> Okay. Try in C++ compiler (e.g. g++). I'm using g++ in this case.
#include <stdio.h>
void main() {
int x = 0;
int y = 1;
x = (y) ? x = 1 : x = 2;
printf("%d\n", x);
}
gcc version: 2.7.2.3
egcc version: 2.91.61 19990216
g++ version: 2.91.61 19990216
20145 16:50:24 jsc@squeak:~% gcc -g -o test test.c
test.c: In function `main':
test.c:7: invalid lvalue in assignment
20146 16:50:28 jsc@squeak:~% egcc -g -o test test.c
test.c: In function `main':
test.c:7: invalid lvalue in assignment
20147 16:50:34 jsc@squeak:~% g++ -g -o test test.cc
20148 16:50:43 jsc@squeak:~% ./test
1
I believe that gcc/egcc are buggy here - your code _should_ be legal.
Also, whatever version of g++ you are using is _also_ buggy - it should be
generating code to produce 1, not 2, but it is not.
> >I suspect there's a compiler bug to do with assignments in a
> > ternary operator as the rvalue of an assignment.
>
> The expression is correct.
The _expression_ is correct - the _compiler_ is buggy.
> > Wrapping the two x='s on
> > the right-hand side generates the expected output.
>
> No. The expected output is X equal to the result of two other expression
> (E1,E2), related to condition (y).
> x= (condition) ? E1 : E2;
I understand how ?: works. What I said was that gcc/egcc would not compile
this line:
x = (y) ? x = 1 : x = 2;
But if you change it like this:
x = (y) ? (x = 1) : (x = 2);
then gcc/egcc don't give you an error. Try the same workaround in your own
compiler, or upgrade it.
--Jeff
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