2012-01-11 174 views
0

在处理中阐述了一些想法之后,我决定将我的MIDI项目移动到C++以便移植到嵌入式平台。我决定使用RtMidi库进行MIDI I/O,但是我有一些麻烦布置代码,我想要它。我用C++还不是很棒。将RtMidi对象传递给函数(C++)

基本上,我想将RtMidiIn对象和RtMidiOut对象传递给我的printMidiPorts函数(代码与RtMidi捆绑的一些示例代码相同)。我知道这与将midiin和midiout初始化为指针有关,但我不完全确定。

这是我的代码:

#include <stdio.h> 
#include <iostream> 
#include <string> 
#include "rtmidi/RtMidi.h" 

using namespace std; 

void printMidiPorts(RtMidiIn midiin, RtMidiOut midiout) 
{ 
    // Check inputs. 
    unsigned int nPorts = midiin->getPortCount(); 
    std::cout << "\nThere are " << nPorts << " MIDI input sources available.\n"; 
    std::string portName; 
    for (unsigned int i=0; i<nPorts; i++) { 
     try { 
      portName = midiin->getPortName(i); 
     } 
     catch (RtError &error) { 
      error.printMessage(); 
      goto cleanup; 
     } 
     std::cout << " Input Port #" << i+1 << ": " << portName << '\n'; 
    } 

    // Check outputs. 
    nPorts = midiout->getPortCount(); 
    std::cout << "\nThere are " << nPorts << " MIDI output ports available.\n"; 
    for (unsigned int i=0; i<nPorts; i++) { 
     try { 
      portName = midiout->getPortName(i); 
     } 
     catch (RtError &error) { 
      error.printMessage(); 
      goto cleanup; 
     } 
     std::cout << " Output Port #" << i+1 << ": " << portName << '\n'; 
    } 
    std::cout << '\n'; 

    // Clean up 
    cleanup: 
    delete midiin; 
    delete midiout; 

} 

int main() 
{ 

    RtMidiIn *midiin = 0; 
    RtMidiOut *midiout = 0; 

    // RtMidiIn constructor 
    try { 
     midiin = new RtMidiIn(); 
    } 
    catch (RtError &error) { 
     error.printMessage(); 
     exit(EXIT_FAILURE); 
    } 

    // RtMidiOut constructor 
    try { 
     midiout = new RtMidiOut(); 
    } 
    catch (RtError &error) { 
     error.printMessage(); 
     exit(EXIT_FAILURE); 
    } 

    printMidiPorts(midiin, midiout); 

    return 0; 
} 

这是我的编译器输出:

lightArray.cpp: In function ‘void printMidiPorts(RtMidiIn, RtMidiOut)’: 
    lightArray.cpp:19: error: base operand of ‘->’ has non-pointer type ‘RtMidiIn’ 
    lightArray.cpp:24: error: base operand of ‘->’ has non-pointer type ‘RtMidiIn’ 
    lightArray.cpp:34: error: base operand of ‘->’ has non-pointer type ‘RtMidiOut’ 
    lightArray.cpp:38: error: base operand of ‘->’ has non-pointer type ‘RtMidiOut’ 
    lightArray.cpp:50: error: type ‘class RtMidiIn’ argument given to ‘delete’, expected pointer 
    lightArray.cpp:51: error: type ‘class RtMidiOut’ argument given to ‘delete’, expected pointer 
    lightArray.cpp: In function ‘int main()’: 
    lightArray.cpp:79: error: conversion from ‘RtMidiIn*’ to non-scalar type ‘RtMidiIn’ req 

任何帮助深表感谢。谢谢!

回答

1

它看起来像在主函数,midiinmidiout是类型RtMidiIn*RtMidiOut*(对象指针),而参数printMidiPorts是类型RtMidiInRtMidiOut(对象)。它看起来像你需要做的是改变签名printMidiPorts

+0

太棒了。工作完美 - 谢谢。 – distorteddisco 2012-01-11 03:45:35

0

你的函数签名是错误的。

这个:void printMidiPorts(RtMidiIn midiin, RtMidiOut midiout) 声明midiin和midiout为常规值,而不是指针。

void printMidiPorts(RtMidiIn *midiin, RtMidiOut *midiout)将是您的功能的正确签名。