Insert integer in an array C++

Question

I wrote the following code to convert string of type 'aaadddbbbccc' to 'a3d3b3c3' :

#include <iostream>
#include <string.h>
using namespace std;

    void stringCompression(char *str,char *newStr){
        int a[256] = {0};
        int newCount = 0;
        for(int i = 0; i < strlen(str) ; i++){
            int j = str[i];
            if (a[j] == 0 && strlen(newStr) <= strlen(str)){
                a[j] = 1 ;
                newStr[newCount] = str[i];
                newCount++;
                int count = 0;
                for (int n = i; n < strlen(str); n++){
                    if(str[i] == str[n]){
                        count = count + 1;
                    }
                }
                newStr[newCount] =(char) count;
                newCount++ ;
            } else if (strlen(newStr) > strlen(str)){
                strcpy(newStr,str);
            }
        }
    }

    int main() {
        char str[] = "abcdabcdabcd";
        char *newStr = new char[strlen(str)+1];
        stringCompression(str,newStr);
        cout << newStr;
        return 0;
    }

My problem is at step

newStr[newCount] =(char) count;

even though it is inserted but the output is not a3b3c3d3 but a*squarebox*b*squarebox*c*squarebox*d*squarebox*. squarebox being 2*2 matrix with one value as the number that is desired. I am using eclipse IDE. . I would really appreciate your help. How can I correct this. Am I using the correct approach?

Thanks in advance.

Assuming count is a simple digit, you may use newStr[newCount] = static_cast<char>('0' + count);. — Jarod42
– Jarod42, Commented Jul 22, 2014 at 7:42
Not related to the problem, but you really don't want to use strlen in the condition of the loop. C style strings are designed for access through the pointer, so your outer level loop should be something like for ( char const* p = str; *p != '\0'; ++ p ), with *p to access the characters. — James Kanze
– James Kanze, Commented Jul 22, 2014 at 10:34
And of course, using the char converted to an int is not going to work for all characters, at least not if char is signed (which is often the case). — James Kanze
– James Kanze, Commented Jul 22, 2014 at 10:35

Community · Accepted Answer · 2017-05-23 12:21:26Z

1

The problem is that

newStr[newCount] =(char) count;

converts the number "count" into the character corresponding to that number according to the ascii table (http://www.asciitable.com/), which is "end of text" for "3", that does not correspond to any number.

You should convert "count" into a string instead. See here for example: Easiest way to convert int to string in C++

However, be aware that it might be longer than one digit, for example if count is "11", it will take two letters in string representation.

edited May 23, 2017 at 12:21

CommunityBot

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answered Jul 22, 2014 at 6:39

Jan Rüegg

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Comments

PankajM · Accepted Answer · 2014-07-22 06:37:03Z

0

Hey you have to use http://www.cplusplus.com/reference/cstdlib/itoa/ to convert integer to char string

answered Jul 22, 2014 at 6:37

PankajM

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1 Comment

James Kanze Over a year ago

Not a very reliable reference, but they do point out that atoi is a sometimes thing---it's not standard, and not all compilers support it.

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