main()
{
char *x="girl";
int n,i;
n=strlen(x);
*x=x[n];
for(i=0;i<n;i++)
{
printf("%s \n",x);
x++;
}
}
What is the output?
Please explain the output.......................
o/p is :
irl
rl
l
4 Answers
The output is undefined behaviour. You modified a string literal.
2 Comments
pmg
... and called a variadic function without a prototype in scope; and didn't return a value from
main (the code is not C99 (which would accept a main without a return) because C99 forbids implicit int).
Lightness Races in Orbit
@Adi: You only changed the word in the string. That doesn't change the UB.
As others have pointed out, the program as written has undefined behaviour.
If you make a small alteration:
char x[] = "girl";
then I believe it is legal, and possible to explain. (EDIT: Actually there are still problems with it. It's int main(), and you should return 0; at the end. You also need to #include <string.h> because you are using strlen, and #include <stdio.h> because you are using printf.)
The line
*x = x[n];
sets x[0] (i.e. *x) to x[4] (which happens to be the string terminator '\0'). Thus the first string to be printed is the empty string, because the very first character is the string terminator.
We then loop through the string, one character at a time, printing the substrings:
irl
rl
l
Comments
While the result is undefined behavior, as DeadMG said, let's assume you declared x as char x[] = "girl".
You assign 4 to n (since the length of the word "girl" is 4), and you assign the value in x[4] to *x (which is x[0]), but this value is '\0' (null terminator)
Now you loop and print the word from x to the next null terminator, but the first time, the first char is the null terminator, so you get nothing. after that you print the word from incrementing index.
g i r l \0
*x = x[4]:
\0 i r l \0
^ ^ ^ ^
it1 it2 it3 it4 // << Where does x points to in each iteration of the for loop
Comments
The code is distictly suspect. *x=x[n] attempts to overwrite the literal "girl", and the effect will vary between platforms and compilers. More correctly it should be declared as:
const char *x = "girl";
and then it will not (should not) compile.
3 Comments
Lightness Races in Orbit
More correctly he should be making it safe to modify the string, not forcing his entire program to break by making the original non-mutability of the string compiler-checked.
Simon Nickerson
Huh? How is it more correct to write code which will not compile?
John Bode
@Simon: it's "more correct" in the sense that it won't allow him to modify a string literal; IOW, the compiler will catch those kinds of errors for him.
char*xtochar x[]to avoid undefined behaviour.