This was probably asked somewhere but I couldn't find it. Could someone clarify why this code compiles and prints out 1?
这可能是在某个地方被问到但我找不到它。有人可以澄清为什么这个代码编译并打印1?
long i = (byte) + (char) - (int) + (long) - 1;
System.out.println(i);
3 个解决方案
#1
42
It's being parsed as this:
它被解析为:
long i = (byte)( +(char)( -(int)( +(long)(-1) ) ) );
where all the + and - operators are unary + or -.
所有+和 - 运算符都是一元+或 - 。
In which case, the 1 gets negated twice, so it prints out as a 1.
在这种情况下,1被否定两次,因此它打印为1。
#2
5
Because both '+' and '-' are unary operators, and the casts are working on the operands of those unaries. The rest is math.
因为'+'和' - '都是一元运算符,而演员正在研究那些一元的操作数。剩下的就是数学。
#3
5
Unary operators and casting :)
一元运算符和铸造:)
+1 is legal
+1是合法的
(byte) + 1 is casting +1 to a byte.
(byte)+ 1将+1转换为一个字节。
Sneaky! Made me think.
偷偷摸摸的!让我思考。
#1
42
It's being parsed as this:
它被解析为:
long i = (byte)( +(char)( -(int)( +(long)(-1) ) ) );
where all the + and - operators are unary + or -.
所有+和 - 运算符都是一元+或 - 。
In which case, the 1 gets negated twice, so it prints out as a 1.
在这种情况下,1被否定两次,因此它打印为1。
#2
5
Because both '+' and '-' are unary operators, and the casts are working on the operands of those unaries. The rest is math.
因为'+'和' - '都是一元运算符,而演员正在研究那些一元的操作数。剩下的就是数学。
#3
5
Unary operators and casting :)
一元运算符和铸造:)
+1 is legal
+1是合法的
(byte) + 1 is casting +1 to a byte.
(byte)+ 1将+1转换为一个字节。
Sneaky! Made me think.
偷偷摸摸的!让我思考。