关于c＃：当看起来没有涉及“long”类型时，不能隐式地将类型’long’转换为’int’

Cannot implicitly convert type 'long' to 'int', when it seems no `long` type involved

另一个跟踪错误的帖子Cannot implicitly convert type 'long' to 'int'

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public int FindComplement(int num) {
uint i = 0;
uint mask = ~i;

while((mask&num) != 0) mask <<= 1;
//return ~mask^num; //<-- error CS0266
return (int)~mask^num; //<--it works with (int)
}

不好意思问了太多问题，我想知道为什么return ~mask^num会引起错误，比如

error CS0266: Cannot implicitly convert type 'long' to 'int'. An explicit conversion exists (are you missing a cast?)

在我的环境中，return ~mask^num;会导致错误，而return (int)~mask^num可以工作。这里似乎没有涉及到long型。

相关讨论

您试图对操作数int和uint执行^操作。没有这样的运算符，因此两个操作数都转换为long，并使用long ^(long, long)运算符。

根据ECMA C_5规范第12.4.7.1节：

Numeric promotion consists of automatically performing certain implicit conversions of the operands of
the predefined unary and binary numeric operators. Numeric promotion is not a distinct mechanism, but
rather an effect of applying overload resolution to the predefined operators. Numeric promotion
specifically does not affect evaluation of user-defined operators, although user-defined operators can be
implemented to exhibit similar effects.

从12.4.7.3：

Binary numeric promotion occurs for the operands of the predefined +, –, *, /, %, &, |, ^, ==, !=, >, <, >=,
and <= binary operators. Binary numeric promotion implicitly converts both operands to a common type which, in case of the non-relational operators, also becomes the result type of the operation. Binary numeric promotion consists of applying the following rules, in the order they appear here:

... (rules that don't apply here)

Otherwise, if either operand is of type uint and the other operand is of type sbyte, short, or int,
both operands are converted to type long.