难道我投malloc的结果？ [英] Do I cast the result of malloc?
问题描述
在<一个href=\"http://stackoverflow.com/questions/571945/getting-a-stack-overflow-exception-when-declaring-a-large-array\">this问题，有人在评论中建议我的的不的投的malloc 的结果，即的：
In this question, someone suggested in a comment that I should not cast the results of malloc, i.e:
int *sieve = malloc(sizeof(int)*length);
而不是：
int *sieve = (int *)malloc(sizeof(int)*length);
为什么会是这种情况？
Why would this be the case?
推荐答案
否;您的不的投出结果，因为：
No; you don't cast the result, since:
- 这是不必要的，因为
无效*自动，安全地提升到在这种情况下任何其他指针类型。 - 它可以隐藏一个错误，如果你忘了，包括
＆LT;＆stdlib.h中GT;。这可能会导致崩溃（或者更糟，的不的导致崩溃，直到后来的方式在code的一些完全不同的一部分）。试想，如果指针和整数为不同大小会发生什么;那么你通过铸造隐藏警告，可能会失去你的返回地址位。 - 它增加混乱到code，强制转换不是很容易阅读（特别是如果该指针类型是长）。
- 这让你重复自己，这是普遍不好。
- It is unnecessary, as
void *is automatically and safely promoted to any other pointer type in this case. - It can hide an error, if you forgot to include
<stdlib.h>. This can cause crashes (or, worse, not cause a crash until way later in some totally different part of the code). Consider what happens if pointers and integers are differently sized; then you're hiding a warning by casting and might lose bits of your returned address. - It adds clutter to the code, casts are not very easy to read (especially if the pointer type is long).
- It makes you repeat yourself, which is generally bad.
作为一个澄清，请注意，我说：你不要投，而不是你不要的需求的投。在我看来，这是一个失败，包括中投，即使你是正确的。有简单地做没有任何好处，但一帮潜在的风险，以及包括中投表示您不知道的风险。
As a clarification, note that I said "you don't cast", not "you don't need to cast". In my opinion, it's a failure to include the cast, even if you got it right. There are simply no benefits to doing it, but a bunch of potential risks, and including the cast indicates that you don't know about the risks.
另外请注意，评论家指出，约有直C，不是C以上的会谈++。我非常坚信C和C ++作为单独的语言。
Also note, as commentators point out, that the above talks about straight C, not C++. I very firmly believe in C and C++ as separate languages.
要进一步增加，你的code不必要的重复类型信息（ INT ），这可能会导致错误。这是更好地取消引用所使用的指针来保存返回值，锁定两个一起：
To add further, your code needlessly repeats the type information (int) which can cause errors. It's better to dereference the pointer being used to store the return value, to "lock" the two together:
int *sieve = malloc(length * sizeof *sieve);
这也是移动长度来前增加知名度，并且丢弃原来的冗余括号的sizeof ;他们的只需要的当参数是一个类型的名字。许多人似乎并不知道（或忽略）这一点，这使得他们的code更详细。请记住：的sizeof 不是一个函数！ :)
This also moves the length to the front for increased visibility, and drops the redundant parentheses with sizeof; they are only needed when the argument is a type name. Many people seem to not know (or ignore) this, which makes their code more verbose. Remember: sizeof is not a function! :)
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