Proper casting

I want to cast a chunk of binary data to an array, how do I do that? I come from a C++ background and haven't been doing this in a while. Its the array thats throwing me off. Casting as "uint8_t*" or "uint8_t[6]" give me wrong type and casting an array errors respectively.

Code:

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void foo(char *rawdata) {
    uint8_t data[6];

    data = (???)rawdata;
}

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Yeah... you can't cast that.

You can do a strcpy() or a memcpy(), though.

data is not a pointer. It's a fixed location in memory representing an array of data.

I thought data was a pointer to the first element to an array of 6 elements, so you can do this:

Code:

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int i;
unsigned int data[6];
data[0] = 1; data[1] = 2; data[2] = 3; data[3] = 4; data[4] = 5; data[5] = 6;
unsigned int *ptr = data;
for(i = 0; i < 6; ++i)
{
        printf("&#37;u", *ptr++);
}

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To solve your problem you should
1. Know the size of the char array.
2. copy the memory into the new array. (so data needs to be a pointer indeed!)

It is a pointer in a sense, but you can't change what it points to. It points to the first element of the array and is const in that regard.

You're just doing the opposite and having a pointer assigned to an array. The OP wants an array assigned to a pointer. Can't work that way.

If you want to know how to cast properly, ask a fisherman :)

Quote:

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Originally Posted by Salem

If you want to know how to cast properly, ask a fisherman :)

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Should be plenty of those around here, what with all the C's around.

Thanks guys, that did the trick. Sorry for the simple question (and the more to come I'm sure). In the past I've just fumbled around with code til I got it to worked, but now I like actually getting an explanation. I appreciate it.

The name of an array decays to a pointer in a non-indexed context. But it's still not an l-value -- you can't assign to it.