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

Please share some reference @reboot

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@ankit3009 You may try to run the above program itself and print sizeof(s).

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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. 

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?

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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

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@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

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.

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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’.

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