Rearranging string letters - c

Question

I was doing a program to copy all string words other than its first 2 words and putting a x at the end of it.
However i cant put x at its end. Please help!!!!
Below is my code.
#include<stdio.h>
#include<string.h>
int main()
{
char a[25], b[25];
int i, j, count = 0, l, k;
scanf("%[^\n]s", a);
i = strlen(a);
if (i > 20)
printf("Given Sentence is too long.");
else
{/* checking for first 2 words and counting 2 spaces*/
for (j = 0; j < i; j++)
{
if (a[j] == ' ')
count = count + 1;
if (count == 2)
{
k = j;
break;
}
}
/* copying remaining string into new one*/
for (j = 0; j < i - k; j++)
{
b[j] = a[j + k];
}
b[j + 1] = 'x';
printf("%s", b);
}
}

Answer 1

you are removing first two index. But you wrote k=j and if you check the current value j there it's 1. so you are updating wrongly k because you removed 2 indexes. So k value should be 2. So checked the below code
/* copying remaining string into new one*/
for (j = 0; j < i - 2; j++)
{
b[j] = a[j + 2];
}
b[j + 1] = 'x';
printf("%s", b);

Answer 2

Your index is off by one. After your second loop, the condition j < i-k was false, so j now is i-k. Therefore, the character after the end of what you copied is b[j], not b[j+1]. The correct line would therefore be b[j] = 'x';.
Just changing this would leave you with something that is not a string. A string is defined as a sequence of char, ending with a '\0' char. So you have to add b[j+1] = 0; as well.
After these changes, your code does what you intended, but still has undefined behavior.
One problem is that your scanf() will happily overflow your buffer -- use a field width here: scanf("%24[^\n]", a);. And by the way, the s at the and doesn't make any sense, you use either the s conversion or the [] conversion.
A somewhat sensible implementation would use functions suited for the job, like e.g. this:
#include<stdio.h>
#include<string.h>
int main(void)
{
// memory is *cheap* nowadays, these buffers are still somewhat tiny:
char a[256];
char b[256];
// read a line
if (!fgets(a, 256, stdin)) return 1;
// and strip the newline character if present
a[strcspn(a, "\n")] = 0;
// find first space
char *space = strchr(a, ' ');
// find second space
if (space) space = strchr(space+1, ' ');
if (space)
{
// have two spaces, copy the rest
strcpy(b, space+1);
// and append 'x':
strcat(b, "x");
}
else
{
// empty string:
b[0] = 0;
}
printf("%s",b);
return 0;
}
For functions you don't know, google for man <function>.

Answer 3

In C strings are array of chars as you know and the way C knows it is end of the string is '\0' character. In your example you are missing at the last few lines
/* copying remaining string into new one*/
for(j=0;j<i-k;j++)
{
b[j]=a[j+k];
}
b[j+1]='x';
printf("%s",b);
after the loop ends j is already increased 1 before it quits the loop.
So if your string before x is "test", it is like
't', 'e', 's', 't','\0' in char array, and since your j is increased more than it should have, it gets to the point just right of '\0', but characters after '\0' doesnt matter, because it is the end, so your x will not be added. Simple change to
b[j]='x';