Passing pointer of char array to function

Question

Below is my function which takes a char array of size 2, passes it to a function and the function should return the same 2 chars back (its a bit of a convoluted process as it's talking to a hardware device). The problem is when I'm passing the char (*in)[2] into the function; When I print the elements it only displays the data in[0] and not in[1]. Not sure why. Any help appreciated.

int function(struct device *d, char (*in)[2], char (*out)[2])
{

    printf("in: %c %c\n", *in[0], *in[1]); // TEST
    uint16_t data = (*in[0] << 8) + *in[1];

    send_request(d, IN, ECHO, data, 0, *out, 2);

    printf("%s\n", *out);

    return 0;
}

Thanks Bob

This: *in[0] likely isn't parsing how you think it is. Likewise for *in[1]. — WhozCraig
– WhozCraig, Commented Oct 2, 2014 at 6:57

2501 · Accepted Answer · 2014-10-02 07:12:50Z

4

I'm assuming you are passing addresses of arrays to you function().

The problem is that [] operator has a higher precedence than * operator.

The correct code would be.

printf("in: %c %c\n", (*in)[0], (*in)[1]);

First you dereference in to get the original array and then you get the element of that array.

answered Oct 2, 2014 at 7:12

2501

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1 Comment

Bob Jane Over a year ago

You posted first so I suppose its only fair to give you the points. And thank you.

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