blocks|key|1849166|text|struct+node+*tempNode+=+(struct+node*)+malloc(sizeof(struct+node));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1849167|您正在为节点指针tempNode动态分配空间，并将其类型转换为struct+node*类型。(指向结构的指针)|unstyled|1849168|struct+node+*tempNode+=+malloc(sizeof(struct+node));|1849169|与上面相同，除了没有显式类型转换(struct+node*)|1849170|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>struct node *tempNode = (struct node*) malloc(sizeof(struct node));
</code></pre>

you're dynamically allocating space to a node pointer tempNode and typecasting it to be a struct node* type. (pointer to structure)

<pre><code>struct node *tempNode = malloc(sizeof(struct node));
</code></pre>

The same as above, except without the explicit typecast (struct node*)

blocks|key|1849229|text|这是没有区别的。由于malloc返回类型void+*，因此可以将其赋给任何指针类型的变量，而无需强制转换。正如John+Bode所提到的，强制转换malloc的结果被认为是不好的做法，因为它可能会掩盖由于具有malloc的其他定义而导致的编译器错误。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1849230|entityMap^0|A|6|K|6|21|6|2W|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]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

There is no difference. Since <code>malloc</code> returns type <code>void *</code>, it can be assigned to a variable of any pointer type without casting. As John Bode mentioned, casting the result of <code>malloc</code> is considered bad practice because it may mask the compiler error that would result from having some other definition of <code>malloc</code>.

blocks|key|1849236|text|既然你的造型是正确的，那么两者之间就没有区别。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1849237|但在C中，这种强制转换通常是不受欢迎的，因为void*可以隐式转换为任何其他(数据)指针类型。|offset|length|style|CODE|1849238|另请参阅以下C常见问题条目：What's+wrong+with+casting+malloc's+return+value?|1849239|第三个在那里：|1849240|struct+node+*tempNode+=+malloc(sizeof+*tempNode);|code-block|syntax|javascript|1849241|这是更好的，因为它使得在类型改变的情况下改变较少。|1849242|entityMap|0|LINK|mutability|MUTABLE|url|http://c-faq.com/malloc/mallocnocast.html^0|0|M|5|0|E|1C|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|14|8|@]|9|@$D|15|E|16|1|17]]|A|$]]|$1|J|3|K|5|6|7|18|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|19|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

Since you are casting it correctly there's no difference between the two. 

But the cast is generally frowned upon in C since a <code>void*</code> can be implicitly converted into any other (data) pointer type.

Also see this C FAQ entry: <a href="http://c-faq.com/malloc/mallocnocast.html" rel="nofollow noreferrer">What's wrong with casting malloc's return value?</a>

A third one is there:

<pre><code>struct node *tempNode = malloc(sizeof *tempNode);
</code></pre>

which is better since it makes for less changes in case the type is changed.

blocks|key|1526514|text|通过调用malloc，分配的内存大小与struct+node的大小相同。它向分配的内存返回一个void*。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1526515|现在的问题是:您的两行代码执行的是完全相同的操作，但是当使用C%2B%2B编译器编译时，第二行代码将导致一个警告。|1526516|C允许将void*隐式转换为不同的指针类型，而C%2B%2B编译器将-至少-警告您，或者给您一个错误。|1526517|entityMap^0|4|6|J|B|1B|5|0|0|4|5|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|K|$]]

With the call to <code>malloc</code>, memory is allocated that has the size of the <code>struct node</code> in bytes. It returns a <code>void*</code> to the allocated memory.

Now the thing is: both of your lines do exactly the same, but the second line will lead to a warning when compiled with a C++ compiler.

C allows implicit conversion of a <code>void*</code> to a different pointer type, whereas the C++ compiler will - at least - warn you, or give you an error.

<pre><code>struct node *tempNode = (struct node*) malloc(sizeof(struct node));

//and

struct node *tempNode = malloc(sizeof(struct node));
</code></pre>

What is the difference between these two lines?

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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struct node *tempNode = (struct node*) malloc(sizeof(struct node));//andstruct node *tempNode = malloc(sizeof(struct node));

问这两行之间的区别是什么？
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问这两行之间的区别是什么？EN