binary'='：找不到带有'TradeRecord *'类型右手操作数的运算符（或者没有可接受的转换） [英] binary '=' : no operator found which takes a right-hand operand of type 'TradeRecord *' (or there is no acceptable conversion)

问题描述

 m_trades = ExtManagerPump-> TradesUserHistory（loginno，from，to，& m_trades_total）; 

这是我的麻烦给我一个解决方案

解决方案

这意味着<$ c $的类型c> m_trades 并且返回类型 TradesUserHistory 不是赋值兼容。如果编译时类型相同，或者返回类型是从 m_trades 的类型派生的，它们将是兼容的。



我解释了什么是错的，但请不要问我们该做什么;这与编写一些乱码并询问某人如何正确编写它是一样的。从编译器的角度来看，这个赋值是乱码，我们不知道你想通过这个不正确的陈述告诉你什么。

-SA

m_trades=ExtManagerPump->TradesUserHistory(loginno,from,to,&m_trades_total); 

This is my trouble give me a solution

解决方案

It means that the type of m_trades and and the return type of TradesUserHistory are not assignment-compatible. They would be compatible if the compile-time types were identical, or the return type was derived from the type of m_trades.

I explained what's wrong, but please don't ask us what to do; it would be the same as writing some gibberish and asking someone how to write it correctly. From the standpoint of the compiler, this assignment is gibberish, and we have no idea what did you want to tell by this incorrect statement.

—SA

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