指針到複雜<double>罪

Question

我嘗試編譯這個簡單的例子：

#include<iostream> 
#include<math.h> 
#include<complex> 

typedef double (*d_sin)(double); 
typedef std::complex<double> (*c_sin)(std::complex<double>); 

int main(void) 
{ 

/* case 1 */ 
std::cout << "sin(pi/2)=" << sin(M_PI/2) << std::endl; 

/* case 2 */ 
d_sin var = &sin; 
std::cout << "sin(pi/2)=" << (*var)(M_PI/2) << std::endl; 

/* case 3 */ 
c_sin var2 = &sin; 
std::cout << "sin(pi/2)=" << (*var2)(M_PI/2) << std::endl; 

return 0; 
} 

和錯誤：

$ g++ main.cpp && ./a.out 
main.cpp: In function ‘int main()’: 
main.cpp:19:15: error: invalid conversion from ‘double (*)(double)throw()’ to ‘c_sin {aka   std::complex<double> (*)(std::complex<double>)}’ [-fpermissive] 
c_sin var2 = &sin; 
     ^

爲什麼這個樣本有錯誤？如何獲得複雜案件的正確行爲？非常感謝。

Answer 1

typedef std::complex<double> (*c_sin)(std::complex<double>); 

到

typedef std::complex<double> (*c_sin)(const std::complex<double> &); 

和

c_sin var2 = std::sin;