How to insert an extra character in a string in C - c

Question

#include <stdio.h>
#include <string.h>
void replace(char a[],char b[]);
int main(void)
{
char string1[100]; char string2[100];
printf("Please enter a string:");
fgets(string1, sizeof(string1), stdin);
replace(string1, string2);
}
void replace(char a[],char b[])
{
int i;
strcpy(b,a);
for(i=0;i<strlen(a);i++)
{
if(a[i]=='a')
{
b[i+1]=a[i];
}
}
printf("%s\n",b);
}
The function of this program is to copy string1 into string2 but if there's any 'a' character in string1, it needed to be double in string2.
For example, if string1 is "aeroplane", the string2 needs to be "aaeroplaane".
However, the code above seem to change the next character after 'a' in string1 instead of simply adding it. For example, if I type "aloha", it will simply give me "aaoha" instead of "aalohaa". I've tried searching online but to no avail. I've been sticking on this problem for hours.

Answer 1

In order to work you must implement your replace function correctly
It should be like:
void replace(char a[],char b[])
{
int idxA = 0, idxB = 0;
for(idxA;idxA<strlen(a);idxA++, idxB++)
{
b[idxB] = a[idxA];
if(a[idxA]=='a')
{
idxB++;
b[idxB]=a[idxA];
}
}
b[idxB] = '\0';
printf("%s\n",b);
}
But you must ensure b has enough space to hold the complete string. For example, you can reserve twice the size of a, the worst case when array a has all a characters

Answer 2

Ok, if we are playing code golf, I would suggest:
void replace (char *a, char *b) {
while ((*b++ = *a))
if (*a++ == 'a')
*b++ = 'a';
}
;)
Mind to provide enough memory in b; if you want to play it safe, it has to be at least strlen(a)*2+1 (the case a consists only of character "a").

Answer 3

For the sake of brevity and elegance, here's a one liner:
void replace (char *a, char *b) { while((*b++=*a)&&(*a++!='a'||(*b++='a'))); }
;)

Answer 4

with
b[i+1]=a[i]
you're changing the next character to be 'a' and overwriting the next character.
What should be done is moving all the characters after b[i+1] one step forward and then changing 'a' in b[i+a]

Answer 5

void replace(char a[],char b[])
{
int i = 0, j = 0;
for(i=0; i<strlen(a); i++, j++)
{
b[j] = a[i];
if(a[i] == 'a')
b[++j] = a[i];
}
}
see if this works.

Answer 6

I wont solve it for you.
But here is what you need to look into.
if(a[i]=='a')
{
b[i+1]=a[i];
}
What it is doing is, if ith index of a array is 'a' then write ith index of a array in i+1th location of array b.
This way you are over-writing the contents of b.
You need to make a place for an additional character there only after shifting the string from that index to right.
It can be done in multiple ways.
Additionally, you might want to allocate more size (ideally twice: To cover the case when you pass a string1 with 100 'a's, the resulting string2 should hold 200 'a's) to your second array than the first one, because of obvious reasons.

Answer 7

You want this:
#include <stdio.h>
#include <string.h>
void replace(char a[], char b[]);
int main(void)
{
char string1[100]; char string2[100];
printf("Please enter a string:");
fgets(string1, sizeof(string1), stdin);
replace(string1, string2);
printf("%s\n", string2); // printf result from replace here
}
void replace(char a[], char b[])
{
int i, j; // using two indexes, one for a and one for b
// no strcpy needed here, copying is done in the for loop below
for (i = 0, j = 0; i < strlen(a); i++, j++)
{
b[j] = a[i]; // copy one char
if (a[i] == 'a') // if it's 'a'
{
b[++j] = a[i]; // copy it again
}
}
b[j] = 0; // terminate the destination string
// we don't printf b here but in main
}
For brevity no buffer overflow checking is done here.
There is still room for improvement though.