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Amazon Interview Question for Software Engineer / Developers






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4
of 6 vote

Tried compiling with Dev c++ (Which internally uses gcc for c files)
- Nothing is printed when for(;0;)
- Infinite loop when for(;1;)

- snk_anindya February 04, 2011 | Flag Reply
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0
of 0 vote

Is that an Amazon question indeed ?

- AV October 31, 2010 | Flag Reply
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0
of 0 vote

body of cycle will not run a single time because condition for cycle working/entering it is constantly false (0 - false).

And that was asked by Amazon`s interviwer?

- Anonymous October 31, 2010 | Flag Reply
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0
of 0 vote

test for attention must be something like that

for(;0;);
printf("\n guess");

- Anonymous October 31, 2010 | Flag Reply
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0
of 0 vote

Amazon didn't ask this, come on..

- Anonymous October 31, 2010 | Flag Reply
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0
of 0 votes

i think so too.

- daksh December 06, 2021 | Flag
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0
of 0 vote

guys cn u guide me abt d written pattern and ques for amazon..
at careercup i cn find a lot of interview questions..bt wat abt d written??

- help needed November 01, 2010 | Flag Reply
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0
of 0 vote

nothing will print because if we don't mention any condition then default condition is true but in 0 indicates false so control doesn't enter into loop

- kondu naresh November 01, 2010 | Flag Reply
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0
of 0 vote

The legendary Dennis Ritchie clearly explains for loop (for (init; cond; inc/dec)) is nothing but a while loop with the following structure:

init;
while (cond)
{
// body
inc/dec;
}

Hence, the condition (0) fails and the body is never executed.

- Vikas Upendra November 29, 2010 | Flag Reply
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0
of 0 vote

@all above

answer comes and it prints guess

- sandy880 January 18, 2011 | Flag Reply
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0
of 0 votes

No, it shows 'No errors or program output.'

- pavithra January 09, 2012 | Flag
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0
of 0 votes

actually no, you are wrong. nothing will get executed but the program will run successfully.

- daksh December 06, 2021 | Flag
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0
of 0 vote

yes, it prints guess

- Anonymous January 31, 2011 | Flag Reply
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0
of 2 vote

I have tried other options too :
->if 0 is replaced by any other no.(i.e. true) then, it prints guess infinite number of times,
->if 0 is replaced by FALSE(defined as 0), then it also prints guess just once
CONSIDERING THESE CASES IT SEEMS THAT :- "for loop without any initialization and incre/decre, it works as do-while loop"
!!!BUT again contrary to this when 0 is replaced by any wrong condition it behaves like while loop(i.e. don't print guess, as expected)
PLEASE SOMEONE HELP WITH SOME EXPLANATION

- Anonymous February 02, 2011 | Flag Reply
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0
of 0 votes

Its because..a for loop will always do the initialisation first and executes the loop's body...ONLY after executing the loops body once,it will check for the condition..

So..in this case ...it executes loop's body once..then it finds 0(ie .false) ,hence it quits but instead of 0,any other number is considered true,hence executes infinite number of times...

- sha February 02, 2011 | Flag
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0
of 0 votes

if for loop follows do-while loop, then it should execute the loop body at least once as for the earlier case even if, 0 is replaced by some wrong condition(like int i=0;
for(;i<0;) printf("\nGUESS");
BUT ITS CONTRADICTORY TO THE ASSUMED CONCLUSION AND DOES NOT PRINT ANYTHING

- Anonymous February 05, 2011 | Flag
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0
of 0 votes

@sha
you are wrong ,during 1 st iteration it is initialized and condition is checked before executing the body

try this code
for(x=1;x>2;x++)
printf("%d",x);

so according to you it should have printed 1 ,but it won't print because condition is also checked before entering body

- Sandy880 February 09, 2011 | Flag
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0
of 0 votes

yeah that's true for( ;0; ) runs the body for one time and even for(int i=0;0; ) runs the body once, but for(int i=0;i<0; ) the body does not gets executed for a single time.
I don't know the reason. If anybody does, please help.

- kriti13796 September 20, 2015 | Flag
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0
of 2 vote

for(;0;)
    printf("\n guess");

#nothing is printed because it is false.

- weijiang2009 February 06, 2011 | Flag Reply
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1
of 1 vote

actually its printing guess once beyond all logical reasoning

- Anonymous February 06, 2011 | Flag
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0
of 0 votes

(1)do the initilization, print once.
(2)and then judge, do nothing and exit

- weijiang2009 February 06, 2011 | Flag
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0
of 0 vote

nothing would print. because before going printf statement thew would exit.due to for(;0;)

- Anonymous April 27, 2011 | Flag Reply
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0
of 0 vote

for(;0;)
printf("\n guess");


Output:
No errors or program output.

and please dont joke this cant be amazom question

- Learn Android: http://learnandroideasily.blogspot.in/ June 15, 2011 | Flag Reply
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0
of 0 vote

Nothing will be printed.....

- Sanchit Choubey November 16, 2012 | Flag Reply
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0
of 0 vote

sha is correct

- Anonymous February 01, 2013 | Flag Reply
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0
of 0 vote

for(;;;)

- Anonymous May 13, 2015 | Flag Reply
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0
of 0 vote

the logic of 0 is to tell the compiler that you have interacted or tested a false condition, thus the output comes out to be [null] or ASCII decimal value '0'.

- daksh December 06, 2021 | Flag Reply
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0
of 0 vote

it is also of the same form as for(; ;) where the empty space represents [null] which in arithmetic conditions will take up its ascii decimal value of 0, hence nothing will get executed and the program will exit out with the return code.

- daksh December 06, 2021 | Flag Reply


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