C ++初级问题：解引用对乘法 [英] C++ beginner question: dereference vs multiply

问题描述

只是进入C ++。当我看到乘法符号（ * ）被用于表示变量的取消引用时，我会不断地被抛弃。

例如：
unsigned char * pixels = vidgrabber.getPixels（）;

这会抛弃别人吗？

谢谢。

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pps我有另一个相当简单的问题，没有得到回答:(这里：

解决方案

C和继承C ++，被操作符淹没，并且内在上下文敏感。你必须习惯它：

如果 * 在要声明的变量名称 （或定义）之前出现 类型修饰符 ，并使该变量成为指针。

如果是 一元前缀操作符 >对于作为表达式一部分的变量， 取消引用（或任何重载的内容）。

如果是 （）

（从这里可以看到 * / code>不是取消引用的一元前缀，而是一个类型修饰符。）

注意 & 几乎类似于 * ，只是它的含义是不同的：它使一个变量引用 -of运算符，或二进制AND。

Just getting into C++. I'm getting constantly thrown off track when I see the symbol for multiply (*) being used to denote the dereferencing of a variable

for example: unsigned char * pixels = vidgrabber.getPixels();

Does this throw other people off? What's the tip for getting my head around this?

Thank you.

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p.s. I have another reasonably simple question, that didn't get answered :( here: beginner question: add/subtract to value rather than just be that value pretty please! and thanks for your time!

解决方案

C, and by inheritance C++, are swamped with operators and are inherently context-sensitive. You will have to get used to it:

If * appears before the name of a variable that is being declared (or defined), it's a type modifier and makes that variable a pointer.
If it is a unary prefix operator for a variable that is part of an expression, it's dereferencing (or whatever it's been overloaded to).
If it is a binary infix operator for two variables that are part of an expression, it's multiplication (or whatever it's been overloaded to).

(From this you can see that the * in your unsigned char * pixel isn't a dereferencing unary prefix, but a type modifier.)

Note that & pretty much resembles *, only it's meaning is different: it makes a variable a reference, is the address-of operator, or the binary AND.

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