c+is+a+char
int+b+=+(int)+c+//+you+tell+your+compiler+to+interpret+c+as+an+int|code-block|syntax|javascript|2094478|entityMap^0|4|4|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|$]]

It's called a cast. It means that it converts b in the (A) type.

<pre><code>char c = 'a'; // c is a char
int b = (int) c // you tell your compiler to interpret c as an int
</code></pre>

blocks|key|2094512|text|不，用C语言将void*转换为char*+(这是一个常见例子，用来解释它们的不同之处！)是隐式的，因此转换是不必要的(然后错误，因为如果您错误地将char更改为int，它可能会隐藏问题)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2094513|引用K&R的“C编程语言第二版”(§A.6.8)：|BOLD|2094514|指向对象的任何指针都可以转换为void*类型，而不会丢失信息。如果结果被转换回原始指针类型，则恢复原始指针。不同于第A.6.6部分中讨论的指针到指针的转换(通常需要显式转换)，指针可以分配给类型为void*的指针，也可以从类型为void*的指针中分配，并且可以与它们进行比较。|blockquote|2094515|请注意“如果结果被转换回原来的指针类型”，因为这是至关重要的:如果不是char*，而是int*，那么它可能是错误的，因为内存对齐。|2094516|根据C99标准(§6.3.2.3)，何时可以进行转换：|2094517|指向空的指针可以转换为指向任何不完整类型或对象类型的指针，也可以从指针转换为指向对象类型的指针。指向任何不完整或对象类型的指针可以再次转换为无效指针和返回指针；结果应与原始指针相比较。|2094518|现在让我们看看什么时候可以隐式(感谢mafso非常快速的搜索)，来自C11+(n1570)+§6.5.4p3|2094519|除6.5.16.1约束允许的情况外，涉及指针的转换应通过显式转换方式指定。|2094520|然后§6.5.16.1|2094521|下列一项应适用：左操作数具有原子的、限定的或不限定的指针类型，并且(考虑到左操作数在lvalue转换后将具有的类型)，一个操作数是指向对象类型的指针，另一个操作数是指向限定或非限定版本的void的指针，左边所指向的类型具有所有由右指向的类型的限定符。|2094522|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/1736833/void-pointers-difference-between-c-and-c^0|7|5|F|5|21|4|28|3|Q|4|0|0|H|6|0|0|Z|5|17|4|0|8|8|0|0|1A|8|0|0|2|9|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|F|3|G|5|6|7|1J|8|@$9|1K|A|1L|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|1M|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1N|8|@$9|1O|A|1P|B|C]|$9|1Q|A|1R|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|1S|8|@$9|1T|A|1U|B|H]]|D|@]|E|$]]|$1|P|3|Q|5|K|7|1V|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|1W|8|@$9|1X|A|1Y|B|H]]|D|@]|E|$]]|$1|T|3|U|5|K|7|1Z|8|@]|D|@]|E|$]]|$1|V|3|W|5|6|7|20|8|@$9|21|A|22|B|H]]|D|@]|E|$]]|$1|X|3|Y|5|K|7|23|8|@]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|24|8|@]|D|@]|E|$]]]|10|$11|$5|12|13|14|E|$15|16]]]]

No, conversion from <code>void*</code> to <code>char*</code> in C (it's a <a href="https://stackoverflow.com/questions/1736833/void-pointers-difference-between-c-and-c">common example</a> to explain where they're different!) is implicit so casting is unnecessary (then wrong because it may hide a problem if you wrongly change <code>char</code> to <code>int</code>).

Quoting "The C Programming Language, 2nd edition" by K&amp;R (§A.6.8):

<blockquote>
 Any pointer to an object may be converted to type void* without loss of information. If the result is converted back to the original pointer type, the original pointer is recovered. Unlike the pointer-to-pointer conversions discussed in Par.A.6.6, which generally require an explicit cast, pointers may be assigned to and from pointers of type void*, and may be compared with them.
</blockquote>

Please note "If the result is converted back to the original pointer type" because is crucial: if instead of <code>char*</code> you had <code>int*</code> then it may be wrong because of memory alignment.

From C99 standard (§6.3.2.3) about when conversion is possible:

<blockquote>
 A pointer to void may be converted to or from a pointer to any incomplete or object type. A pointer to any incomplete or object type may be converted to a pointer to void and back again; the result shall compare equal to the original pointer.
</blockquote>

Now let's see when can be implicit (thanks to mafso for very quick search), from C11 (n1570) §6.5.4p3:

<blockquote>
 Conversions that involve pointers, other than where permitted by the constraints of 6.5.16.1, shall be specified by means of an explicit cast.
</blockquote>

Then §6.5.16.1:

<blockquote>
 One of the following shall hold: [...] the left operand has atomic, qualified, or unqualified pointer type, and (considering the type the left operand would have after lvalue conversion) one operand is a pointer to an object type, and the other is a pointer to a qualified or unqualified version of void, and the type pointed to by the left has all the qualifiers of the type pointed to by the right
</blockquote>

blocks|key|857398|text|情况正好相反:在C%2B%2B中，强制转换是绝对必要的，但在C中则不是必要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|857399|这个将在C中编译，但是不在C%2B%2B|offset|length|857400|static+void+hexdump(const+void*+pv,+int+len)+{
++const+unsigned+char*+p+=+pv;
++...
}|code-block|syntax|javascript|857401|entityMap|0|LINK|mutability|MUTABLE|url|http://ideone.com/kXnkIU|1|http://ideone.com/QnXZPs^0|0|2|6|0|B|5|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@$D|W|E|X|1|Y]|$D|Z|E|10|1|11]]|A|$]]|$1|F|3|G|5|H|7|12|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]|S|$5|N|O|P|A|$Q|T]]]]

You've got it the other way around: the cast is absolutely necessary in C++, but it is not necessary in C.

This <a href="http://ideone.com/kXnkIU" rel="nofollow">will compile in C</a>, but <a href="http://ideone.com/QnXZPs" rel="nofollow">not in C++</a>:

<pre><code>static void hexdump(const void* pv, int len) {
 const unsigned char* p = pv;
 ...
}
</code></pre>

blocks|key|857413|text|在我的实践中，一开始不进行强制转换，尽管有编译器警告，在测试pass之后，添加强制转换以消除警告。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|857414|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|$]]

In my practice, no casting at first, though there's compiler warnings, after testing pass, casting is added to eliminate warnings.

I am reading some open-source C code and encountering <code>A a = (A) b;</code> type conversion many times. For example,

<pre><code>static void hexdump(const void* pv, int len)
{
 const unsigned char* p = (const unsigned char*) pv;
 // some other code
}
</code></pre>

<code>A a = (A) b;</code> code happens mainly when <code>b</code> is a pointer, <code>void *</code> pointer most often. I have C++ background. I think in C++, the assignment operator takes<code>=</code> care of the type conversion automatically? Because it already know that <code>a</code> is of type <code>A</code>.

Is the explicit type conversion necessary in C?

Is explicit type conversion necessary in A a = (A) b;

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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我正在阅读一些开源C代码，并多次遇到A a = (A) b;类型转换.例如,static void hexdump(const void* pv, int len){  const unsigned char* p = (const unsigned char*) pv;  // some other code}A a...

问是A_a= (A) _b中必需的显式类型转换；
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问是A_a= (A) _b中必需的显式类型转换；EN