奇怪的是"无法推断'T'的模板参数"错误

Question

错误在此代码中:

//myutil.h
template <class T, class predicate>
T ConditionalInput(LPSTR inputMessage, LPSTR errorMessage, predicate condition);    

//myutil.cpp
template <class T, class Pred>
T ConditionalInput(LPSTR inputMessage, LPSTR errorMessage, Pred condition)
{
        T input
        cout<< inputMessage;
        cin>> input;
        while(!condition(input))
        {
                cout<< errorMessage;
                cin>> input;
        }
        return input;
}

...

//c_main.cpp 
int row;

row = ConditionalInput("Input the row of the number to lookup, row > 0: ",
"[INPUT ERROR]: Specified number is not contained in the range [row > 0]. "
"Please type again: ", [](int x){ return x > 0; });

错误是:

Error   1       error C2783: 'T ConditionalInput(LPSTR,LPSTR,predicate)' :
could not deduce template argument for 'T' c_main.cpp        17      1

我一直在努力工作几个小时,但似乎无法找到解决方案.我认为错误可能是微不足道的,但在类似情况下我找不到其他人遇到错误.非常感谢!

编辑:Frederik Slijkerman所做的更正解决了一个问题,但创造了另一个问题.这次错误是:

Error   1   error LNK2019: unresolved external symbol "int __cdecl ConditionalInput<int,class `anonymous namespace'::<lambda0> >(char *,char *,class `anonymous namespace'::<lambda0>)" (??$ConditionalInput@HV<lambda0>@?A0x109237b6@@@@YAHPAD0V<lambda0>@?A0x109237b6@@@Z) referenced in function _main

请耐心帮我解决这个问题.

Answer 1

C++无法推断出函数的返回类型.它只适用于它的参数.你必须明确地打电话ConditionalInput<int>(...).