2

Count number (0-9) occurrences in a string from a txt file in C

I created the part that reads the string from the file and saves it as the variable "line"

My idea is to create a table with 10 elements that at the beginning are 0 and for example if I found 2 numbers of 0 it changes the table[1] in 2 or if it founds 1 number of 9 it changes table[10] to 1 but couldn't implement it

Output of the string "456 890 111111" should be:

Number 1 appears 6 times
Number 1 appears 1 times
...

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


int main()
{
    char line[255];
    int table[10];
    FILE *f = fopen("input.txt", "r");

    fgets(line, 255, f);
    printf("String read: %s\n", line);

    fclose(f);
    return 0;
}

Updated code:

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


int main()
{
    char line[255];
    unsigned int table[10] = { 0 };
    FILE *f = fopen("input.txt", "r");

    fgets(line, 255, f);
    printf("String read: %s\n", line);

    int n=0;

    for ( char *p = line; *p != '\0'; p++ )
    {
        if ( '0' <= *p && *p <= '9' ) {
            ++table[*p - '0'];
        }

    }


    fclose(f);
    for (int i=0; table[i] < 10; i++) {
        printf("Number", i, "apears", table[i], "times");
    }

    return 0;
}
designalix
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  • Does it help to know that `table[c - '0']` will give you the right table entry for any `c` that is a single ascii char representing a decimal digit? So you just need to iterate over each char in `line` and use that to increment the appropriate `table` entry each time. Don't forget to initialise the array contents, `int table[10] = { 0 };`. Also be sure to validate `c` before using it in the index calculation. – kaylum Mar 09 '22 at 21:20
  • Note that in C, indexes are `0`-based, not `1`-based, so valid indexes into an array of `10` elements are `0` to `9`, not `1` to `10`. – Andreas Wenzel Mar 09 '22 at 21:43
  • Are you supposed to read a single line from the file, or several lines? – Andreas Wenzel Mar 09 '22 at 22:06

1 Answers1

4

As the count of a digit can not be negative it is better to declare the array with the element type unsigned int. Also you need to initialize the array with zeroes.

For example

unsigned int table[10] = { 0 };

when in a loop you can count digits in the string. For example

for ( const char *p = line; *p != '\0'; ++p )
{
    if ( '0' <= *p && *p <= '9' )
    {
        ++table[*p - '0'];
    }
}

All digits as characters are represented sequentially without other intermediate symbols.

Pay attention to that indices in arrays start from 0.

After you updated the code in the question then this for loop

for (int i=0; table[i] < 10; i++) {
    printf("Numarul", i, "apare de", table[i], "ori");
}

is incorrect. For starters this condition table[i] < 10 makes no sense. And read the documentation about printf.

At least the loop should look like

for ( int i = 0; i < 10; i++ ) 
{
    printf( "Numarul %d apare de %u ori\n", i, table[i] );
}
Vlad from Moscow
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