为什么当用作格式参数时，NSInteger变量必须转换为long？ [英] Why does an NSInteger variable have to be cast to long when used as a format argument?

问题描述

NSInteger myInt = 1804809223;
NSLog(@"%i", myInt); <==== 

上述代码产生错误：

Values of type "NSInteger" should not be used as format arguments: add an explicit cast to 'long' instead.

正确的 NSLog 消息实际上是 NSLog（@％lg，（long）myInt）; 为什么要将myInt的整数值转换为long

The correct NSLog message is actually NSLog(@"%lg", (long) myInt); Why do I have to convert the integer value of myInt to long if I want the value to display?

推荐答案

如果您在OS X（64位）上编译，会收到此警告，因为在该平台上 NSInteger 定义为 long ，是一个64位整数。另一方面，％i 格式是用于32位的 int 。

You get this warning if you compile on OS X (64-bit), because on that platform NSInteger is defined as long and is a 64-bit integer. The %i format, on the other hand, is for int, which is 32-bit. So the format and the actual parameter do not match in size.

由于 NSInteger 是32位元或64位元，因此格式和实际参数大小不符。位，根据平台，编译器建议
将转换添加到 long 。

Since NSInteger is 32-bit or 64-bit, depending on the platform, the compiler recommends to add a cast to long generally.

更新：由于iOS 7现在也支持64位，您可以在为iOS编译
时获得相同的警告。

Update: Since iOS 7 supports 64-bit now as well, you can get the same warning when compiling for iOS.

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