Converting a defined structure into double (complex numbers) in c++
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Im trying to find my way around defining structures, and coded this to define a complex number. Ive already had success working with them, adding multiplying e.g., but I dont know how to get the absolute value of my complex number converted into a double, my compiler keeps telling me it doesnt know how to convert it, even though the complex struct is literally made of 2 doubles..
#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;
struct Complex_Number {
double re, im;
};
void print(const Complex_Number& w) //my print function for complex numbers
{
cout << '(' << w.re << " + i*" << w.im << ')';
}
double abs(const Complex_Number& z )
{
z = z.re + z.im;
abs = sqrt((z.re*z.re)+(z.im*z.im));
return abs;
}
int main()
{
return 0;
}
c++ struct double complex-numbers
asked Nov 11 at 19:39
Henryk5161
43
add a comment |
up vote
0
down vote
favorite
Im trying to find my way around defining structures, and coded this to define a complex number. Ive already had success working with them, adding multiplying e.g., but I dont know how to get the absolute value of my complex number converted into a double, my compiler keeps telling me it doesnt know how to convert it, even though the complex struct is literally made of 2 doubles..
#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;
struct Complex_Number {
double re, im;
};
void print(const Complex_Number& w) //my print function for complex numbers
{
cout << '(' << w.re << " + i*" << w.im << ')';
}
double abs(const Complex_Number& z )
{
z = z.re + z.im;
abs = sqrt((z.re*z.re)+(z.im*z.im));
return abs;
}
int main()
{
return 0;
}
c++ struct double complex-numbers
asked Nov 11 at 19:39
Henryk5161
43
1
absis the name of the function - i.e. a function pointer - you need to create a double variable.
– Neil Butterworth
Nov 11 at 19:42
Seestd::complex.
– Thomas Matthews
Nov 11 at 20:22
As @ThomasMatthews mentions, you should check out your localstd::complex<double>to see how the interfaces can be defined by your standard library.
– Bo R
Nov 11 at 20:44
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Im trying to find my way around defining structures, and coded this to define a complex number. Ive already had success working with them, adding multiplying e.g., but I dont know how to get the absolute value of my complex number converted into a double, my compiler keeps telling me it doesnt know how to convert it, even though the complex struct is literally made of 2 doubles..
#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;
struct Complex_Number {
double re, im;
};
void print(const Complex_Number& w) //my print function for complex numbers
{
cout << '(' << w.re << " + i*" << w.im << ')';
}
double abs(const Complex_Number& z )
{
z = z.re + z.im;
abs = sqrt((z.re*z.re)+(z.im*z.im));
return abs;
}
int main()
{
return 0;
}
c++ struct double complex-numbers
asked Nov 11 at 19:39
Henryk5161
43
Im trying to find my way around defining structures, and coded this to define a complex number. Ive already had success working with them, adding multiplying e.g., but I dont know how to get the absolute value of my complex number converted into a double, my compiler keeps telling me it doesnt know how to convert it, even though the complex struct is literally made of 2 doubles..
#include <iostream>
#include <cstdlib>
#include <cmath>
using namespace std;
struct Complex_Number {
double re, im;
};
void print(const Complex_Number& w) //my print function for complex numbers
{
cout << '(' << w.re << " + i*" << w.im << ')';
}
double abs(const Complex_Number& z )
{
z = z.re + z.im;
abs = sqrt((z.re*z.re)+(z.im*z.im));
return abs;
}
int main()
{
return 0;
}
c++ struct double complex-numbers
c++ struct double complex-numbers
asked Nov 11 at 19:39
Henryk5161
43
asked Nov 11 at 19:39
Henryk5161
43
asked Nov 11 at 19:39
Henryk5161
43
asked Nov 11 at 19:39
Henryk5161
43
asked Nov 11 at 19:39
Henryk5161
43
43
1
absis the name of the function - i.e. a function pointer - you need to create a double variable.
– Neil Butterworth
Nov 11 at 19:42
Seestd::complex.
– Thomas Matthews
Nov 11 at 20:22
As @ThomasMatthews mentions, you should check out your localstd::complex<double>to see how the interfaces can be defined by your standard library.
– Bo R
Nov 11 at 20:44
add a comment |
1
absis the name of the function - i.e. a function pointer - you need to create a double variable.
– Neil Butterworth
Nov 11 at 19:42
Seestd::complex.
– Thomas Matthews
Nov 11 at 20:22
As @ThomasMatthews mentions, you should check out your localstd::complex<double>to see how the interfaces can be defined by your standard library.
– Bo R
Nov 11 at 20:44
1
1
abs is the name of the function - i.e. a function pointer - you need to create a double variable.– Neil Butterworth
Nov 11 at 19:42
abs is the name of the function - i.e. a function pointer - you need to create a double variable.– Neil Butterworth
Nov 11 at 19:42
See
std::complex.– Thomas Matthews
Nov 11 at 20:22
See
std::complex.– Thomas Matthews
Nov 11 at 20:22
As @ThomasMatthews mentions, you should check out your local
std::complex<double> to see how the interfaces can be defined by your standard library.– Bo R
Nov 11 at 20:44
As @ThomasMatthews mentions, you should check out your local
std::complex<double> to see how the interfaces can be defined by your standard library.– Bo R
Nov 11 at 20:44
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
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I don't know what you intend with z = z.re + z.im; The second part just needs to be assigned to a variable or returned directly. So write...
double abs(const Complex_Number& z ) {
double ret = sqrt((z.re*z.re)+(z.im*z.im));
return ret;
}
or
double abs(const Complex_Number& z ) {
return sqrt((z.re*z.re)+(z.im*z.im));
}
Assigning something to the "function name" as in your code, i.e. abs = sqrt((z.re*z.re)+(z.im*z.im)), is not valid C++ code.
answered Nov 11 at 19:44
Stephan Lechner
25.3k21839
1
Whoah thanks for that.. i have spent too much time today that i literally blocked myself from seeing this obvious one.. i really appreciate your help!
– Henryk5161
Nov 11 at 19:51
1
Was just waiting for the option to be enabled, thx :)
– Henryk5161
Nov 11 at 20:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I don't know what you intend with z = z.re + z.im; The second part just needs to be assigned to a variable or returned directly. So write...
double abs(const Complex_Number& z ) {
double ret = sqrt((z.re*z.re)+(z.im*z.im));
return ret;
}
or
double abs(const Complex_Number& z ) {
return sqrt((z.re*z.re)+(z.im*z.im));
}
Assigning something to the "function name" as in your code, i.e. abs = sqrt((z.re*z.re)+(z.im*z.im)), is not valid C++ code.
answered Nov 11 at 19:44
Stephan Lechner
25.3k21839
1
Whoah thanks for that.. i have spent too much time today that i literally blocked myself from seeing this obvious one.. i really appreciate your help!
– Henryk5161
Nov 11 at 19:51
1
Was just waiting for the option to be enabled, thx :)
– Henryk5161
Nov 11 at 20:17
add a comment |
up vote
1
down vote
accepted
I don't know what you intend with z = z.re + z.im; The second part just needs to be assigned to a variable or returned directly. So write...
double abs(const Complex_Number& z ) {
double ret = sqrt((z.re*z.re)+(z.im*z.im));
return ret;
}
or
double abs(const Complex_Number& z ) {
return sqrt((z.re*z.re)+(z.im*z.im));
}
Assigning something to the "function name" as in your code, i.e. abs = sqrt((z.re*z.re)+(z.im*z.im)), is not valid C++ code.
answered Nov 11 at 19:44
Stephan Lechner
25.3k21839
1
Whoah thanks for that.. i have spent too much time today that i literally blocked myself from seeing this obvious one.. i really appreciate your help!
– Henryk5161
Nov 11 at 19:51
1
Was just waiting for the option to be enabled, thx :)
– Henryk5161
Nov 11 at 20:17
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I don't know what you intend with z = z.re + z.im; The second part just needs to be assigned to a variable or returned directly. So write...
double abs(const Complex_Number& z ) {
double ret = sqrt((z.re*z.re)+(z.im*z.im));
return ret;
}
or
double abs(const Complex_Number& z ) {
return sqrt((z.re*z.re)+(z.im*z.im));
}
Assigning something to the "function name" as in your code, i.e. abs = sqrt((z.re*z.re)+(z.im*z.im)), is not valid C++ code.
answered Nov 11 at 19:44
Stephan Lechner
25.3k21839
I don't know what you intend with z = z.re + z.im; The second part just needs to be assigned to a variable or returned directly. So write...
double abs(const Complex_Number& z ) {
double ret = sqrt((z.re*z.re)+(z.im*z.im));
return ret;
}
or
double abs(const Complex_Number& z ) {
return sqrt((z.re*z.re)+(z.im*z.im));
}
Assigning something to the "function name" as in your code, i.e. abs = sqrt((z.re*z.re)+(z.im*z.im)), is not valid C++ code.
answered Nov 11 at 19:44
Stephan Lechner
25.3k21839
answered Nov 11 at 19:44
Stephan Lechner
25.3k21839
answered Nov 11 at 19:44
Stephan Lechner
25.3k21839
answered Nov 11 at 19:44
Stephan Lechner
25.3k21839
25.3k21839
1
Whoah thanks for that.. i have spent too much time today that i literally blocked myself from seeing this obvious one.. i really appreciate your help!
– Henryk5161
Nov 11 at 19:51
1
Was just waiting for the option to be enabled, thx :)
– Henryk5161
Nov 11 at 20:17
add a comment |
1
Whoah thanks for that.. i have spent too much time today that i literally blocked myself from seeing this obvious one.. i really appreciate your help!
– Henryk5161
Nov 11 at 19:51
1
Was just waiting for the option to be enabled, thx :)
– Henryk5161
Nov 11 at 20:17
1
1
Whoah thanks for that.. i have spent too much time today that i literally blocked myself from seeing this obvious one.. i really appreciate your help!
– Henryk5161
Nov 11 at 19:51
Whoah thanks for that.. i have spent too much time today that i literally blocked myself from seeing this obvious one.. i really appreciate your help!
– Henryk5161
Nov 11 at 19:51
1
1
Was just waiting for the option to be enabled, thx :)
– Henryk5161
Nov 11 at 20:17
Was just waiting for the option to be enabled, thx :)
– Henryk5161
Nov 11 at 20:17
add a comment |
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absis the name of the function - i.e. a function pointer - you need to create a double variable.– Neil Butterworth
Nov 11 at 19:42
See
std::complex.– Thomas Matthews
Nov 11 at 20:22
As @ThomasMatthews mentions, you should check out your local
std::complex<double>to see how the interfaces can be defined by your standard library.– Bo R
Nov 11 at 20:44