How to add a number to a string at the end in a c program?

Question

I need to write a function which will take a string as input and outputs as below

input : aaabbdd output : a3b2d2 input : aaaaaaaaaaaaaaaabbccc output : a16b2c3

basically I have to append the count to every character. I should not use itoa() to convert int to string

I wrote the logic. But I got struck at appending the number to the string. For example, if count is 16, how can I add the number 16 to end of a string?

My logic is as given below.

#include <stdio.h> void str1(char *str) { int i, j, cnt; int len = strlen(str); char *nstr = (char *) malloc(len * sizeof(char)); int k = 0; cnt = 1; for(i = 0, j = 1; i < len - 1;) { if(str[i] == str[j]) { j++; cnt++; continue; } else { if(cnt == 1) { nstr[k++] = str[i]; } else { nstr[k++] = str[i]; nstr[k++] = cnt; // GOT STUCK HERE } i = j; j = i + 1; cnt = 1; } } printf("\n%s\n", nstr); } main() { char str[] = "aaaaaaaaaaaaaaaabbcdd"; str1(str); }

why do you want to store it, you are only asked to output it? — Pieter21
– Pieter21, Commented Sep 8, 2014 at 18:37
@Pieter21 : How can I output it? Are you suggesting that just print character and number instead of storing it in a buffer and printing entire string at once? — kadina
– kadina, Commented Sep 8, 2014 at 18:40

Sergey Kalinichenko · Accepted Answer · 2014-09-08 18:25:36Z

You can implement itoa yourself. The logic is as follows:

If the number is zero, append zero (in your case, this should never happen, but in general case this is obviously possible)
Prepare a temporary buffer for your output. The size depends on the number of bits in the integer that you are printing
Store the last digit by computing n % 10
Divide the number by ten using integer division
Continue the last three steps until the remaining part is zero
Append the reversed sequence of digits from the temporary buffer to your output

This is only one way of implementing the logic. Other ways are possible as well - for example, you could build a lookup table of powers of ten, and compute each digit using a combination of integer division and taking the remainder.

Pieter21 · Accepted Answer · 2014-09-08 18:59:47Z

All the code you need, decorated with comments:

if (len == 0) return; /* initialize */ char c = str[0]; int count = 1; /* include terminating '\0', and that will resolve itself! */ for (i = 1; i <= len; i++) { if (str[i] == str[i-1]) { /* continue sequence */ count++; } else { /* end sequence */ printf("%c%d", c, count); /* start new sequence */ c = str[i]; count = 1; } } printf("\n"); /* flush buffer */

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