Casting a char[x][x] to char *[]?

[homer_3]

So I have some code like

static void myfunc(char * val[])
{
   if(*(val[0]) == 'x')
   //do stuff
}

char foo[30][30];
//populate foo
myfunc(foo);

This won't compile. I tried myfunc((char *[])foo); but that wouldn't compile either. I could only get it to compiile with myfunc((char **)foo);. But this doesn't run because the the value at val[0] gets treated like a pointer. So I tried val[0][0], but the same thing happened.

Is this because I'm casting to char ** when I pass it?

[laserlight]

The problem is that val is a pointer to a pointer, but foo is an array of arrays. When an array of arrays is passed as an argument, it is converted to a pointer to an array, not a pointer to a pointer.

Can you change the parameter of myfunc to accommodate foo? If not, one possibility is to create an array of pointers such that each pointer points to a corresponding inner array of foo. Passing this array of pointers will then yield a pointer to a pointer that is exactly what myfunc requires.

[EddieK]

You can always do something like this:

static void MyFunction(char *val) {
	if (val[0] == 'x') {
		//...
	}
}

char foo[50][50];
//...

MyFunction(*foo);

[homer_3]

Can you change the parameter of myfunc to accommodate foo? If not, one possibility is to create an array of pointers such that each pointer points to a corresponding inner array of foo. Passing this array of pointers will then yield a pointer to a pointer that is exactly what myfunc requires.

No, I can't change myfunc's params. Too much other code uses it. I was hoping to avoid malloc, but it looks like I'll have to do char *foo[30] and then malloc each index.

[laserlight]

You don't have to use malloc if you are willing to create a parallel array of pointers.