Pointers Question

**kwikness** · 10-30-2007

Hi I was messing around with pointer in my program and ran into a problem. The compiler was giving me an initialization error for:

Code:

void swap(char *a[10], char *b[10])
{
        char *temp[10] = *b;

        *b = *a;
        //*a = *temp;
        printf("\n\n\n--->%s\n\n\n\n",*b );
}

But when changed to:

Code:

void swap(char *a[10], char *b[10])
{
        char *temp[10];
        *temp = *b;

        *b = *a;
        //*a = *temp;
        printf("\n\n\n--->%s\n\n\n\n",*b );
}

It works fine. Why can't I initialize this pointer on the same line I declare it? There must be a way.

**MacGyver** · 10-30-2007

If you don't know what the difference is between those two lines, you should pause and make sure you understand it.

The first one attempts to set the array itself to contain the value of what b points to, or to be slightly pedantic, the first element of the array of b. The second example assigns the first element of temp to be that of the first element of the array b.

**kwikness** · 10-30-2007

I understand, but how would I accomplish both the initialization and declaration in one line?

**MacGyver** · 10-30-2007

What are you even trying to do? Are you trying to copy the contents of the array b into temp? If so, neither example actually is close to doing that.

If you just want to set one element of temp to be equal to one element of b, I would recommend doing it on a separate line. I don't know of a way off hand to do it all on one line.

**kwikness** · 10-30-2007

I want to accomplish what these statements do:

Code:

        char *temp[10];
        *temp = *b;

But on one line.

As in: I want to initialize a char pointer and assign it the value of *b

**Salem** · 10-31-2007

So what are you trying to swap?

Swap a and b themselves - seems kind of pointless, since the change won't affect the caller.

Swap each pair of elements of a and b respectively. In which case the temp variable would just be
char *temp = b[0]; // or *b if you're so inclined.

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