CODEWITHSUNDEEP
crealloc
yogi
yogi

Reputation: 23

Allocating Memory to String at runtime

I am writing a program to count the occurrence of '2' followed by '1' in a sting. I dynamically allocated string

Code is:

#include <stdio.h>
#include<stdlib.h>

    int penalty_shoot(char* s){
        int count=0,i=0;
        while(s[i]!='\0'){
           if(s[i]=='2')
                if(s[i+1]=='1')
                    count++;
        i++;
        }
        return count;
    }

    int main() {
        int t;
        int i=0;
        scanf("%d",&t);           //t is for number of test cases.
        while(t--){
            char *str, c;
            str = (char*)malloc(1*sizeof(char));
            while(c = getc(stdin),c!='\n')
            {
                str[i] = c;
                i++;
                str=realloc(str,i*sizeof(char));
            }
            str[i] ='\0';
            printf("%s\n",str);
            printf("%d\n",penalty_shoot(str));

            free(str);
            str=NULL;
            i=0;
        }
        return 0;
    }

Input is :

3
101201212110
10101
2120

I am facing 2 problems:

1) I feel the dynamic allocation is not working fine.I wrote the code for dynamic allocation seeing various codes on stackoverflow . (Can anyone suggest some changes.)

2) The code isn't reading '2120' as the 3rd input. (why is it so ?)

Upvotes: 0

Views: 97

Answers (2)

vishal-wadhwa
vishal-wadhwa

Reputation: 1031

Three errors:

  1. Not checking for EOF:

    Change while(c = getc(stdin),c!='\n') to while(c=getc(stdin),c!='\n'&&c!=EOF)

  2. Reallocating with the wrong number of bytes:

    Change str=realloc(str,i*sizeof(char)); to str=realloc(str,(i+1)*sizeof(char));

    After taking one character input we increment i (i++), so the next character will be stored at the ith position. Now, in order to store the character at ith position, the length of the character array must be i+1. So, we realloc with i+1.

    Just for the sake of brevity, as suggested by Basile, you might as well do this:

    Change str=realloc(str,(i+1)*sizeof(char)); to str=realloc(str,i+1);

    Why? Because sizeof char is 1 byte

  3. Not consuming the \n after inputting t:

    Change scanf("%d",&t); to scanf("%d ",&t); or scanf("%d\n",&t);

    scanf("%d ",&t); or scanf("%d\n",&t);.

    Either of them works. Why, you ask? Read this explanation taken from another SO answer here:

    An \n - or any whitespace character - in the format string consumes an entire (possibly empty) sequence of whitespace characters in the input. So the scanf only returns when it encounters the next non-whitespace character, or the end of the input stream.

Tested here.

Upvotes: 2

Joseph Asaf Gardin
Joseph Asaf Gardin

Reputation: 480

you can use scanf("%d ", &t); when user input to test then just before second while loop, which condition should be c != '\n' write c = getchar(); and then make sure you create a char variable, i called mine clear, that receives 0 so when you loop after initiating your string you write c = clear; and under it c = getchar() again. and when you use realloc make sure you make it bigger by (i+1) since char is only the size of 1 byte.

we create the clear variable in order to clear the buffer.

it worked for me. make sure you insert the string all at once.

Upvotes: 0

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