In printf, the cursor goes to left to right. so
x++ + ++x + x++
5 + 7 +7 = 19.
( Here x=5, then we put 5 in place of first x, after that x will be
incremented(post increment) that means x is 6 and we take second x which is
pre-increment and hence once again x will be incremented that means x is 7
it is compiler dependant da..the evaluation of this kind of expressions
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I ran this code..
int main() { int x=5;
printf(%d,(x++ + ++x + x++));
}
The output printed was 18 instead of 19.. Should it not be 19?
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it sud b 19 naaa
x++ = 5
++x = 7
x++ = 7
5+7+7
On Sat, Aug 28, 2010 at 4:35 PM, jagadish jagadish1...@gmail.com wrote:
I ran this code..
int main() { int x=5;
printf(%d,(x++ + ++x + x++));
}
The output printed was 18 instead of 19.. Should it not be 19?
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You received this message
output is 18
1. control goes to ++x =6 ,x=6
2. control goes to both x++, i.e. both x++ are evaluated together, therefore
x++ + ++x + x++= 6 +6+6 =18 x=7
On 28 August 2010 17:05, jagadish jagadish1...@gmail.com wrote:
I ran this code..
int main() { int x=5;
printf(%d,(x++ + ++x + x++));
The output is undefined. Depends on the compiler. + is not a sequence point
which may result in undefined behavior
On Sat, Aug 28, 2010 at 5:05 PM, jagadish jagadish1...@gmail.com wrote:
I ran this code..
int main() { int x=5;
printf(%d,(x++ + ++x + x++));
}
The output printed was 18
