char array through function

Center For Modern Dance



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Last edited on
Just copy/paste the declaration of your variable.
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const int num=11;
const int length =100;

// Always prototype your functions.
void searchDirectory(char directory[num][length]);

int main()
{
char directory[num][length] = 
{   "Becky Warren, 555-1223",
    "JoeLooney, 555-0097",
    "Geri Palmer, 555-8787",
    "Lynn Presnell, 555-1212",
    "Holly Gaddis, 555-8878",
    "Sam Wiggins, 555-0998",
    "Bob Kain, 555-8712",
    "Tim Haynes, 555-7676",
    "Warren Gaddis, 555-9037",
    "Jean James, 555-4949",
    "Ron Palmer, 555-2783"};

searchDirectory( directory);

}

void searchDirectory(char directory[num][length])
{
cout<<"The Array is passed \n";
}


These declarations have exactly the same meaning.
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// The left-most dimension can be left empty
void searchDirectory(char directory[][length]);

// The left-most dimension can be written in pointer notation.
void searchDirectory(char (*directory)[length]);

If you do decide to go with the pointer notation, the ( ) are very important.
char (*directory)[length] and char *directory[length] are very different things.

Because arrays do not contain their length, it is customary to also pass the length.
So
void searchDirectory(char directory[][length], int num);


The rule 'arrays decay to pointers' is not a recursive statement. It only applies to the left-most dimension. So array[X][Y] becomes (*array)[Y], not **array.


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