将负Integer.MIN转换为正long [英] Converting negative Integer.MIN to positive long

问题描述

我正在尝试使用以下代码将负Integer.MIN转换为正长值：

I am trying to convert negative Integer.MIN to positive long value using below code :

long lv = -1 * value_int;

除了v​​alue_int = Integer.MIN_VALUE的值外，上面的代码工作正常。在Integer.MIN_VALUE的情况下，lv的值始终为Integer.MIN_VALUE;

Above code works perfectly except when value of value_int = Integer.MIN_VALUE. In Integer.MIN_VALUE case, value of lv is always Integer.MIN_VALUE;

以下代码工作在所有情况下包括Integer.MIN_VALUE

Below code work in all cases including Integer.MIN_VALUE

long lv = value_int;
lv = -1 * value_int;

任何想法为什么long lv = -1 * value_int在Integer.MIN_VALUE情况下不起作用？

Any idea why long lv = -1 * value_int does not work in Integer.MIN_VALUE case?

推荐答案

1 （前面是一元 - 运算符，在本例中是一个 int 字面值。由于 value_int 也是 int ，因此乘法是以 int - 哪个溢出，然后才提升为长。如果您使用 long 字面值，则乘法将作为long进行，您将得到正确的结果：

1 (preceded by the unary - operator, in this case) is an int literal. Since value_int is also an int, the multiplication is done as an int - which overflows, and only then promoted to a long. If you use a long literal instead, the multiplication will be done as a long, and you'd get the correct result:

long lv = -1L * value_int;
// Here-----^

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