Replacing three 'a' in with a single '*' in a string - c

Question

So my program should get input from an user and store it in an array. After that if the input string includes three 'a's in a row it should be replaced with a single '*'. However I can't seem to get it right. It only replaces the first a with a *. I tried to replace the following 2 a with a blank but the output looks funny.
For this exercise I have to use putchar() and getchar().
Thank you in advance.
#include <stdio.h>
char c;
char buffer[256];
int counter= 0;
int i;
int main()
{
while ((c = getchar()) != '\n') {
buffer[counter] =c;
counter++;
if (counter >255) {
break;
}
}
for(i=0; i<256; i++) {
if(buffer[i]== 'a'&&buffer[i+1]=='a'&&buffer[i+2]=='a')
{
buffer[i]= '*';
buffer[i+1]=' ';
buffer[i+2]=' ';
}
putchar(buffer[i]);
}
putchar('\n');
return 0;
}

Answer 1

So my program should get input from an user and store it in an array.
After that if the input string includes three 'a's in a row it should
be replaced with a single '*'. However I can't seem to get it right.
You almost got it! Just move index by 2 to and continue.
#include <stdio.h>
char c;
char buffer[256];
int counter= 0;
int i;
int main(void)
{
while ((c = getchar()) != '\n') {
buffer[counter] =c;
counter++;
if (counter >= 255) {
break;
}
}
buffer[counter] ='\0';
for(i=0; i<256; i++) {
if(buffer[i]== 'a'&&buffer[i+1]=='a'&&buffer[i+2]=='a')
{
buffer[i]= '*';
putchar(buffer[i]);
i = i + 2;
continue;
}
putchar(buffer[i]);
}
putchar('\n');
return 0;
}
Test:
123aaa456aaa78
123*456*78

Answer 2

In string you must assign a end of character at the end and that is call null character \0 or just a numeric 0. Correct your code like below:-
while ((c = getchar()) != '\n') {
buffer[counter] =c;
counter++;
if (counter >=255) {
break;
}
}
buffer[counter] ='\0';// or buffer[counter] =0;
To avoid side effect in a string array always set all its value with 0 first:-
char buffer[256];
memset(buffer, 0, sizeof(buffer));

Answer 3

If you want to change the number of characters, you will need to create a different buffer to copy the output to.
If you really just want to output to the console, you could just write every character until you hit your matching string.
#include <stdio.h>
char c;
char buffer[256];
char output[256];
int counter= 0;
int i, j;
int main()
{
while ((c = getchar()) != '\n') {
buffer[counter] = c;
counter++;
if (counter >255) {
break;
}
}
buffer[counter] = 0;
for(i=0, j=0; i<256; i++, j++) {
if(buffer[i] == 'a' && buffer[i+1] == 'a'&& buffer[i+2] == 'a')
{
output[j]= '*';
i += 2;
}
else
output[j] = buffer[i];
putchar(output[j]);
}
putchar('\n');
return 0;
}

Answer 4

There are multiple problems in your code:
there is no reason to make all these variables global. Declare them locally in the body of the main function.
use int for the type of c as the return value of getchar() does not fit in a char.
you do not check for EOF.
your test for buffer overflow is off by one.
you do not null terminate the string in buffer. You probably make buffer global so it is initialized to all bits 0, but a better solution is to set the null terminator explicitly after the reading loop.
to replace a sequence of 3 characters with a single one, you need to copy the rest of the string.
You can use a simple method referred as the 2 finger approach: you use 2 different index variables into the same array, one for reading, one for writing.
Here is how it works:
#include <stdio.h>
int main() {
char buffer[256];
int c;
size_t i, j, counter;
for (counter = 0; counter < sizeof(buffer) - 1; counter++) {
if ((c = getchar()) == EOF || c == '\n')
break;
buffer[counter] = c;
}
buffer[counter] = '\0';
for (i = j = 0; i < counter; i++, j++) {
if (buffer[i] == 'a' && buffer[i + 1] == 'a' && buffer[i + 2] == 'a') {
buffer[j] = '*';
i += 2;
} else {
buffer[j] = buffer[i];
}
}
buffer[j] = '\0'; /* set the null terminator, the string may be shorter */
printf("modified string: %s\n", buffer);
return 0;
}