Xcode4 架构x86_64的未应答未定义符号

我得到了错误，全部是：“架构x86_64的未定义符号”

发生此错误的最简单类是名为Peaking的类，该类计算均衡

这是我的类定义峰值

#ifndef _PEAKING_
#define _PEAKING_

#include "IPlugBase.h"
#include "IGraphics.h"

class Peaking
{
public:
  Peaking()
  {
      a = new double[3];
      b = new double[3];

      previousx1=0;
      previouspreviousx1=0;
  }
  ~Peaking() {}

  double* a;
  double* b;

  void ComputeParams(double G, double fc, double Q, double fs);
  double ComputeAudio(double x);

    double previouspreviousx1;
    double previousx1;


private:

};

#endif

这里是实现

#include <math.h>
#include "defines.hpp"
#include "Peaking.h"

void Peaking::ComputeParams(double G, double fc, double Q, double fs)
{

double K = tan((M_PI * fc)/fs);
double V0 = pow(10,(G/20.));

//Invert gain if a cut
if (V0 < 1)
    V0 = 1/V0;

double a1, a2, b0, b1, b2;

////////////////////////////
//   BOOST
////////////////////////////
double K2 = pow(K,2.);

double divQ = 1/Q;
double z1 = (divQ*K);
double divz2 = 1/(1 + z1 + K2);
double z3 = ((V0*divQ)*K);
double divz4 = 1/(1 + z3 + K2);


if( G > 0 )
{
    b0 = (1 + z3 + K2) * divz2;
    b1 =        (2 * (K2 - 1)) * divz2;
    b2 = (1 - z3 + K2) * divz2;
    a1 = b1;
    a2 = (1 - z1 + K2) * divz2;
}
////////////////////////////
//    CUT
////////////////////////////
else
{
    b0 = (1 + z1 + K2) * divz4;
    b1 = (2 * (K2 - 1)) * divz4;
    b2 = (1 - z1 + K2) * divz4;
    a1 = b1;
    a2 = (1 - z3 + K2) * divz4;
}

a[0] = 1;
a[1] = a1;
a[2] = a2;

b[0] = b0;
b[1] = b1;
b[2] = b2;

}


double Peaking::ComputeAudio(double x)
{
    // E:\Programmation\#Audio\#EQ\dfilt_df2.gif
    double srg = a[2];

    double x1 = x - a[1] * previousx1 - a[2] * previouspreviousx1;

    double y = x1 * b[0] + previousx1 * b[1] +  previouspreviousx1 * b[2];

    previouspreviousx1 = previousx1;
    previousx1 = x1;


    return y;


}

如果我将函数“double Peaking:：ComputeAudio（double x）”的核心放在类中，一个错误就会消失

如果我将Vumeter的虚拟函数“draw”的主体放入，另一个错误就会消失所以我不能在.cpp文件中放入一些主体函数，为什么？

谢谢你的帮助

---

Jeff

好的，我发现：“构建阶段/编译源代码”面板中没有一些.cpp文件

现在它开始工作了

我对在OSX上开发相对较新，因此这个noob错误

---

Jeff

我想说清楚，你确定你的cpp文件正在编译吗？我没有说我必须把所有的函数都放在.h文件中，只是有些函数而已，所以我使用WDL-OL在各种架构（应用程序、AU、VST、AAX等）中创建音频插件，到目前为止，应用程序工作正常，但我在尝试编译AU插件时遇到了问题

Undefined symbols for architecture x86_64:
  "OnOff::OnOff(IPlugBase*, int, int, int, IBitmap*, IBitmap*, IChannelBlend::EBlendMethod)", referenced from:
      AutoEQ::AutoEQ(IPlugInstanceInfo) in AutoEQ.o
  "Peaking::ComputeAudio(double)", referenced from:
      AutoEQ::traiterLearning(int) in AutoEQ.o
      AutoEQ::ProcessDoubleReplacing(double**, double**, int) in AutoEQ.o
  "Peaking::ComputeParams(double, double, double, double)", referenced from:
      AutoEQ::traiterLearning(int) in AutoEQ.o
  "FFTOoura::rdft(int, int, double*, int*, double*)", referenced from:
      AutoEQ::traiterLearning(int) in AutoEQ.o
  "vtable for ProgressBar", referenced from:
      ProgressBar::ProgressBar(IPlugBase*, int, int, int, IBitmap, IBitmap, IChannelBlend::EBlendMethod) in AutoEQ.o
  NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.
  "vtable for EQClass", referenced from:
      EQClass::EQClass(IPlugBase*, int) in AutoEQ.o
  NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.
  "vtable for VUmetre", referenced from:
      VUmetre::VUmetre(IPlugBase*, int, int, int, IBitmap, IBitmap, IChannelBlend::EBlendMethod) in AutoEQ.o
  NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.
  "vtable for MyButton", referenced from:
      MyButton::MyButton(IPlugBase*, int, int, int, IBitmap*, IChannelBlend::EBlendMethod) in AutoEQ.o
  NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)