populating 2-D array

#include <stdio.h>
#include <stdlib.h>

void getA(int MatrixA[3][3]);

int main(void)
{

int MatrixA[3][3];

getA(MatrixA[3][3]);

return 0;

}

void getA(int MatrixA[3][3])
{

for (int y=0; y<3; y++)
{

for (int x=0; x<3; x++)
{

int num;
printf("enter nums");
scanf("%d", &num);
MatrixA[y][x]= num;

}

}

}

when I compile this code, I get the warning
"warning: passing argument 1 of 'getA' makes pointer from integer without a cast"

what does that mean and how can I overcome it?

In C array indexes start at zero so MatrixA[3][3] element doesn't exist. In fact you have walked off the end of the array ie out-of-bounds. Normally you pass the array name, which is a pointer to its first element, as an argument to a function for manipulation.

Delete the [3][3] in your line of code that calls the function. Leave the rest as is.

The function prototype says "I'm going to receive an int array address" (cause that's what arrays turn into when passed to a function).

But the sending line of code is saying "I'm sending you the int at MatrixA[3][3]".

So you're literally trying to change an integer (since it's an int array), into a pointer, without a cast.

And no, don't cast it, please. :p

As ItCbitC pointed out, not only is it wrong, but MatrixA[3][3] is not an element *in* the MatrixA array. Highest row and column is 2 (0,1,2).

thank you!

Quote:

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Originally Posted by edster

thank you!

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You're very welcome.

<< but if you don't put code tags (use the # icon in the editing window) around your program next time, the grumps will be after you. >>

It's amazing how many ways you can try to set this up, and still have it wrong. Tuck your program away for reference, because you're sure to want it later.