为什么 NSInteger 变量在用作格式参数时必须强制转换为 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); <==== 

上面的代码产生了一个错误:

The code above produces an error:

'NSInteger' 类型的值不应用作格式参数；改为向long"添加显式转换

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

更正后的NSLog 消息实际上是NSLog(@"%lg", (long) myInt);.如果要显示值，为什么必须将 myInt 的整数值转换为 long?

The corrected 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

如果你在 OS X(64 位)上编译你会收到这个警告，因为在那个平台上 NSInteger 被定义为 long 是一个 64 位整数.另一方面，%i 格式用于 int，它是 32 位的.所以格式和实际参数大小不匹配.

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 位还是 64 位，根据平台不同，编译器推荐通常向 long 添加一个强制转换.

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，因此您在编译时会收到相同的警告适用于 iOS.

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

这篇关于为什么 NSInteger 变量在用作格式参数时必须强制转换为 long?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！