C++ 中此示例的函数重载中的歧义解释；
我正在阅读 Stroustrup 的书，其中有关重载和相关歧义的部分。
有一个例子如下：
void f1(char);
void f1(long);
void k(int i)
{
f1(i); //ambiguous: f1(char) or f1(long)
}
正如评论所述，调用是不明确的。 为什么？
本书的前一节阐述了基于形式参数和实际参数匹配的 5 条规则。那么上面的函数调用不应该属于关于“促销”的规则 2 吗？ 我的猜测是“i”应该提升为 long，仅此而已。
根据评论，似乎 int 到 char 的转换（降级？）也符合规则 2？
I'm reading Stroustrup's book, the section on overloading and related ambiguities.
There's an example as follows:
void f1(char);
void f1(long);
void k(int i)
{
f1(i); //ambiguous: f1(char) or f1(long)
}
As the comment states, the call is ambiguous.
Why?
The previous section in the book stated 5 rules based on matching formal and actual parameters. So shouldn't the above function call come under rule 2, regarding "promotions"?
My guess was that 'i' should be promoted to a long, and that's that.
As per the comment, it seems that a int to char conversion (a demotion?) also comes under rule 2?
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从上面的 int 得到的任何东西都不再是促销了。任何小于 int 到 int 的值都是升级（极少数情况除外 - 见下文）
因此，如果您更改为以下内容，它就会变得明确，选择第一个
注意，这仅在
int< 的平台上才正确/code> 可以存储所有unsigned Short的值。在情况并非如此的平台上，这不会是促销，并且调用是不明确的。在此类平台上，unsigned int类型将成为升级目标。浮点也会发生同样的事情。将
float转换为double是一种提升，但将double转换为long double则不是一种提升。在这种情况下，C++ 与 C 不同，其中double到long double同样是一种提升（但是，它没有重载）。Anything goint from int above isn't a promotion anymore. Anything going less than int to int is a promotion (except for rare cases - see below)
So if you change to the following it becomes non-ambiguous, choosing the first one
Notice that this is only true on platforms where
intcan store all values ofunsigned short. On platforms where that is not the case, this won't be a promotion and the call is ambiguous. On such platforms, the typeunsigned intwill be the promotion target.Sort of the same thing happens with floating points. Converting
floattodoubleis a promotion, butdoubletolong doubleisn't a promotion. In this case, C++ differs from C, wheredoubletolong doubleis a promotion likewise (however, it doesn't have overloading anyway).int 可以转换为 char，int 也可以转换为 long。
因此，从这个意义上说，它是不明确的，因为编译器无法判断您正在调用哪个。
An int can be converted to a char, and a int can also be converted to a long.
So in that sense it is ambiguous, as the compiler cannot tell which you're calling.