Thread: ternary operator

I don't know what kind of compiler you're using:

Code:

itsme@itsme:~/C$ cat cond.c
#include <stdio.h>

int main(void)
{
  int a = 0;
  int b;

  a<10?b=100:b=200;

  return 0;
}

Code:

itsme@itsme:~/C$ gcc -Wall cond.c -o cond
cond.c: In function `main':
cond.c:8: warning: use of conditional expressions as lvalues is deprecated
cond.c:8: error: invalid lvalue in assignment
itsme@itsme:~/C$

It doesn't work no matter what a starts out as.

Originally Posted by quzah

Actually you're wrong. It ends up:

(a < 10) = ...
0 = ...
or
1 = ...

That's why it's wrong. Because "a < 10" evaluates to either a 1 or a 0, and you're trying to assign a value to that. Which you can't. Because that's not a valid lvalue.


Quzah.

I'm sorry but that makes no sense to me. A conditional evaluates to either of it's two posible conditions. And the conditional is called before assignment opperator.

The reason it returns an invalid lvalue, is because it is returning the value contained in b, not b itself. Thus you cannot derefference the value.

It's still 200=... or the previous value of b = ...

PS, if you use a?b=200:b=100, it still won't work.

Originally Posted by citizen

While itsme's explaination is easier to understand but goes the wrong way, it is still correct.
1. Evaluate c - d. if that is nonzero (true) then e() is called if a + b is true. If c - d is zero (false) call i() if a + b is true.
2. Evaluate a + b. if it is true, then step 1 matters. If it's false, goto 3.
3. Evaluate j * k. if it is true run f(), otherwise run g().

Code:

#include <stdio.h>

int f(void) { return putchar('f'); }
int g(void) { return putchar('g'); }
int e(void) { return putchar('e'); }
int i(void) { return putchar('i'); }

int main(void) {
   int a = 2, b = 3;
   int c = 1, d = -1;
   int j = 0, k = 5;

   (a + b)? (c - d)? e() : i() : (j * k)? f() : g();

   return 0;
}

Playing with that makes it pretty clear.

Trouble with testing it is that it does not tell you what is undefined. Does c-d get called before j*k? always or just on my compiler?

But perhalps I should run a few tests. It will atleast tell me which functions and opperations do not get called.

Originally Posted by abhijith gopal

2. correct me if i am wrong..

this is my interpretation of the code
(a<10?b=100:b)=200;

the value of "b" is returned... now since there is an intializing statement the value 200 is equated against the value of b which was returned previously...

I'm sorry but that makes no sense to me. A conditional evaluates to either of it's two posible conditions. And the conditional is called before assignment opperator.

The reason it returns an invalid lvalue, is because it is returning the value contained in b, not b itself. Thus you cannot derefference the value.

It's still 200=... or the previous value of b = ...

PS, if you use a?b=200:b=100, it still won't work.

Guess i was right according to King Mir's Explanation...

but the statement does NOT flag an error when the second operand is not enclosed within parenthesis.. ie., a<10?b=100:(b =200); // for a less than ten

Well, quzah seems to disagree with me, so give him a chance to reply before you take my word on it.

In either case, a<10?b=100:(b =200) is perfectly valid, because the result of the conditional is not used as an lvalue for the assignment opperator.

Originally Posted by citizen

Always. This is because, on every operator table ever written, the ternary operator is associated right to left, and c - d is the rightmost operator. j * k is actually the else clause in the a + b evaluation, so it will always be evaluated last.

I get it now!!!

But you're wrong. a + b gets evaluated first.
Then either "(c - d)? e() : i()" or "(j * k)? f() : g()"

Notice how "(c - d)? e() : i()?f():g()" is never evaluated. That's what is ment by the operator being read right to left.

And all it takes is a useless example of "a?b?c:d:e" to get me all confused. That example does not show anything.