Because what Salem linked to seems to contradict Dave. I assumed Dave's explanation was right, but I did not want to throw out Salem's points.Originally Posted by itsme86
But the more important question I have is just below that. Namely in what order is the following evaluated "a+b?c-d?e():i():j*k?f():g()".
Last edited by King Mir; 07-09-2006 at 09:18 PM.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
1) It evalues a+b
2) If it's true it evalues c-d. Otherwise skip to step 4
3) If c-d evalues true, it calls e(). Otherwise it calls i(). Done.
4) It evaluates j*k.
5) If it's true it calls f(). Otherwise it calls g(). Done.
It's all very simple...
If you understand what you're doing, you're not learning anything.
And that's what they call reading right to left huh?
Any reference will tell you that conditionals are evaluated right to left, including some linked earlier in this thread.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
Nobody said anywhere earlier in this thread that conditionals are evaluated right to left except you.
If you understand what you're doing, you're not learning anything.
READ THE LINK.
I mean come on, I told you that Dave's post disagreed with Salem's links.
Besides I can find pleny of other tables just like that one.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
You did not read any of them.
You are confusing operator associativity with evaluation order, which is exactly what Salem warned you against.Code:?: a ? e1 : e2 Expression e1 if a is nonzero; Expression e2 if a is zero
Briefly stated, the second link in Salem's post merely warns that in an expression like
int d = f() - g() * h();
The results of g() and h() will be multiplied before they are subtracted from the return value of f(). However, the compiler is free to call the functions in any order it wants. This is always true, but when you are controlling program flow the compiler is very rigorous in what happens.
I get all that exept:
What does it mean by the following?
In a more complicated example, associativity rules specify that b ? c : d is evaluated first in the following example:
a ? b ? c : d : e;
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
No, that's actually correct. We all did it backwards, but Dave's post explains how the ternary operator works. Salem's explains how two ternary operators get evaluated if they are in the same expression: the rightmost chunk of logic gets evaluated first, and I am thoroughly, embarrassed.
I am sure that both Dave's and Salem's post are correct in there own way, I just don't get how they go together.
I am still confused about the line "a+b?c-d?e():i():j*k?f():g()".
Perhalps I am the dunce looking for fire today . . . (see my sig)
Last edited by King Mir; 07-09-2006 at 10:34 PM.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
hi everybody..
i dont know whether the question i posted was clear enough or not...my question still remains unanswered... Let me restate the question..
I am AWARE that
can be written asCode:a<10?b=100:b=200;
but...Code:b=a<10?100:200;
I want a justification as to why
is giving an "invalid lvalue" error...Code:a<10?b=100:b=200;
Wat i am looking for is a "Lexical analysis" of the above statement...
that is how exactly the compiler is tokenizing the statement.. because the statement is logically correct
hope that was clear enough...
Regards
Abhijith
I though you wanted to know what happends with "a<10?b=100:(b=200).
Any way, in the case of "a<10?b=100:b=200;" it is the same as
It should be clear why that gives an invalid lvalue.Code:(a<10?b=100:b)=200;
Last edited by King Mir; 07-09-2006 at 11:29 PM.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
While itsme's explaination is easier to understand but goes the wrong way, it is still correct.I am sure that both Dave's and Salem's post are correct in there own way, I just don't get how they go together.
I am still confused about the line "a+b?c-d?e():i():j*k?f():g()".
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().
Playing with that makes it pretty clear.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; }
I though you wanted to know what happends withAny way, in the case of "a<10?b=100:b=200;" it is the same asCode:"a<10?b=100:(b=200).
Code:(a<10?b=100:b)=200;
It should be clear why that gives an invalid lvalue.
i am not completely convinced with it..
two doubts
1. If i am not mistaken the address of "b" is returned & is attempted to be initialized to 200.
But this error doesnt happen if the staement is true..
ie., if a<10 is true. in which case b=100 should be evaluated.. the error doesn occur...
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...
Actually you're wrong. It ends up:I though you wanted to know what happends with "a<10?b=100b=200).
Any way, in the case of "a<10?b=100:b=200;" it is the same as
It should be clear why that gives an invalid lvalue.Code:(a<10?b=100:b)=200;
(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.
Hope is the first step on the road to disappointment.
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.
thanks.. but... here's one more doubt...
say a=0
a<10?b=100:b=200;
how come this statement works n "b" gets initialized to 100... in this case parenthesis is not required. i tried it out... it seems to mandate the parenthesis for the third operand...
Regards
Abhijith
