NVIDIA Interview Question for Software Engineer in Tests


Country: United States
Interview Type: Phone Interview




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7
of 7 vote

The result of above code is undefined and may be different in different compilers.Because as per the c++ standers if we modify a variable more then one time in a single statement the result is undefined and based on compiler implementation.

- Guru June 22, 2012 | Flag Reply
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0
of 0 votes

U seem to be correct but i tried same code with same compiler ( cfree ) on xp and win7 .... and i am getting different answers. Is it also machine/os dependent ???

- saj1919 June 22, 2012 | Flag
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0
of 0 votes

@Guru
Thank you for pointing it out to me. I forgot to take compiler and machine dependency into account which reminds me of a similar question based on determining the size of the object of a class.
There cannot be one predictable output of the above code.

- Omkar June 22, 2012 | Flag
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1
of 1 vote

i never understand why ppl ask such questions in interviews.
i mean what are they trying to test that they had some wasted time that they come up with weird looking expression and want others to waste there time tooo.

- guest October 21, 2012 | Flag
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2
of 2 vote

The processing follows as:

5+7=12
7+9+9+10=35
12+12+14+14=52

- pulkit June 22, 2012 | Flag Reply
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0
of 0 vote

i got answer as
12
35
52
but
When i ran this program on machine i shocked to get the answer 12,32,50.
i am writing here what concept i used.
If we do some operation in printf statement on same element then we have to use stack.
so in first printf statement:-
structure of stack is
++i (top of stack) => 6
i++ => 6
and after the execution of this value of i is 7 and i++ + ++i is 6+6 = 12

in second printf statement
structure of stack is
i++ <- (top of stack) = > 7
i++ =>8
++i => 10
i++ (lower index of stack) => 10
and after the execution of this value of i is 11 and i++ + ++i + i++ + i++is 10 + 10 + 8 + 7 = 35
in third printf statement
structure of stack is
i++ <- (top of stack) = > 11
++i =>13
i++ => 13
++i (lower index of stack) => 15
and after the execution of this value of i is 15 and ++i + i++ + ++i + i++ is 15 + 13 + 13 + 11 = 52

but When i ran this program on machine i shocked to get the answer 12,32,50.
i dont know what concept they used ..most probably side effect..
let me know if some have answer of this question with proper explanation.
thanks in advance ..!!!

- Gupta June 21, 2012 | Flag Reply
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0
of 0 votes

Check your code.
My complier is giving the correct output as
12
35
52

- Ashish August 05, 2012 | Flag
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0
of 0 votes

it is just a compiler problem..

- Priya Singh September 13, 2012 | Flag
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0
of 0 votes

i think v cannot use stack for this operation ...
coz v are actually modifying the memory contents
say
i=4 ,,, i++ in the sense whatsoever the case may be i will be 5

if i think in this way im getting 12 and 32 as answers
52 for the second case ... but here 50 !!! ....

- bharath November 24, 2012 | Flag
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0
of 0 vote

same here even i got the ans as
12
35
52

- dekontj June 21, 2012 | Flag Reply
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0
of 0 vote

When I compile the code on gcc compiler and run it in a Unix environment I am getting the answer as 12,32 and 50.
One way to break down the problem is as follows:
First printf: During this stage the ++i causes i to become 6 so the addition is 6+6 = 12 and then i++ is executed causing i to be 7.
Second printf: Here again the ++i causes i to be 8 and so the addition is 8+8+8+8 = 32 and then i++ is executed three times causing i to be 11
Third printf: This is somewhat tricky part. Here the first ++i causes i to be 12 and hence i remains 12 till the second ++i which causes it to become 13 so in conclusion the addition becomes: 12 + 12 + 13 + 13 = 50 and now after this statement we have the two i++ which cause i to become 15 in the end.
This is the only explanation i can think of. Also you can verify this by inserting alternate printf statements in your existing code to print values of i at different stages.

- Omkar June 22, 2012 | Flag Reply
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1
of 1 vote

@Omkar: According to your Logic, In First printf ans should be 5+6=11 . Because for postfix ++ operator, you are are not increamenting i's value in the same printf (as you have done in second and third printf)

- X June 27, 2012 | Flag
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0
of 0 vote

please give me the solution :

we have an array of integers:
arr[] ={1,2,3,4,5}
need to write a program in C which will display all possible combinations
{1,2,3,4,5,12,13,14,15,123,124,....}

- Alok November 06, 2012 | Flag Reply
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0
of 0 vote

please give me the solution :

we have an array of integers:
arr[] ={1,2,3,4,5}
need to write a program in C which will display all possible combinations
{1,2,3,4,5,12,13,14,15,123,124,....}

- Alok November 06, 2012 | Flag Reply
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0
of 0 vote

VS2010: 12;32;52 (seems it does "++i" first, then uses the result in each row; "i++" is obliviously done after each printf)

- maxus July 17, 2013 | Flag Reply
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0
of 0 vote

VS2010: 12;32;52 (seems it does "++i" first, then uses the result in each row; "i++" is obliviously done after each printf)

- maxus July 17, 2013 | Flag Reply
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0
of 0 vote

VS2010: 12;32;52 (seems it does "++i" first, then uses the result in each row; "i++" is obliviously done after each printf)

- maxmah July 17, 2013 | Flag Reply
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0
of 0 vote

every time it will be "i" will be changed from right most to left most..
try to print
i=5;
printf("%d%d",i++,++i);
output=6 6

- Ganeshdip Dumbare August 20, 2013 | Flag Reply
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0
of 0 vote

Answer is undefined behavior. This is because the '+' between i++ and ++i is NOT a sequence point. A sequence point is a point in code where side effects like ++ of all previous computations is completed. Example of sequence points are ; (end of statement), ? in ?: operator, logical operators like &&, ||, etc.

- Anonymous August 28, 2013 | Flag Reply
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0
of 0 vote

{ int i,j,k; i=2; j=4; k=i++>j&2; printf("%d\n",k); if(++k && ++i<--j|| i++) { j=++k; } printf(" %d %d %d",i,-j--,k); getch();}

- Anonymous September 12, 2014 | Flag Reply
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0
of 0 vote

void main()
{
char s[]="oracle is the best";
char t[40];
char *ss,*tt;
while(*tt++=*ss++);
printf("%s",t);
getch();
}

- Anonymous September 12, 2014 | Flag Reply
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0
of 0 vote

Int x = 10 , y = 5 , z = 0 ;
If (x >y//x==y)
Z = ++x+__y;
System . Out . Println ( z+ " " + x + " " + y ) ;

- Anonymous June 13, 2020 | Flag Reply
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-1
of 1 vote

when i first solved the problem i got 12,35,52 only .
but ans is 12 ,32,50
here is the explanation :
1)i++ + ++i = 6+6 =12, due to post increment value .
after this step value of i=7.
2) i++ + ++i + i++ + i++ = 8 + 8 + 8 + 8..
value of i will become 8 when it reaches ++i, so the effect of 1st two i's will be 8+8.
but due to post increment , the value of next i++ will be 8 not 9.
value of i after this step =12.
3) ++i + i++ + ++i + i++= 12 + 12 + 13 + 13 =50
similar explanation for this also.

- dekontj July 22, 2012 | Flag Reply


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