如何处理 Xcode 警告“没有以前的函数原型......"? [英] How to handle the Xcode warning "no previous prototype for function..."?

问题描述

此警告在某些第三方库中大量出现.

This warning is popping up a bunch in some third party libraries.

有没有办法在不修改代码的情况下处理它(例如忽略警告)?

Is there a way to handle it without modifying the code (e.g. ignore the warning)?

如果我必须修改代码来修复它，我该怎么做?

If I have to modify the code to fix it how do I do it?

这是导致警告的代码块之一:

Here's one of the code blocks that's causing a warning:

BOOL FBIsDeviceIPad() {
 #if __IPHONE_OS_VERSION_MAX_ALLOWED >= 30200
  if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad) {
   return YES;
  }
 #endif
  return NO;
}

推荐答案

通常对于这样的警告，您可以在文件顶部定义一个函数原型，例如:

Usually with warnings like this you can just define a function prototype at the top of your file, for instance:

BOOL FBIsDeviceIPad();

但是在 C 中的一个方法在大括号之间没有任何内容，即 () 实际上意味着有任意数量的参数.相反，定义应该变成 (void) 来表示无参数:

But in C a method with nothing between the braces, i.e. () actually implies there are an arbitrary number of parameters. Instead the definition should become (void) to denote no parameters:

BOOL FBIsDeviceIPad(void);

...

BOOL FBIsDeviceIPad(void) {
#if __IPHONE_OS_VERSION_MAX_ALLOWED >= 30200
  if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad) {
   return YES;
  }
#endif
  return NO;
}

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