c ++将const对象引用传递给函数

问题描述

错误:将const QByteArray"作为QByteArray&"的this"参数传递；QByteArray::append(const QByteArray&)' 丢弃限定符 [-fpermissive]

因为在作为函数参数传递时使对象成为 const 是一种约定，所以我已经这样做了.但现在我收到一个错误！！，我不想使函数保持不变，因为我必须将 qbyte 数组中的数据转换为 short，然后将其附加到另一个数组.

since it is a convention to make objects const while passing as function arguments i have done it.but now i am getting an error!!,i dnt want to make the function constant as i have to convert data in qbyte array into short and then append it another array.

QByteArray ba((const char*)m_output.data(), sizeof(ushort));
    playbackBuffer.append(ba);

我真的需要把这个数组传入playbackbuffer;
它给我一个关于 playbackBuffer.append(ba);

I really need to pass this array into playbackbuffer;
It is giving me an error on playbackBuffer.append(ba);

请帮忙
提前致谢

please help
thanks in advance

推荐答案

这意味着您正在对 const 成员调用非常量成员函数.据推测，您的 append 函数修改了字节数组.使用 const 引用，您不应该修改.

This means you are calling a non-const member function on a const member. Presumably, your append function modifies the byte array. With a const reference, you shouldn't be modifying.

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