#clanguage

Question

int i;
i=5;
i=i++;
printf("%d",i);

What is output of this program??

Is it undefined behavior ?? Why??

Answer 1

That is what is called as undefined behavior of C .

i = i++

Here, we are modifying the same value twice , which is not allowed .

Hence, if we talk about this,

a[i] = i++

which is incrementing , i.e, changing i and even using it at the same time, which is really not allowed.

Hence, it all depends on the compiler.

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Run this simple code 

i = 5;

i = i++;

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some compilers will give 5 and some give 6.

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Refer ==> http://www.geeksforgeeks.org/sequence-points-in-c-set-1/

http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points

Comments

@Kapil, check this.. i think its typo right ??

Run this simple code  i = 10; i = i++; some compilers will give 5 and some give 6.

Answer 2

its output should be 6 and it is not undefined beahviour

i=i++ is equvt to i=i+1 ;

in next line i is printed as 6 so no problem

PS undefined behaviour comes into picture when i is incremented more than once in single statement

like i=++i++  + i++ etc

Comments

Plz see this...

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even then given  above statement should not have undefined behaviour

( otherwise how looping will work) there is difference between i=i+1 and i=i++ +1 and a[i]=i++;

Why doesn't this code: a[i] = i++; work?

The subexpression i++ causes a side effect--it modifies i's value--which leads to undefined behavior since i is also referenced elsewhere in the same expression. There is no way of knowing whether the reference will happen before or after the side effect--in fact, neither obvious interpretation might hold; see question 3.9. (Note that although the language in K&R suggests that the behavior of this expression is unspecified, the C Standard makes the stronger statement that it is undefined

 Arjun have a look

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i = i++; i = (i = j) + 2;

Both are undefined behaviour. Both modifies i two times without a sequence point in between.

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thanks got it

Answer 3

Modifying the same variable more than once  within a single expression is Undefined behaviour.(diffrent compiler do it differently).

Here Value of i can be {5 or 6}

 

Some other Undefined behaviour:-

1. f()+g();

2.a[i]=i++;

3.y=x++ * ++x

Note f(x),g(x) is not undefined behaviour bcz ,(comma) operator introduces a sequence point that f(x) then g(x)