Printing an array of strings until array == NULL

Hi,

I have a weird problem that I have no idea how to solve.

Here's the code and the output: C code - 19 lines - codepad

The jist of my problem is that:

Code:

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while(array[i][SL] > 0){

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works OK, but displays one less value of array[i][SL] than it should. So what I've done is edit the code like so:

Code:

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while(array[i-1][SL] > 0){

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But I am aware that this is simply masking the real problem. Please help!

Post your code here... don't be making us chase it.

Code:

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#include <stdio.h>


#define AL 100
#define SL 32


int main(){
    char array[AL][SL] = { "at", "be", "funzey", "zebra" };
    int i,used = 0;
    //i want to find out how much of the array is in use.
    for(i=0;i<AL;i++){
        if(array[i][SL]!=NULL){
            printf("Slot #%i = %s\n",i,array[i]);
            used++;
        }
    }
    printf("Slots used: %i\n",used);
    printf("Slots remaining: %i\n",AL-used);
    return 0;
}

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OUTPUT:

Quote:

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Slot #0 = at
Slot #1 = be
Slot #2 = funzey
Slots used: 3
Slots remaining: 97

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Codepad is pretty nice, though. You can edit my code, compile it and run it without leaving the page.

Uh, it might not be intentional, but it looks like your posted code is completely formatted wrongly. If you are using some external bbcode/HTML based syntax highlighting tool, ditch it and just post your "raw" code as syntax highlighting will be done for you.

fixed

Quote:

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Originally Posted by mikemhz

Code:

---

#include <stdio.h>


#define AL 100
#define SL 32


int main(){
    char array[AL][SL] = { "at", "be", "funzey", "zebra" };
    int i,used = 0;
    //i want to find out how much of the array is in use.
    for(i=0;i<AL;i++){
        if(array[i][SL]!=NULL){
            printf("Slot #%i = %s\n",i,array[i]);
            used++;
        }
    }
    printf("Slots used: %i\n",used);
    printf("Slots remaining: %i\n",AL-used);
    return 0;
}

---

Codepad is pretty nice, though. You can edit my code, compile it and run it without leaving the page.

---

Our job is to help you fix your own code... we are not here to fix it for you.

Ok mate, just saying. Who knows, you might need to play around with it to come up with a solution yourself... I thought I was doing you a favour, not asking you to do my homework for me...

Thank you for being here none-the-less. You are a valued human being. I will hopefully owe you 20 minutes of not asking at another forum for helping me.

The thing is, array[i][SL] is not a pointer. Furthermore, it is actually out of bounds of each individual array, but because they are all part one a 2D array, it is not out of bounds of the 2D array except for the last array.

Rather, compare array[i][0] to '\0', since you are searching for the first empty string.

Good shout. Thanks :)

Would you mind clarifying what an array of strings is, other than a pointer?

I understand that int array[5]; is used like a pointer...

e.g.

Code:

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func(int * a){
    stuff();
}

int main(){
    int array[5];
    func(array); //rather than func(&array);
}

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Quote:

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Originally Posted by mikemhz

Would you mind clarifying what an array of strings is, other than a pointer?

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An array of strings is not a pointer :)

Quote:

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Originally Posted by mikemhz

I understand that int array[5]; is used like a pointer...

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When an array is passed as an argument, it is converted to a pointer to its first element.

Look at these two lines.

Quote:

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Code:

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        if(array[i][SL]!=NULL){
            printf("Slot #%i = %s\n",i,array[i]);

---

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The second line happily passes a pointer to the string you want to output and all is well. So why is the first line doing something different with the array?