“Initializer元素不是编译时常量”。为什么？ [英] "Initializer element is not a compile-time constant" why?

问题描述

我有这段代码：

- (NSString *) calculate: (uint) position {
    static NSArray * localArray = [NSArray arrayWithArray: self.container.objects ];
    // some un related code
    return obj;
}

编译器抱怨说：Initializer元素不是编译时常量 。它发生在我向localArray添加static时。但为什么？

The compiler complains saying: "Initializer element is not a compile-time constant". It happened when I added "static" to localArray. But why?

推荐答案

因为 [NSArray arrayWithArray：self.container.objects] 不是编译时常量，它是必须在运行时计算的表达式。在C和Objective-C中， static 函数内部的变量必须使用编译时常量进行初始化，而C ++和Objective-C ++则更宽松，允许非编译时常量。

Because [NSArray arrayWithArray: self.container.objects ] isn't a compile-time constant, it's an expression that must be evaluated at runtime. In C and Objective-C, static variables inside functions must be initialized with compile-time constants, whereas C++ and Objective-C++ are more lenient and allow non-compile-time constants.

将代码编译为Objective-C ++，或者将其重构为以下内容：

Either compile your code as Objective-C++, or refactor it into something like this:

static NSArray *localArray = nil;
if (localArray == nil)
    localArray = [NSArray arrayWithArray: self.container.objects ];

这与编译器为<$ c $引发的代码非常相似c> static 变量用非编译时常量初始化（实际上，它会使用第二个全局标志来指示值是否已初始化，而不是使用像<$ c $这样的标记值c> nil here;在这种情况下，我们假设 localArray 永远不会 nil ）。如果需要，可以查看编译器的反汇编。

Which is fairly similar to the code that the compiler would generate under the hood for a static variable initialized with a non-compile-time constant anyways (in actuality, it would use a second global flag indicating if the value was initialized, rather than using a sentinel value like nil here; in this case, we are assuming that localArray will never be nil). You can check out your compiler's disassembly for that if you want.

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