Why does the output have random characters at the end? - c

Question

I was trying to solve a problem in toph.co platform. I have made a code. It is not working as it should be. Giving me the correct output in some cases. But when I am trying a case with 'o' it is entering in a loop. And maybe the problem is there. But I am unable to find that out.
Problem link: https://toph.co/p/better-passwords
Please help me to solve the problem in my code. I am using c programming language. As I am new in programming, I am not getting the point that is wrong.
I have tried it in many ways. Modifying the code again and again. Now I am becoming mad.
#include <stdio.h>
#include<string.h>
int main()
{
char s[40], i, j;
int len;
gets(s);
len= strlen(s);
for(i=0; i<=len; i++)
{
if(s[i] == 's')
{
s[i]= '$';
}
else if(s[i] == 'i')
{
s[i]= '!';
}
else if(s[i] == 'o')
{
s[i]= '(';
for(j=len; j>i; j--)
{
s[j]=s[j-1];
len= strlen(s);
}
s[i+1]=')';
}
else
{
continue;
}
len= strlen(s);
}
if (s[0]>='a' && s[0]<='z')
{
s[0]= s[0]- 32;
}
int new_len=strlen(s);
s[new_len]='.';
puts(s);
return 0;
}
I expect the output Un$()ph!$t!cated. , but it is showing Un$()ph!$t!cated.....'Many unwanted charecter'.....

Answer 1

Your string null byte is missing after manipulations, easiest way to avoid such problems - is to initialize whole character array into zero bytes:
char s[40] = {0}, // was char s[40], uninitialized !
Also do you notice a compile message "warning: the `gets' function is dangerous and should not be used" ?
gets() is dangerous, because it is not protected from buffer overflow - try to run your program with very long string exceeding your buffer s capacity and you will get a crash :
* stack smashing detected *: terminated Aborted (core dumped)
Use fgets() instead of gets(), like so :
fgets(s, 39, stdin);
if (s[strlen(s)-1] == '\n') s[strlen(s)-1] = '\0'; // deleting newline character
Notice, that we read here BUFFER_SIZE - 1 characters here, i.e.- by 1 char less, than your buffer is able to hold (40), because if we enter long string which is of full buffer size,- then your code extending string will smash stack once again. You need to be serious about buffer overflows.

Answer 2

As most commentators wrote: you don't put the end-of-string character '\0' behind the extended string. Strings in C are marked by this extra character at the end. It is needed to find the end of a string.
Example: A string "abc" is actually a sequence of 4 characters, 'a', 'b', 'c', and '\0'. Its length is 3, and its s[3] contains '\0'.
Your loop:
for(j=len; j>i; j--)
{
s[j]=s[j-1];
len= strlen(s);
}
In the first turn of the loop the character at s[len] is replaced by the character at s[len-1]. This overwrites the original '\0' with the last (visible) character of your string.
If you change the loop into:
for (j = len; j > i; j--)
{
s[j + 1] = s[j];
}
the '\0' will be copied in the first run.
Note 1: Move the assignment to len behind the loop.
Note 2: Make sure that your variable s is big enough for all inputs.
Note 3: See the whitespace I use? Adopt a good code style and stick to it.

Answer 3

There are a couple of places in your code where you are overwriting the "\0".
These are:
s[j]=s[j-1];
And
s[new_len]='.';
So the rectified code would be....
int main()
{
char s[40], i, j;
int len;
gets(s);
len= strlen(s);
for(i=0; i<=len; i++)
{
if(s[i] == 's')
{
s[i]= '$';
}
else if(s[i] == 'i')
{
s[i]= '!';
}
else if(s[i] == 'o')
{
s[i]= '(';
for(j=len+1; j>i; j--) //j needs to be one more than the length
{
s[j]=s[j-1];
len= strlen(s);
}
s[i+1]=')';
}
else
{
continue;
}
len= strlen(s);
}
if (s[0]>='a' && s[0]<='z')
{
s[0]= s[0]- 32;
}
int new_len=strlen(s);
s[new_len]='.';//add a '.'
s[new_len+1] = '\0'; //add the terminating character
puts(s);
return 0;
}