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过载问题

complex-numbers segmentation-fault c++

haster8558 · 技术社区 · 7 年前

我试图创建一组类来处理复数。我已经看到有一组用于复数的类,但因为我正在学习C++,所以我认为创建一个基本实现是一个好主意。当我试图使运算符“/”过载时,出现了问题。我遇到了一个segfault,我无法理解问题是否在于我对部门的执行:

复杂的水电站:

#include <iostream>
#include <cstdlib>

class Complex {
  float real;
  float imm;
public:
  Complex(float new_real = 0,float new_imm = 0) {this->real = new_real;this->imm = new_imm;}
  void set(float new_real,float new_imm) {this->real = new_real; this->imm = new_imm;}
  float get_real(void) const { return this->real;}
  float get_imm(void) const { return this->imm;}
  Complex conj(void) const {Complex tmp; tmp.set(this->real,-1.0 * this->imm); return tmp;}
  friend std::ostream& operator<<(std::ostream& os, const Complex& cpx) {os << "Real: " << cpx.real << " Imm: " << cpx.imm << std::endl; return os; }
  friend Complex operator*(const Complex& lhs,const Complex& rhs);
  friend Complex operator+(const Complex& lhs,const Complex& rhs);
  friend Complex operator+(const Complex& lhs,const float& rhs);
};

复杂的cpp:

#include "complex.hpp"


Complex operator*(const Complex& lhs,const Complex& rhs)
{
  float real_part = (lhs.real * rhs.real) - ( lhs.imm * rhs.imm);
  float imm_part = (lhs.real * rhs.imm) + ( lhs.imm * rhs.real);
  Complex result;
  result.set(real_part,imm_part);
  return result;
}

Complex operator+(const Complex& lhs,const Complex& rhs)
{
  float real_part = lhs.real + rhs.real;
  float imm_part = lhs.imm + rhs.imm;
  Complex result;
  result.set(real_part,imm_part);
  return result;
}

Complex operator+(const Complex& lhs,const float& rhs)
{
  float real_part = lhs.real + rhs;
  float imm_part = lhs.imm;
  Complex result;
  result.set(real_part,imm_part);
  return result;
}

Complex operator/(const Complex& lhs,const Complex& rhs)
{
  Complex numerator(0,0);
  numerator = rhs * rhs.conj();

  Complex denominator(0,0);
  denominator = lhs * rhs.conj();

  Complex result;
  float real_numerator = numerator.get_real();
  result = denominator / real_numerator;
  return result;
}

Complex operator/(const Complex& lhs,const float& rhs)
{
  float real_part = lhs.get_real() / rhs;
  float imm_part = lhs.get_imm() / rhs;
  Complex result;
  result.set(real_part,imm_part);
  return result;
}

2复数除法的全部思想是将分子共轭的分子和分母相乘,以便分子上只有一个实数。我想说清楚:

(a+ib)/(c+id)=((a+ib)/(c+id))*((c-id)/(c-id))=((a+ib)*(c-id))/(c^2+d^2)

现在,当我尝试这样做时:

主要的cpp:

int main(int argc, char *argv[])
{
  Complex x(4,8);
  Complex y(3,7);
  Complex result = x / y;
  result = x / 6;
  return 0;
}

我有一个错误,我不明白:

(gdb) break main
Breakpoint 2 at 0x401c56: file equalization_main.cpp, line 49.
(gdb) r
The program being debugged has been started already.
Start it from the beginning? (y or n) y
`/home/campiets/workspace/frontend/dfe_equalizer_fe/dev/view/src_c/test' has changed; re-reading symbols.
Starting program: /home/campiets/workspace/frontend/dfe_equalizer_fe/dev/view/src_c/test 

Breakpoint 2, main (argc=1, argv=0x7fffffffbf08) at equalization_main.cpp:49
49    Complex x(4,8);
(gdb) n
50    Complex y(3,7);
(gdb) n
51    Complex result = x / y;
(gdb) n

Program received signal SIGSEGV, Segmentation fault.
0x0000000000401e64 in Complex::Complex (this=<error reading variable: Cannot access memory at address 0x7fffff3feff8>, new_real=<error reading variable: Cannot access memory at address 0x7fffff3feff4>, 
    new_imm=<error reading variable: Cannot access memory at address 0x7fffff3feff0>) at complex.hpp:38
38    Complex(float new_real = 0,float new_imm = 0) {this->real = new_real; this->imm = new_imm;}

有什么想法吗?

2 回复 | 直到 7 年前

Sid S 7 年前

Complex operator/(const Complex& lhs,const Complex& rhs)
{
  ...
  Complex denominator...;
  ...
  float real_numerator = ...;
  result = denominator / real_numerator;
  ...
}

这是无限递归。

因为编译器没有看到 operator/(const Complex &lhs, const float &rhs) ,它转换 float 参数到 Complex 因此得到递归。

最简单的解决方案是声明或定义 操作员/(const Complex和lhs、const float和rhs) 之前 operator/(const Complex &lhs, const Complex &rhs) .

不过,我倾向于将操作符实现为类成员。这会产生更简单的源代码,也解决了问题。

R Sahu 7 年前

功能

Complex operator/(const Complex& lhs,const Complex& rhs) { ... }

导致堆栈溢出,因为行

result = denominator / real_numerator;

最终被解释为:

result = denominator / Complex(real_numerator);

您可以通过定义或声明

Complex operator/(const Complex& lhs, const float& rhs)

在它之前。

如果更改代码以使用:

Complex operator/(const Complex& lhs,const float& rhs)
{
   return Complex(lhs.get_real()/rhs, lhs.get_imm()/rhs);
}

Complex operator/(const Complex& lhs,const Complex& rhs)
{
   ...
}

您的程序将正常工作。

简化上述内容的建议 operator/ 作用

如果添加以下成员函数

float magnitude_square() const { return (real*real + imm*imm); }

然后您可以使用

Complex operator/(const Complex& lhs,const Complex& rhs)
{
   return (lhs * rhs.conj())/rhs.magnitude_square());
}