GATE CSE 2004 | Question: 33

Question

Consider the following C program segment:

char p[20]; int i;
char* s = "string";
int length = strlen(s);
for(i = 0; i < length; i++)
    p[i] = s[length-i];
printf("%s", p);

The output of the program is:

• A. gnirts
• B. string
• C. gnirt
• D. no output is printed

Comments

Note: Here  $p[i] = s[length-i]$  is allowed but,  $s[i] = s[length-i]$ is not allowed as 

char* s = "string"; 

“string” is stored in a shared read-only section. 

---

Note to self: 

The default value in the character array is null ( \0 ).

String functions use this null character to know when to stop the function.

#include <stdio.h>

int main()
{
    char arr[5] = {'r','a','h','u','l'};
    printf("%s\n",arr);

    return 0;
}

 

The code above prints $rahul$ but $\%s$ expects a null terminated string. 

Please note the above case is coincedental and doesn't always works, so this is undefined behavior in $C$.

Good Read : c - How should character arrays be used as strings? - Stack Overflow

---

@Abhrajyoti00 yes, because the string literal is used in this question, if it would have been a character array then it would be modifiable.

Answer 1

Here,

$p[0] = s[length] = $ '\0'; //compiler puts a '\0' at the end of all string literals

Now, for any string function in C, it checks till the first '\0' to identify the end of string. So, since the first char is '\0', printf %s, will print empty string. If we use printf("%s", p+1); we will get option (C) with some possible garbage until some memory location happens to contain "\0". For the given code, answer is (D).

Comments

Sir,

\0 indicates  end of the string but how printf() knows that it should print only till \0?

---

Here, p[20] is not initialized. So, it contains garbage values and in the for loop we are replacing the garbage value.

Program on different IDE

https://ideone.com/Yi8Ed8

https://ide.geeksforgeeks.org/gTUVz8OHoU

---

@Arjun sir

if we take int p[20]={1,2,3};

then if we write

for(int i=0;i<20;i++) printf(“%d ”,p[i]);  ------------→  1 2 3 0 0 …...17^th 0’s 

for(int i=0;i<40;i++) printf(“%d ”,p[i]);  ------------→  1 2 3 0 0 …...17th 0’s + 20^th (garbage value)

Is this correct ????

 

Answer 2

s	t	r	i	n	g	\0
s[0]	s[1]	s[2]	s[3]	s[4]	s[5]	s[6]

note: string literals always ends with a null character (\0)

 

s is a pointer to string literal and it is pointing to the first element of "string" (ie s)

length of the string is 6 so strlen(s) will return 6 and  int length will be 6 

since length is 6 the for loop will run from 0 to 5 

 

for(i = 0; i < length; i++)
    p[i] = s[length-i];

this will become 

for(i = 0; i < 6; i++)
    p[i] = s[6-i];

for i=0 --> p[0] = s[6-0] =s[6] --> s[6] is a null character (\0) so we will add it in our char array p at p[0]

for i=1 --> p[1] = s[6-1] =s[5] --> s[5] is g and we will add it at p[1]

for i=2 --> p[2] = s[6-2] =s[4] --> s[4]=n

for i=3 --> p[3] = s[6-3] =s[3] --> s[3]=i

for i=4 --> p[4] = s[6-4] =s[2] --> s[2]=r

for i=5 --> p[5] = s[6-5] =s[1] --> s[1]=t

for i=6 --> the condition i<6 fails

so the array p will look like 

\0	g	n	i	r	t
p[0]	p[1]	p[2]	p[3]	p[4]	p[5]

printf("%s", p); here the %s is used to print a string and 

it prints from the first element of the string untill it finds the null character 

Here the first element is null (\0) itself so it will print nothing. Hence the correct ans 

D. no output is printed

 

 

Comments

Sukriya 💖

Answer 3

no output

Comments

Strlen(s)  gives 6 as null is not counted then how can null be the first char in p and hence no output.

---

Somewhere it is stored like

characters	‘s'	‘t’	‘r’	‘i’	‘n’	‘g’	‘\0’
index	0	1	2	3	4	5	6

now strlen() will give number of char except null so it will return 6.

in program p[0]=s[length-0]  so p[0] =s[6-0]='\0’.

---

because S[6] is element stored at 6th position which is null character.

Answer 4

Declaration	Is '\0' added automatically?	Valid C String?	Internal Storage
c char s[] = "hello";	Yes	Yes	h e l l o \0
c char *p = "hello";	Yes	Yes	h e l l o \0
c char s[10] = "abc";	Yes	Yes	a b c \0 0 0 ...
c char s[] = {'h','e','l','l','o'};	No	No	h e l l o
c char s[] = {'h','e','l','l','o','\0'};	Already given manually	Yes	h e l l o \0
c char s[5] = "hello";	No space for '\0'	No	h e l l o
c char s[6] = "hello";	Yes	Yes	h e l l o \0

Quick Memory Trick--->

Syntax	'\0' Added?
Using " " (double quotes)	Yes
Using { } manually	No (unless you add it)