GATE IT 2008 | Question: 49

Question

What is the output printed by the following C code?

# include <stdio.h>
int main ()
{
    char a [6] = "world";
    int i, j;
    for (i = 0, j = 5; i < j; a [i++] = a [j--]);
    printf ("%s\n", a);
}

• A. dlrow
• B. Null string
• C. dlrld
• D. worow

Comments

why for loop is terminated here 

 for (i = 0, j = 5; i < j; a [i++] = a [j--]);

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Notice the semicolon

for (i = 0, j = 5; i < j; a [i++] = a [j--]);

Consider it as 

for (i = 0, j = 5; i < j; a [i++] = a [j--])
{
;

}

so this for loop does nothing except a[0] = a[5]

---

This for loop also does a[1] = a[4] and a[2] = a[3] and then it terminates.

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This question just blew my mind ,really awesome ,because most of us never thought about who would place null character in the beginning of the string .so mostly in hurry ,no one would thought that this will be null string 

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This question shows the beauty of GATE like never thought such an easy question would be such a tricker one

Answer 1

Char $a[6] = \begin{array}{|l|l|l|l|l|l|} \hline \text{w} &  \text{o} &  \text{r} &  \text{l} &  \text{d} &  \text{\0}\\\hline \end{array}$ 

After the loop executes for the first time, 

$a[0] = a[5]$

$a[0] =$`\0`

Next two more iterations of the loop till $i < j$ condition becomes false, are not important for the output as the first position is '\0';

printf(‘’%s’’, $a$);

printf function for format specifier '%s' prints the characters from the corresponding parameter (which should be an address) until "\0" occurs. Here, first character at $a$ is "\0" and hence it will print NOTHING. 

So, option (B).                  

Comments

@ bharti if possible can u plz explain stepwise manual execution of this bcoz i am not able to catch this.

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@admin you pasted the wrong code in Ideone that is why it was giving an error.

the printf is outside the for loop and you placed it inside 

try this https://ideone.com/Si6VP9 its working fine
 

---

@admin

It runs fine.

https://ideone.com/uuogzo

Answer 2

char a[6] = "WORLD"  which be stored like this 

W

O

R

L

D

\0

Key of this question : there is semicolon (;) after the for loop  , so printf statement will only execute when control will come out of the loop or loop condition will be false.

Iteration 1 : i=0,j=5; 0<5 (condition is true ) ;

                  a[0++] = j[5--] 

/o

o

r

l

d

\0

 ( here , R.H.S will execute first and j[5--](=null) will be assigned to a[0++] .and then j=5 will be decremented once, bcz post decrement here , same with a[0++] . i will be incremented by one after this statement . )

iteration 2: i=1,j=4 ; 1<4 ( condition is true )

                   a[1++] = j[4--]

\o

d

r

l

d

\o

iteration 3 :  i=2,j=3; 2<3 (condition is true)

                      a[2++]=j[3--]

\o

d

l

l

d

\o

iteration 4 : i=3,j=2; 3<2 (condition false )

                    so control will go to the printf statement , and the command inside printf statement is saying print the string present on this address of a  (base address of array) upto null.

but , here first string is null at a[0] . so control will not move further and the output will be null .

so option b is the correct answer.

Comments

@Arjun

https://ideone.com/enLGEJ

Please check this. I have made small edit in the code by making j=4 and removing semicolon of the for loop.

Please explain the output.

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thank you for good explaination bharti

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@bharti  thanks for this line

Key of this question : there is semicolon (;) after the for loop  , so printf statement will only execute when control will come out of the loop or loop condition will be false.

Answer 3

$a[6]=w|o|r|l|d|\setminus0$

$int\ i,j;$

$for(i=0,j=5;i<j;a[i++]=a[j--])$ $;$

$0<5 //condition\ True$

Because of the semi-colon it will not execute the statement instead it will do increment/decrement.

$a[0]=a[5]$

$i=1,j=4$

$1<4 //condition\ True$

$a[1]=a[4]$

$i=2,j=3$

$2<3 //condition\ True$

$a[2]=a[3]$

$i=3,j=2$

$3<2 //condition\ False$

$printf("\%s\setminus n",a)$

Correct Answer (B): Null string

Answer 4

We can consider like this

for(i=0, j=5 ; i<j ; i++, j--)

{

a[i] = a[j];

}

Therefore a[0] = null char which means end of string. Hence Ans will be Null String.

Thank you for reading.