这段代码有什么不对，数字溢出警告(what's wrong this this code, Numeric overflow warning)

这段代码有什么不对，数字溢出警告(what's wrong this this code, Numeric overflow warning) fillColor = (fillColor & (0xFF << 24)) | (colorRGB & 0xFFFFFF);

以上是我的代码，我认为这应该没问题。 但Android Studio给了我一个警告：

表达式中的数字溢出少...（Ctrl + F1）

该检查检查在计算期间溢出的表达式，即：a = 1.0 / 0.0;

我找不到代码有什么问题，请帮忙。

fillColor = (fillColor & (0xFF << 24)) | (colorRGB & 0xFFFFFF);

Above is my code, and I think this should be OK. But the Android Studio give me a warning saying :

Numeric overflow in expression less... (Ctrl+F1)

This inspection checks for expressions which overflow during computation, i.e.:       a = 1.0/0.0;

I can not find what is the matter with the code, please help.

最满意答案

0xFF << 24等于0xFF000000 ，“技术上”不适合作为有符号值的int（例如，改变<< 24到23并看到错误消失）。

这个“hack”将值设置为long，但然后将其转换为int。 相同的位最终会出现在值中，但是演员特别告诉编译器您很乐意以这种方式截断该值。

(int)((long)0xFF << 24);

所以为了特别修复你的代码，我将0xFF << 24扩展为long（注意尾随L），然后将其转换为int。

fillColor = (fillColor & (int)0xFF000000L) | (colorRGB & 0xFFFFFF);

另请参见为什么Java能够将0xff000000存储为int？

0xFF << 24 equals 0xFF000000 which "technically" doesn't fit into an int as a signed value (eg. change << 24 to 23 and see the error go away).

This "hack" sets up the value as a long, but then casts it to an int. The same bits end up in the value but the cast specifically tells the compiler you're happy to truncate the value in this way.

(int)((long)0xFF << 24);

So to fix your code in particular, i have expanded 0xFF << 24 as a long (note the trailing L) and then cast it to an int.

fillColor = (fillColor & (int)0xFF000000L) | (colorRGB & 0xFFFFFF);

See also Why is Java able to store 0xff000000 as an int?