operations on complex numbers in C99

[David Marsh]

I understand that C99 supports a complex type and complex
arithmetic. There is nothing about it in the FAQ and online
searches turned up very little except synopses. Can anyone point
me toward sources or give examples which show how to:

-declare a complex variable
-assign the real and imaginary parts
-perform the basic +,-,*,/ operations on complex numbers

Thanks,
David

[Old Wolf]

On Aug 7, 9:59 am, Keith Thompson <[e-mail] rote:

jacob navia <[e-mail] mcomp.frwrites:

Some compilers like lcc-win32 or gcc use
double _Complex myvar = 5.778 + 45.87i;
but that is not standard.

and is therefore irrelevant unless you want to make your code
gratuitously non-portable.

It's useful to know this sort of thing, for when
you have to maintain someone else's code.

[Keith Thompson]

Old Wolf <ol*****@inspir e.net.nzwrites:

On Aug 7, 9:59 am, Keith Thompson <[e-mail] rote:

>jacob navia <[e-mail] mcomp.frwrites:

Some compilers like lcc-win32 or gcc use
double _Complex myvar = 5.778 + 45.87i;
but that is not standard.

and is therefore irrelevant unless you want to make your code
gratuitously non-portable.

It's useful to know this sort of thing, for when
you have to maintain someone else's code.

A fair point. It's pretty obvious what '5.778 + 45.87i' means, but I
suppose not everyone would recognize that it's an extension.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[jacob navia]

David Marsh wrote:

jacob navia wrote:

>David Marsh wrote:

>>I understand that C99 supports a complex type and complex arithmetic.
There is nothing about it in the FAQ and online searches turned up
very little except synopses. Can anyone point me toward sources or
give examples which show how to:

-declare a complex variable

double _Complex myvar = 5.778 + 45.87*I;
(standard notation)

Some compilers like lcc-win32 or gcc use
double _Complex myvar = 5.778 + 45.87i;
but that is not standard.

>>-assign the real and imaginary parts

creal(myvar)=8 .0;
cimag(myvar)=6 .9;

>>-perform the basic +,-,*,/ operations on complex numbers

(myvar+2)/(myvar-4)

Nothing special

Thanks, Jacob, but I'm having a problem with the assignments using
creal() and cimag(). It works OK if the commented out lines are used
instead. What am I doing wrong?

David

#include <stdio.h>
#include <complex.h>

int main(void)
{
//double _Complex z1 = 1.2 + .5 * I;
//double _Complex z2, z3;
double _Complex z1, z2, z3;

creal(z1) = 1.2; /* error */
cimag(z1) = .5; /* error */
z2 = csqrt(z1);
z3 = z1 + z2;
printf("sqrt = %g %gi\n", creal(z2), cimag(z2));
printf("z1+z2 = %g %gi\n", creal(z3), cimag(z3));

return 0;
}

C:\WINDOWS\Desk top\data>lcc cxtest.c
Error cxtest.c: 10 the left hand side of the assignment can't be
assigned to
Error cxtest.c: 11 the left hand side of the assignment can't be
assigned to
2 errors, 0 warnings

The correct syntax must be:

z1 = 1.2 + cimag(z1)*I;

[those who know me have no need of my name]

in comp.lang.c i read:

>creal() and cimag() are functions, which therefore do not yield
lvalues. You can't assign to them.

that is not a general restriction on functions, which may certainly return
a modifiable lvalue. though creal and cimag do not.

--
a signature

[Robert Gamble]

On Aug 8, 10:35 am, those who know me have no need of my name <not-a-
[e-mail] wrote:

in comp.lang.c i read:

creal() and cimag() are functions, which therefore do not yield
lvalues. You can't assign to them.

that is not a general restriction on functions, which may certainly return
a modifiable lvalue.

Can you provide such an example?

Robert Gamble

[Richard Heathfield]

Robert Gamble said:

On Aug 8, 10:35 am, those who know me have no need of my name <not-a-
[e-mail] wrote:

>in comp.lang.c i read:

>creal() and cimag() are functions, which therefore do not yield
lvalues. You can't assign to them.

that is not a general restriction on functions, which may certainly
return a modifiable lvalue.

Can you provide such an example?

#include <stdio.h>

char *foo(char *s)
{
printf("%s\n", s);
return s;
}

int main(void)
{
char array_of_modifi able_lvalues[] = "Hello World";
*foo(array_of_m odifiable_lvalu es) = 'C';
puts(array_of_m odifiable_lvalu es);
return 0;
}

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Richard Tobin]

In article <Lo************ *************** ***@bt.com>,
Richard Heathfield <rj*@see.sig.in validwrote:

>>creal() and cimag() are functions, which therefore do not yield
lvalues. You can't assign to them.

>>that is not a general restriction on functions, which may certainly
return a modifiable lvalue.

>Can you provide such an example?

>char *foo(char *s)
{
printf("%s\n", s);
return s;
}

The value returned by foo() is a pointer to a modifiable lvalue, not a
modifiable lvalue.

-- Richard
--
"Considerat ion shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

[Harald van =?UTF-8?B?RMSzaw==?=]

Richard Heathfield wrote:

Robert Gamble said:

>On Aug 8, 10:35 am, those who know me have no need of my name <not-a-
[e-mail] wrote:

>>in comp.lang.c i read:

creal() and cimag() are functions, which therefore do not yield
lvalues. You can't assign to them.

that is not a general restriction on functions, which may certainly
return a modifiable lvalue.

Can you provide such an example?

#include <stdio.h>

char *foo(char *s)
{
printf("%s\n", s);
return s;
}

int main(void)
{
char array_of_modifi able_lvalues[] = "Hello World";
*foo(array_of_m odifiable_lvalu es) = 'C';
puts(array_of_m odifiable_lvalu es);
return 0;
}

foo does not return a lvalue, especially a modifiable one. It returns a
value ("rvalue") of type pointer to char. You can use this value to
construct a lvalue, but if that is good enough to claim the function itself
returns a lvalue, what's preventing foo in

int foo(void) {
return 0;
}
int main(void) {
int arr[] = { 1 };
arr[foo()] = 0;
return arr[0];
}

from being considered as returning a lvalue?

[Richard Heathfield]

Richard Tobin said:

In article <Lo************ *************** ***@bt.com>,
Richard Heathfield <rj*@see.sig.in validwrote:

>>>creal() and cimag() are functions, which therefore do not yield
lvalues. You can't assign to them.

>>>that is not a general restriction on functions, which may certainly
return a modifiable lvalue.

>>Can you provide such an example?

>>char *foo(char *s)
{
printf("%s\n", s);
return s;
}

The value returned by foo() is a pointer to a modifiable lvalue, not a
modifiable lvalue.

Good point. I cannot, then, see any way that a function may return a
modifiable lvalue. I await education with eager delight.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Robert Gamble]

On Aug 8, 12:24 pm, Richard Heathfield <[e-mail].i nvalidwrote:

Robert Gamble said:

On Aug 8, 10:35 am, those who know me have no need of my name <not-a-
[e-mail] wrote:

in comp.lang.c i read:

creal() and cimag() are functions, which therefore do not yield
lvalues. You can't assign to them.

that is not a general restriction on functions, which may certainly
return a modifiable lvalue.

Can you provide such an example?

#include <stdio.h>

char *foo(char *s)
{
printf("%s\n", s);
return s;

}

int main(void)
{
char array_of_modifi able_lvalues[] = "Hello World";
*foo(array_of_m odifiable_lvalu es) = 'C';
puts(array_of_m odifiable_lvalu es);
return 0;

}

As I fully suspect you already know, foo does not return a modifiable
lvalue, it returns an rvalue which when dereferenced produces an
lvalue.

Robert Gamble