C++ int num = *(int *)number; 这有什么作用？
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int num = *(int *)number; What does this do?
提问by Charles Ma
I was looking at some c++ code, and I saw this:
我在看一些 C++ 代码，我看到了这个：
int num = *(int *)number;
I had never seen this before? it was in a function labeled as such:
我以前从未见过这个？它在一个标记为这样的函数中：
void *customer(void *number){ }
What does that even do? Is there a different way to show this?
那到底有什么作用呢？有没有不同的方式来展示这一点？
Thanks, this isn't homework btw I was just confused at what this does?
谢谢，这不是作业顺便说一句，我只是对它的作用感到困惑？
回答by Charles Ma
The (int *) part casts the variable number to a pointer to an int, then the * in front dereferences it to an int.
(int *) 部分将变量 number 转换为指向 int 的指针，然后前面的 * 将其取消引用为 int。
回答by Steve Jessop
The function takes a void*, but somehow it knows (perhaps it's required in some documentation somewhere) that the pointer it's given actually points to an int.
该函数采用 a void*，但不知何故它知道（可能在某处的某些文档中需要它）它给出的指针实际上指向int.
So, (int*)numberis "the original pointer, converted to an int*so that I can read an intfrom it", and *(int*)numberis the int value that it points to.
所以，(int*)number是“原始指针，转换为 anint*以便我可以从中读取 an int”，*(int*)number是它指向的 int 值。
回答by Bill K
The correct answers are already here, but can I tell you a trick that generally helped me when I had to use C a lot?
正确答案已经在这里，但我能告诉你一个在我不得不大量使用 C 时通常对我有帮助的技巧吗？
It's how you pronounce "*" in your head--and there are two parts.
这就是你在头脑中发音“*”的方式——有两个部分。
The common part is when it is part of a type--and everybody probably says "pointer" when they read that, which is great. So (int *) is an int pointer--or I'll even reverse it in my head to read "pointer to an int" which seems to help a little.
常见的部分是当它是类型的一部分时——每个人在阅读时可能都会说“指针”，这很好。所以 (int *) 是一个 int 指针——或者我什至会在我的脑海中反转它来阅读“指向 int 的指针”，这似乎有点帮助。
The thing that helps a lot for me is whenever you see * in your code--read it as "what is pointed to by".
对我有很大帮助的事情是每当您在代码中看到 * 时——将其读作“指向的内容”。
If you follow this pattern, then:
如果你遵循这个模式，那么：
int num = *(int *)number;
is an integer variable "num" gets assigned the value: what is pointed to by an int pointer, number. It just translates itself.
是一个整数变量“num”被分配了值：int 指针所指向的值，number。它只是自己翻译。
Sometimes you have to mess with the phrasing a little, but since I got into that habit I've never had a big problem reading pointer code.
有时你不得不稍微弄乱措辞，但自从我养成这个习惯后，我在阅读指针代码时从来没有遇到过大问题。
I believe I also read & as "The address of" in C, but I think it's been overloaded in C++ if I recall correctly.
我相信我在 C 中也将 & 读为“的地址”，但如果我没记错的话，我认为它在 C++ 中被重载了。
回答by DVK
The function accepts a void pointer (thus void *). In order to dereference it to a variable of a certain type (e.g. int) - which is what teh first "*" does - you need to cast it to a pointer to an actual type - in this case to an int pointer via (int *) cast
该函数接受一个 void 指针（因此 void *）。为了将它解引用到某种类型的变量（例如 int）——这是第一个“*”所做的——你需要将它转换为指向实际类型的指针——在这种情况下，通过 (int *） 投掷
回答by ta.speot.is
I'm assuming customeris used like this:
我假设customer是这样使用的：
int lookup = 123;
customer_key *key = customer(&lookup);
// do something with key here
In which case, the code in customer is typecasting the void *to an int *and then dereferencing it (getting its value). It has to typecast first because void *basically means "pointer to something", which allows you to pass in any type you want. Without the typecast the compiler doesn't know if you want to read a char(usually 1 byte), a short(usually 2 bytes) or an int(usually 4 bytes). The typecast removes the ambiguity.
在这种情况下， customer 中的代码正在将 the 类型转换void *为 anint *然后取消引用它（获取其值）。它必须首先进行类型转换，因为void *基本上意味着“指向某物的指针”，它允许您传入任何您想要的类型。如果没有类型转换，编译器不知道您是否要读取 a char（通常为 1 个字节）、a short（通常为 2 个字节）或 an int（通常为 4 个字节）。类型转换消除了歧义。
Note using void *for the argument is probably not the best, since you could do:
注意void *用于参数可能不是最好的，因为你可以这样做：
double lookup = 69.0f;
customer_key *key = customer(&lookup);
And this will compile, but won't look up customer 69 (a doubleis not an int!).
这将编译，但不会查找客户 69（adouble不是int！）。
The use of void *may be intentional, the code may be able to determine (hopefully safely) between pointers and an argument like: (void *)3- which would be a special case.
的使用void *可能是有意的，代码可能能够（希望安全地）确定指针和参数之间，例如：(void *)3- 这将是一种特殊情况。
