我不明白这一点 [英] I dont understand this

问题描述

您好我不明白这一点：



i有一个变量：



PULONG尝试;



为什么在我学习的代码中，我发现了这样的东西？ ：



试试[2]



i不明白因为PULONG不是和阵列。



谢谢

Hi i dont understand this :

i have one variable :

PULONG try;

why in the code that im studing, i found something like this? :

try[2]

i dont understand becuause PULONG isnt and array.

Thanks

推荐答案

为了使用索引器，类不必是一个数组：想一下泛型列表 - 您可以使用类似数组的索引器语法访问单个元素：

A class doesn't have to be an array in order to use indexers: think of a generic List - you can access individual elements using array-like indexer syntax:

List<String^>^ dinosaurs = gcnew List<String^>();
dinosaurs->Add("Tyrannosaurus");
dinosaurs->Add("Amargasaurus");
dinosaurs->Add("Mamenchisaurus");
dinosaurs->Add("Deinonychus");
dinosaurs->Add("Compsognathus");
for each(String^ dinosaur in dinosaurs )
{
    Console::WriteLine(dinosaur);
}
Console::WriteLine("\ndinosaurs[3]: {0}", dinosaurs[3]);

（取自MSDN示例： http://msdn.microsoft.com/en-us/library/6sh2ey19（v = vs.110）.ASPX CS-保存-lang = 1& cs-lang = cpp＃code-snippet-3 [ ^ ]）



因此，如果您的类派生自支持索引器的类，或实现索引器运算符

operator [] 然后完全允许索引。

(Taken from MSDN example: http://msdn.microsoft.com/en-us/library/6sh2ey19(v=vs.110).aspx?cs-save-lang=1&cs-lang=cpp#code-snippet-3[^])

So if your class is derived from a class that supports indexers, or implements the indexer operator
operator[] then indexes are perfectly allowed.

假设您使用 C （尝试是 C ++ 保留字，你不能用它来命名变量）和 PULONG 以这种方式声明

Assuming you are using C (try is a C++ reserved word, you cannot use it to name a variable) and PULONG is declared this way

typedef unsigned long * PULONG;

那么你可以写一下，例如

Then you may write, for instance

PULONG try = malloc(100*sizeof(unsigned long));
if ( ! try ) { /* handle error */}
try[2] = 5;
//...

free(try);

编译器不会抱怨。

And the compiler won't complain.

在C / C ++语言，表达式 x [y] 相当于 *（x + y），即内容地址（x + y）。使用此语法的正常（*）方式是 x 是一个数组或指针， y 是一个内存偏移量或数组索引。两种解释都是可能的，实际上是可互换的。



（*）：一个鲜为人知的后果是你实际可以交换 x 和 y 在这样的表达式中，不改变含义： y [x] 将被接受编译器，iff x [y] 是一个有效的表达式，两者的含义是相同的！

In the C/C++ language, the expression x[y] is equivalent to *(x+y) , that is, the contents of address (x+y) . The normal(*) way to use this syntax is that x is an array or a pointer, and y is a memory offset or array index. Either interpretation is possible, and in fact interchangible.

(*): A less known consequence is that you can actually swap x and y in such an expression, without changing the meaning: y[x] will be accepted by the compiler, iff x[y] is a valid expression, and the meaning of both is the same!

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