how to see if there are 1 or 2 poker pairs in a hand in C

Question

I am trying to develop a C program that checks if there are 1 or 2 pairs in a 5 card poker hand.

I am using a 5x3 array where every line is a card (the 3rd column being for the \0 character). Every time I execute the code it always shows the "two pairs" print.

I want to make sure that each letter (i, j, a, b) representing each line is different. Any help?

P.S.: This is for a university/college project, I have only started programming a few months ago from absolute scratch, so any detailed explanations on my mistakes would be very much appreciated :)

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

char (cards[5][3])=
{
    "5S", "6D", "4H", "KD", "5C"
};

int main ()
{
    pair (cards[5][3]);
    return 0;
}

void pair (char (arg[n][0]))
{
    int i,j,a,b;

    if (i!=j!=a!=b)
    {
        if ((arg[i][0]==arg[a][0])&&(arg[b][0]!=arg[j][0]))
        {
            printf("2 -> pair");
        }

        if ((arg[i][0]==arg[a][0])&&(arg[b][0]==arg[j][0]));
        {
            printf("3 -> two pairs");
        }

        if ((arg[i][0]!=arg[a][0])&&(arg[b][0]!=arg[j][0]))
        {
            printf("there is no pair");
        }
    }
    else
    {
        printf("there is no pair");
    }
}

Answer 1

The posted code has several issues, both logical and syntactical, some have been pointed out in the comments.

Just to pick one, consider this line

if ((arg[i][0]==arg[a][0])&&(arg[b][0]==arg[j][0]));
{
     // This body will never be executed           ^
}

I'd suggest to restart from scratch and to proceed in small steps. See, for instance, the following minimal implementation

// Include all the needed header files, not the unneeded ones.
#include <stdio.h>

// Declare the functions prototype before their use, they will be defined after.
int count_pairs(int n, char const cards[][3]);
// Always specify the inner size,         ^  when passing a multidimensional array
void show_score(int n_pairs);
int have_the_same_value(char const *a, char const *b);

int main (void)
{
    char hand[5][3] = {
    //       ^^^^^^ You could omit the 5, here
        "5S", "6D", "4H", "KD", "5C"
    };

    int n_pairs = count_pairs(5, hand);
    // Always pass the size   ^ if there isn't a sentinel value in the array

    show_score(n_pairs);
    return 0;
}

// This is a simple O(n^2) algorithm. Surely not the best, but it's
// a testable starting point. 
int count_pairs(int n, char const cards[][3])
{
    // Always initialize the variables.
    int count = 0;

    // Pick every card...
    for (int i = 0; i < n; ++i)
    {
        // Compare (only once) with all the remaining others.
        for (int j = i + 1; j < n; ++j)
        { //         ^^^^^
            if ( have_the_same_value(cards[i], cards[j]) ) {
                ++count;
            }
        }
    }
    return count;
}

int have_the_same_value(char const *a, char const *b)
{
    return a[0] == b[0];
}

// Interpret the result of count_pairs outputting the score
void show_score(int n_pairs)
{
    switch (n_pairs)
    {
        case 1:
            printf("one pair.\n");
            break;
        case 2:
            printf("two pairs.\n");
            break;
        case 3:
            printf("three of a kind.\n");
            break;
        case 4:
            printf("full house.\n");
            break;
        case 6:
            printf("four of a kind.\n");
            break;
        default:
            printf("no pairs.\n");
    }
}

Note that my count_pairs function counts every possible pair, so if you pass three cards of the same kind, it will return 3 (given AC, AS, AD, all the possible pairs are AC AS, AC AD, AS AD).

How to correctly calculate all the poker ranks is left to the reader.