Hi,
Recently I was executing a code in C. The program is:
And the output was nos. 1 to 65536.Code:#include<stdio.h> int main() { short int i =0; for(i<=5 && i>=-1;++i;i>0) printf("%u\n",i); printf("\n"); return 0; }
(Given that short int is 2 bytes in length).
Can any one explain the logic behind such output.I am unable to analyse.
Whoever wrote that for loop is either trying to submit something to The International Obfuscated C Code Contest, or he doesn't know how for loops work.
The first and last parts of the for statement do absolutely nothing relevant, so that leaves us with:
This will end the loop as soon as i becomes negative, i.e. when the short int wraps around.Code:++i
Hi Memloop,
Can you explain how is this:-
"This will end the loop as soon as i becomes negative, i.e. when the short int wraps around."
as you have mentioned in your post.
The loop you've posted looks like this (the other statements don't actually do anything):Code:for(INITIALIZATION EXPRESSION; LOOP CONDITION; INCREMENT EXPRESSION)
As you can see, the only active part of the for loop is the loop condition. In C, any negative integer value evaluates to boolean false, so the loop will terminate when ++i becomes negative. That happens when i wraps around, i.e. it goes something likes this depending on how large it is:Code:for(; ++i;)
Code:1, ..., SHORT_MAX, SHORT_MIN, SHORT_MIN + 1, ...
Last edited by Memloop; 01-24-2010 at 09:50 AM.
This is INCORRECT. The only false value in C is 0. The loop terminates and ends when ++i would equal 0, not a negative.Originally Posted by Memloop
C programming resources:
GNU C Function and Macro Index -- glibc reference manual
The C Book -- nice online learner guide
Current ISO draft standard
CCAN -- new CPAN like open source library repository
3 (different) GNU debugger tutorials: #1 -- #2 -- #3
cpwiki -- our wiki on sourceforge
Despite a blooper that MK27 has explained, Memloop's analysis is generally correct. That said, I believe that your code snippet actually results in undefined behaviour as it results in signed integer overflow to result in 0 and thus terminate the loop.
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