的的std :: find_if 函数的返回值类型等于迭代器的类型，传过来的参数。在你的情况，因为你正在传递的std :: reverse_iterator的＆LT; XYZ *＆GT; 取值为参数，返回类型为的std :: reverse_iterator的＆LT; XYZ *＆GT; 。这意味着

The std::find_if function has a return type equal to the type of iterator passed in as a parameter. In your case, since you're passing in std::reverse_iterator<xyz*>s as parameters, the return type will be std::reverse_iterator<xyz*>. This means that

found = std::find_if(std::reverse_iterator<xyz*>(end),
                     std::reverse_iterator<xyz*>(begin),
                     findit);

将无法编译，因为发现是 XYZ * 。

要解决这个问题，你可以试试这个：

To fix this, you can try this:

std::reverse_iterator<xyz*>
rfound = std::find_if(std::reverse_iterator<xyz*>(end),
                      std::reverse_iterator<xyz*>(begin),
                      findit);

这将解决编译错误。但是，我认为在这行你两个次级错误：

This will fix the compiler error. However, I think that you two secondary errors in this line:

if (found != std::reverse_iterator<xyz*>(end));

首先，请您有如果语句后一个分号，因此如果语句的身体会不管条件是否为真来评价。

First, note that you have a semicolon after the if statement, so the body of the if statement will be evaluated regardless of whether the condition is true.

二，请注意，的std :: find_if 如果没什么predicate匹配，则返回第二个迭代器作为一个定点。因此，这种测试应

Second, note that std::find_if returns the second iterator as a sentinel if the nothing matches the predicate. Consequently, this test should be

if (rfound != std::reverse_iterator<xyz*>(begin))

由于 find_if 将返回的std :: reverse_iterator的＆LT; XYZ *＆GT;（开始）如果元素不找到。

because find_if will return std::reverse_iterator<xyz*>(begin) if the element is not found.

希望这有助于！