1.6k views

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:

1. gnirts
2. string
3. gnirt
4. no output is printed
edited | 1.6k views
0
is it "length-I" or "length -i".please clarify!!
0
its corrected. its length-i

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).

edited by
+1
@arjun sir : i got the answer . but you have written if we write print("%s,p+1) ; will get option c

This is possible only if after t we have null else it will print from p[1] to p[19] ryt ?
+6
Yes, you are correct. But it doesn't stop at p[19] - it keeps on printing until some memory location has value 0 ('\0').
+1
yes i had this on my mind . Yeah :D
0
Sir,

Please let me know why the answer for printf("%s", p+1) is option c) and not option a). I'm confused because when i=5, (ie., i<6 holds) we get to print the character 's' too, right?
0

divya

length is string= 6  it stores as string% end with endmarker%

for loop will make it = %gtrin [s is left out since we can store 6 letters, (a) is answer when sizeof is used intead of length]

Now print(p+1) = grint  [ see here end marker is not present so after reaching index 20 it will halt.

+1
Thank you :)
0
error as i not declarēd
0
You got an eagles eye no one noticed that.
no output
edited