String a = "576055795";
long b = 10*Integer.parseInt(a);
long c = 10*Long.parseLong(a);
System.out.println(b); //Prints 1465590654
System.out.println(c); // Prints 5760557950
Why are they different?
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试):
问题:
回答1:
Integer.parseInt() returns an int, which is a signed 32-bit integer. 10 is also an int; multiplying 576055795 by 10 as ints overflows and yields an int, which is then promoted to a long.
Long.parseLong() returns a long, which is a signed 64-bit integer. Multiplying it by 10 yields a long with no overflow.
