Technical - main(){int a[]={0 1 2 3};int *b;b=&a[2];printf("%d",b[-1]);}TECHNICAL INTERVIEW

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Technical | Question 56 of 157

main()
{
int a[]={0 1 2 3};
int *b;
b=&a[2];
printf("%d",b[-1]);
}

TECHNICAL INTERVIEW

Total Answers and Comments: 10

Last Update: December 14, 2008

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November 18, 2005 03:56:19

Suhas

Member Since: November 2005 Contribution: 24

RE: main(){int a[]={0 1 2 3};int *b...

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February 19, 2006 12:12:36

swaant

RE: main(){int a[]={0 1 2 3};int *b...

the ans

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February 19, 2006 12:12:37

swaant

RE: main(){int a[]={0 1 2 3};int *b...

the ans is

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February 19, 2006 12:13:35

swaant

RE: main(){int a[]={0 1 2 3};int *b...

ans is 1

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June 18, 2006 13:34:54

purani86

Member Since: June 2006 Contribution: 10

RE: main(){int a[]={0 1 2 3};int *b...

error.since array index cannot be negative

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September 01, 2006 11:05:09

Prabhu.T.S

RE: main(){int a[]={0 1 2 3};int *b...

Declaration syntax error in fn. main()

Elements in the array should be seperated by comma while declaration.

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October 05, 2007 15:59:39

bhoji_005

RE: main(){int a[]={0 1 2 3};int *b...

main()
{
int a[] {0 1 2 3};
int *b;
b &a[2];
printf(" d" b[-1]);
}
the ans is :error
why because of b[-1] indicates subscript of -1.....
there is no negetive subscrpts.......

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March 21, 2008 21:29:22

araghav_23

Member Since: March 2008 Contribution: 2

RE: main()n{nint a[]={0 1 2 3};nint *b;nb=&a[2];nprintf("%d",b[-1]);n}nTECHNICAL INTERVIEW

its a fact that the array index cannot be negative but i ran the program on both linux compiler and turboc for both of which i got '1' as the output

can anyone interpret this and explain me the concept plzzzzzzzzz

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March 30, 2008 12:09:37

divinegeek

Member Since: March 2008 Contribution: 1

RE: main()n{nint a[]={0 1 2 3};nint *b;nb=&a[2];nprintf("%d",b[-1]);n}nTECHNICAL INTERVIEW

answer will be 1.........
because a[2] 2
and when it will calculate b[-1] answer is 1............

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December 13, 2008 14:40:00

#10

Bhoomika sharma

Member Since: December 2008 Contribution: 40

RE: main()n{nint a[]={0 1 2 3};nint *b;nb=&a[2];nprintf("%d",b[-1]);n}nTECHNICAL INTERVIEW

It's logical and easy. See we have a[] i.e. assigned with a set of values. Now a pointer b is been referred to the base address of 2nd position of the array's location i.e. the pointer b is a pointer of arrays which store the second location of array as its first location in the pointer's base address i.e. b[0] contains a[2] value as its value thus the previous value is been stored as its b[-1] value.

So frnd ans is thus : 1

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