将RtMidi对象传递给函数(C++)
问题描述:
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!
答
它看起来像在主函数,midiin
和midiout
是类型RtMidiIn*
和RtMidiOut*
(对象指针),而参数printMidiPorts
是类型RtMidiIn
和RtMidiOut
(对象)。它看起来像你需要做的是改变签名printMidiPorts
。
答
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。
太棒了。工作完美 - 谢谢。 – distorteddisco 2012-01-11 03:45:35