How to determine length of pointer array

Hello,

Just a quick question, already googled etc..

If I have a pointer array lets say:

Code:

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char * fruit[] = {"apple", "pear", "orange", "grape", "banana"};

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Its just the program I am witting uses a pointer array which can change in size and I need to know how many variables are inside the array after it has changed size. For example there is 5 fruits in the above array.

I don't want to declare the array as a set length.

Thanks

Code:

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sizeof fruit/sizeof fruit[0];

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you mentioned changing size. if that's the case then you have to keep the size in a separate variable. the sizeof trick won't work on dynamic arrays and it only gives you the capacity on non-dynamic arrays.

Quote:

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Originally Posted by itCbitC

Code:

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sizeof fruit/sizeof fruit[0];

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that won't work because the fruits are not all the same size! [edit: I'm wrong about that, but this would work:] You could add an extra NULL pointer at the end:

Code:

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char * fruit[] = {"apple", "pear", "orange", "grape", "banana", NULL};

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And as long as you keep one there, you could use a function that will iterate thru one element at a time until fruit[X]==NULL.

But Meldreth's idea is more efficient; you just need an extra variable and to make sure it is always incremented and decremented appropriately.

Quote:

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that won't work because the fruits are not all the same size!

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they're all the size of a char pointer. sizeof doesn't reach into the pointer and see how long the string it points to is. ;) it works as-is as long as it's ok to always get 5 as a result. i don't think that's what Dr.Zoidburg wants.

>> that won't work because the fruits are not all the same size!
Yes they are, and yes it will.

The definition of fruits makes it so. Each fruit is the size of a pointer and the size of the array is the sum of all the pointers. Dividing through gives you the number of elements as long as the array was defined in the same scope.

Thanks for the lightening fast replies. I'll give them a go now.

Quote:

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Originally Posted by whiteflags

Yes they are, and yes it will.

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You and Meldreth are right, sorry. Perhaps if someone had written:

Code:

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sizeof fruit/sizeof *char

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Quote:

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Originally Posted by MK27

Perhaps if someone had written:

Code:

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sizeof fruit/sizeof *char

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so what do you think is the type of fruit[0] :D
the compiler auto adds a NULL to the end of (const char* type) arrays so no need to do so explicitly.

Quote:

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Originally Posted by itCbitC

so what do you think is the type of fruit[0] :D
the compiler auto adds a NULL to the end of (const char* type) arrays so no need to do so explicitly.

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Do you mean to say that this:

Code:

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char * fruit[] = {"apple", "pear", "orange", "grape", "banana"};

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actually looks this way:

Code:

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char * fruit[] = {"apple", "pear", "orange", "grape", "banana", NULL};

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I would say not. Try this:

Code:

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char * fruit[] = {"apple", "pear", "orange", "grape", "banana"};
char * veg[] = { "tomato", "cabbage", "celery" };
char *dummy[] = { NULL };
int main()
{
    char **ptr = fruit;
    while(*ptr)
    {
      printf("%s is a fruit\n", *ptr);
      ptr++;
    }
}

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You'd sure to see tomato, cabbage and celery printed as fruits.

A character array in double quotes, where there is no fixed size, will have a NUL-character at the end. But any other arrays will have nothing more in them than what you put there - unless you specify a size, and it is an initalized array, the rest of the array is filled with zeros.

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Mats

Ah! yes you're absolutely right and I'm totally wrong. My bad I got the auto NULL terminated string confused with an array of pointers to char.

Quote:

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Originally Posted by itCbitC

Ah! yes you're absolutely right and I'm totally wrong. My bad I got the auto NULL terminated string confused with an array of pointers to char.

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Glad to have cleared that one out...

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Mats

way to go matsp...