Changing the value of elements in an array in C

Question

My program crashes once it reaches this function in my code. I printed the arrays in the function that are going in and commented out everything else and they are getting passed correctly. I can't figure out why it crashes here.

The arrays are numbers in ascending order.
Maybe my loops or if statements are not right? I think maybe I'm not correctly changing the value in the array at index "d" when it reaches attackerArmy[d] = '0'; //represent 0 as defeated ?

long int defeatedArmies(long int* attackerArmy, long int* defenderArmy, long int size){

    long int i,j,defeated = 0,d;

    for(i=0;i<size;i++){
        for(j=0;j<size;j++){

        //loop for the highest attack value that is lower than defender
            if(attackerArmy[j] < defenderArmy[i])
                d = j; // save the index where it is highest

            if(attackerArmy[j] > defenderArmy[i])
            //no point in checking unnecessary values
                break;
        }
        attackerArmy[d] = '0'; //represent 0 as defeated
    }

    for(i=0;i<size;i++){
        if(attackerArmy[i] == 0) //count how many defeated armies happened
            defeated++;
    }
return defeated;
}

Mohit Jain · Accepted Answer · 2016-03-02 04:49:37Z

3

Problem
if attackerArmy[j] >= defenderArmy[i] is true, d remains uninitialized causing the undefined behavior when you access its value in attackerArmy[d] = '0';.

Possible fix
Initialize d when you declare it.

ex:

long int d = -1L;
...
if(d != -1L) attackerArmy[d] = '0';

answered Mar 2, 2016 at 4:49

Mohit Jain

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4 Comments

Hispazn Over a year ago

ah yes, didn't think about that. there's one fix. it doesn't crash so far. But now when I try printing the new array (attackerArmy) where some things could be zeroed out after it leaves the first two loops. It is filled with a bunch of different numbers that weren't in either array before

Mohit Jain Over a year ago

@Hispazn This sounds like a different problem. Is that new number 48 (0x30)? Did you mean '0' and not 0 or '\0' in line attackerArmy[d] = '0';?

Hispazn Over a year ago

yes :O and can you explain to me the difference between '0' and 0 ? I know \0 is just null character, yes?

Mohit Jain Over a year ago

@Hispazn '0' is character literal for printable character 0 and its value in ascii table is not zero. (Just like 'a' is printable character a and its ascii values in non-zero). 0 OTOH is integer literal with value 0. '\0' is a nul cahracter literal with value 0 in ascii table.

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