CODEWITHSUNDEEP
carrayscharactercounting
user2023827
user2023827

Reputation: 21

is there a better way to find how many times a character appears in an array or string in C?

I'm still learning C and I've been trying to figure out the best way to count occurrences of characters in an array.

I plan on separating it into functions and expanding on it a lot but so far the best working code I've come up with is a bigger version of this:

#define SIZEONE 7
#define SIZETWO 3


int main(void)
{
    int arrOne[SIZEONE] = {97, 97, 98, 99, 99, 99, 99};
    char arrTwo[SIZETWO] = {'a', 'b', 'c'};
    int arrThree[SIZETWO] = {0};
    int countOne = 0;
    int countTwo = 0;
    int countThree = 0;
    
    for(countOne = 0; countOne < SIZEONE; countOne++)
    {
        for(countTwo = 0; countTwo < SIZETWO; countTwo++)
        {
            if(arrOne[countOne] == arrTwo[countTwo])
            {
                arrThree[countTwo] = arrThree[countTwo] + 1;
            }
        }
    }
    
    for(countThree = 0; countThree < SIZETWO; countThree++)
    {
        printf("%c ",arrTwo[countThree]);
    }
    
    countThree = 0;
    printf("\n");
    
    for(countThree = 0; countThree < SIZETWO; countThree++)
    {
        printf("%d ",arrThree[countThree]);
    }
    return 0;
}

From this I should get something that looks like:

a b c

2 1 4

I'm just wondering if there is a simpler way to do this that someone can point me towards or give me an example of before I start using this method.

Upvotes: 0

Views: 216

Answers (3)

Klas Lindb&#228;ck
Klas Lindb&#228;ck

Reputation: 33273

If you use counting sort you get less code:

long count[1u << CHAR_BIT]; 

char *text = "The string we want to count characters in";
long i;

// Clear count array
memset(count, 0, sizeof(count));

// Count characters
for (i = strlen(text) - 1; i >= 0; i--) {
  count[(unsigned char)text[i]]++;
}

// Print occurance:
for (i = 0; i < 1u << CHAR_BIT; i++) {
  if (count[i] > 0) {
    printf("%4c", i);
  }
}
printf("\n");
for (i = 0; i < 1u << CHAR_BIT; i++) {
  if (count[i] > 0) {
    printf("%4ld", count[i]);
  }
}
printf("\n");

Upvotes: 0

SHR
SHR

Reputation: 8313

best way is to define a counting array of 256 (or 127 for ascii only) zero it and for each occurrence increment to counting array. 

void count_chars(char* arr)
{
  int counting[256],i;
  memset(counting,0,sizeof(counting));
  for(i=0; arr[i];++i){
    ++counting[(unsigned char)arr[i]];
  }
  for(i=0; i<256;++i){
    if(counting[i]){
      printf("%c - %d\n", (char)i, counting[i]);
    }
  }
}

Upvotes: 0

Gabriel L.
Gabriel L.

Reputation: 5014

You can try to insert this function as an example for all array sizes :

int findOccurences(const char *array, const int array_size, const char ch_to_find)
{
    int found = 0;
    for(int i = 0; i < array_size; ++i)
    {
        if(array[i] == ch_to_find) found++;  
    }
    return found;
}

It's a better practice to name your variables with a significant name. This will be easier to read for you and for others that can read your code.

Upvotes: 1

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