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typedef const char* Argv[3+ 8];
我在这一点上的猜测[3+8]是创建一个 Argv 类型的匿名数组(它的类型为 const char*)。如果我是对的,匿名数组部分基本上是没有意义的。
[3+8]
它定义Argv为“11 个 const char* 数组”的同义词。
Argv
如同
const char* somevar[11];
或 IOW
Argv somevar;