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我有一个Note和一个Track班级，两者都是*generator成员。当我创建新Note对象时，我想将成员链接到的generator成员Note，Track但我不知道如何做到这一点。

#include <iostream>
using namespace std;

class Generator {
public:
    virtual float getSample(int sample)=0;
};

class Note {
public:

    Generator *generator; // THIS IS WHAT IS CAUSING ME TROUBLE
    Note(Generator *aGen){
        generator = aGen;
    }
};

class Synth  : public Generator{
public:
    virtual float getSample(int sample);
    int varA;
    int varB;
    Synth(){
        varA = 5;
        varB = 8;
    }
};


float Synth::getSample(int sample){
    varA = sample;
    varB = 3;

    return 0;
}

class Track {
public:
    Generator *generator;
    Track(){
        generator = new Synth();
    }
};

int main() {
    cout << "!!!Hello World!!!" << endl; // prints !!!Hello World!!!
    Track track = Track();
    cout << "test" << endl;
    return 0;
}

我想过做这样的事情，但它不起作用：

Track track = Track();
Note n = Note(&track.generator);

错误

prog.cpp: In function ‘int main()’: 
prog.cpp:48:35: error: no matching function for call to ‘Note::Note(Generator**)’ 
prog.cpp:48:35: note: candidates are: 
prog.cpp:13:5: note: Note::Note(Generator*) 
prog.cpp:13:5: note: no known conversion for argument 1 from ‘Generator**’ to ‘Generator*’ prog.cpp:9:7: note: Note::Note(const Note&) 
prog.cpp:9:7: note: no known conversion for argument 1 from ‘Generator**’ to ‘const Note&’ prog.cpp:48:10: warning: unused variable ‘n’ [-Wunused-variable] - See more at: http://ideone.com/E38ibe#sthash.V3QMcYJQ.dpuf

活的例子在这里。

4

2 回答 2

1

track.generator已经是一个指针Generator，你不需要获取它的地址。

就留在身边

Node n = Node(track.generator); // without & operator

更新代码：http: //ideone.com/fAA4JX

于 2013-03-03T13:25:18.003 回答
0

正如编译器告诉你的那样，这一行：

Note n = Note(&track.generator);

尝试构造 aNote并将 a 提供Generator**给它的构造函数（因为track.generator有 type Generator*，所以&track.generator有 type Generator**）。

但是，您的Note类构造函数接受 a Generator*，而不是 a Generator**。只需这样做（注意，这里不需要复制初始化，而是使用直接初始化）：

Track track;
Note n(track.generator);
于 2013-03-03T13:25:38.130 回答