如何在C中比较两个指针的值

Question

My function below always returns true. I assume is because I'm comparing pointers. How can I compare the values and not just the pointers.

struct Card {
    const char *suit;
    const char *face;
};

...
struct Card hand[HAND_SIZE];
...

//Determine whether the hand contains a pair.
bool hasPair(struct Card wHand[]) {
    bool result = false;
    for (unsigned i = 0; i < HAND_SIZE; ++i) {
        for (unsigned j = 0; j < HAND_SIZE; ++j) {
            if(wHand[i].face == wHand[j].face && wHand[i].suit == wHand[j].suit) {
                result = true;
            }
        }
    }
    return result;
}

Answer 1

The biggest problem with this algorithm is that you are comparing the card with itself: you start both i and j at zero, so you get false positives for pairing each card with itself.

A simple fix to this is to start j at i+1 , ensuring that only different cards get compared.

Since setting result to true is a one-way street, consider returning true as soon as you find a match:

for (unsigned i = 0; i < HAND_SIZE; ++i) {
    for (unsigned j = i+1; j < HAND_SIZE; ++j) {
        if(wHand[i].face == wHand[j].face && wHand[i].suit == wHand[j].suit) {
            return true;
        }
    }
}
return false;

Note: This assumes that face and suit are set to string constants in the same code, which may be fragile if you link multiple object files. A much safer option for this would be using an enum for face and suit representation, and making an array of string representations for each enum value:

enum Face {
    Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King, Ace
};
enum Suit {
    Spades, Clubs, Diamonds, Hearts
};
struct Card {
    enum Face face;
    enum Suit suit;
};

Answer 2

result is set to true in the first (and, as it happens, in every) iteration of your i loop, given that at some point in each iteration, j==i and therefore you're comparing the same array element with itself. You should only execute your if statement when i!=j .

To your original point though, you're not comparing pointers - you are comparing the actual values. The array [ ] square brackets serve to dereference the pointer;
wHand[i] does the same thing as *(wHand + i) .