why it prints nothing?? please clarify me this doubt

Question

#include <stdio.h>
#include<string.h>

int main()
{
char p[20];

char *s = "string";

int length = strlen(s);

int i;

for (i = 0; i < length; i++) {

     p[i] = s[length - i];

printf("%s",p);
}
return 0;
}

Answer 1

If you begin at i = 1, it leaves p[0] uninitialized and won't copy the entire string. p[0] may contain 0 or some other garbage value (if it is zero, the result is the same as a null-termination).

So you must subtract 1 from the length.

You then add the null-terminator when you're done copying.

Answer 2

Because it copies the null-byte at the end of the string to the beginning, so printf() believes you passed it an empty string.

Make it p[i] = s[length - i - 1];

and put the printf() statement outside the for-loop.

Answer 3

C Lover
Array indexing begins at 0 and s[5] is the last character in the string, but strlen(s) returns 6.

s[6] contains the null-terminating byte. So you must subtract one from the length.

The string is terminated at the first iteration:
p[0] = s[6 - 0] where s[6] = '\0';

Answer 4

why it prints nothing?? please clarify me this doubt
Gen2oo
#include <stdio.h>
#include<string.h>

int main()
{
char p[20];

char *s = "string";

int length = strlen(s);

int i;

for (i = 1; i < length; i++) {

     p[i] = s[length - i];

printf("%s",p);
}
return 0;
}

I gave i=1 so s[length-] must not be null right but still prints nothing

Answer 5

Gen2oo I tired i=1 also .. which is s[length-1] or s[5] ...s[5] is g ..but still prints nothing

Answer 6

Gen2oo but if I give char s[] instead of char *s it is working fine it is printing