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Can anyone please explain the following code:
The output is : 1 -1 2Code:#include<stdio.h>
void main ()
{
int m = m <= m > 0, n = --m;
n = n++ && m-12;
printf("%d %d %d", m,n, ++m - --n);
}
I want to understand how the above code works.
Can anyone explain how the operator assignment takes place?
Where did you get this from? It is unnecessarily convoluted and results in undefined behaviour.
It's a problem from the book on testing C aptitude....Quote:
Originally Posted by laserlight
Can u plz tell how it works.
It doesn't work. At least, if I were interviewing a person and decided to see if the person had some aptitude in C by showing the candidate this code, anything less than a shriek of disgust from the candidate would make me doubt that the person knows C.Quote:
Originally Posted by nowshek
The first line of main is a total mess that may result in undefined behaviour, but I don't know for sure. The second line is okay because of the sequence point introduced by &&, but it is still dumb. Anyway, I am certain that this line results in undefined behaviour:
m and n are accessed, but at the same time they are modified by side effects.Code:printf("%d %d %d", m,n, ++m - --n);
The behaviour, as laserlight said, is undefined.
The word "undefined" in the C standard essentially means that "how it works" can be anything, from reformatting your hard drive, displaying a cartoon of little green mice chasing spacemen on your screen, or printing the value 42 over and over.
Any code that modifies a variable twice between sequence points - which is what your code does - has undefined behaviour.
And main() returns int, not void.
If you want to demonstrate C aptitude, learn the meaning of "undefined" as it is defined in the C standard, and learn that main() returns int.
but the following code compiles and runs without error............Quote:
Originally Posted by grumpy
That is one possible outcome of undefined behaviour.Quote:
Originally Posted by nowshek
Suppose you murder your neighbour in cold blood. You are never caught. Does that mean that it was legal for you to murder your neighbour?
if it is undefined behaviour then the output of the code must change...but it give 1 -1 2.Quote:
Originally Posted by laserlight
That is not true. It is entirely possible that given a piece of code that results in undefined behaviour, every current compiler with every possible configuration produces the same behaviour... until one of the compilers has a new release, or a new player enters the market. Or, the compiler that you use always compiles your program to produce the behaviour that you have come to expect, except when you compile your program on Friday the 13th.Quote:
Originally Posted by nowshek
No, that would be erratic behaviour.Quote:
Originally Posted by nowshek
Undefined behaviour means that it may not do what you expect... but in a given compiler/OS it usually surprises you in the same way each time.
Or when you're attempting to demonstrate it to your Boss...Quote:
Originally Posted by laserlight
For some extra information on undefined behavior (and unspecified and implementation-defined), read this: Question 11.33.
One potential manifestation of undefined behaviour is erratic behaviour. Ever tried to track down a bug in a function that increments a static pointer twice in a statement, and only yields an abnormal termination - in other code that dereferences that pointer - only after the third time it is called?Quote:
Originally Posted by CommonTater
To quote the 1999 C standard Undefined behavior is "behavior, upon use of a nonportable or erroneous program construct or of erroneous data, for which this International Standard imposes no requirements". (the American origin of the standard is obvious in the spelling it uses).
The standard, immediately goes on in the next clause with this: "NOTE Possible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message)."
If I ever write a compiler for educational purposes or otherwise.... I'll make a point by doing all of those...and some more creatively defined behaviours.Quote:
Originally Posted by grumpy
Just as well I used the word "chasing" in my previous post rather than the first word that came to mind .......