Passing pointers of arrays in C

Question

So I have some code that looks like this:

int a[10];
a = arrayGen(a,9);

and the arrayGen function looks like this:

int* arrayGen(int arrAddr[], int maxNum)
{
    int counter=0;
    while(arrAddr[counter] != '\0') {
        arrAddr[counter] = gen(maxNum);
        counter++;
    }
    return arrAddr;
}

Right now the compilier tells me "warning: passing argument 1 of ‘arrayGen’ makes integer from pointer without a cast"

My thinking is that I pass 'a', a pointer to a[0], then since the array is already created I can just fill in values for a[n] until I a[n] == '\0'. I think my error is that arrayGen is written to take in an array, not a pointer to one. If that's true I'm not sure how to proceed, do I write values to addresses until the contents of one address is '\0'?

Answer 1

@jeffD Passing the index would be the preferred way, as there's no guarantee you won't hit other '\0's before your final one (I certainly was when I tested it).

Answer 2

The way your using it arrayGen() doesn't need to return a value. You also need to place '\0' in the last element, it isn't done automatically, or pass the index of the last element to fill.

Answer 3

I'm not sure what you are trying to do but the assignment of a pointer value to an array is what's bothering the compiler as mentioned by Charlie. I'm curious about checking against the NUL character constant '\0'. Your sample array is uninitialized memory so the comparison in arrayGen isn't going to do what you want it to do.

The parameter list that you are using ends up being identical to:

int* arrayGen(int *arrAddr, int maxNum)

for most purposes. The actual statement in the standard is:

A declaration of a parameter as "array of type" shall be adjusted to "qualified pointer to type", where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation. If the keyword static also appears within the [ and ] of the array type derivation, then for each call to the function, the value of the corresponding actual argument shall provide access to the first element of an array with at least as many elements as specified by the size expression.

If you really want to force the caller to use an array, then use the following syntax:

void accepts_pointer_to_array (int (*ary)[10]) {
    int i;
    for (i=0; i<10; ++i) {
        (*ary)[i] = 0; /* note the funky syntax is necessary */
    }
}

void some_caller (void) {
    int ary1[10];
    int ary2[20];
    int *ptr = &ary1[0];
    accepts_pointer_to_array(&ary1); /* passing address is necessary */
    accepts_pointer_to_array(&ary2); /* fails */
    accepts_pointer_to_array(ptr);   /* also fails */
}

Your compiler should complain if you call it with anything that isn't a pointer to an array of 10 integers. I can honestly say though that I have never seen this one anywhere outside of various books (The C Book, Expert C Programming)... at least not in C programming. In C++, however, I have had reason to use this syntax in exactly one case:

template <typename T, std::size_t N>
std::size_t array_size (T (&ary)[N]) {
    return N;
}

Your mileage may vary though. If you really want to dig into stuff like this, I can't recommend Expert C Programming highly enough. You can also find The C Book online at gbdirect.

Answer 4

Hmm, I know your question's been answered, but something else about the code is bugging me. Why are you using the test against '\0' to determine the end of the array? I'm pretty sure that only works with C strings. The code does indeed compile after the fix suggested, but if you loop through your array, I'm curious to see if you're getting the correct values.

Answer 5

The basic magic here is this identity in C:

*(a+i) == a[i]

Okay, now I'll make this be readable English.

Here's the issue: An array name isn't an lvalue; it can't be assigned to. So the line you have with

a = arrayGen(...)

is the problem. See this example:

int main() {
    int a[10];

    a = arrayGen(a,9);

    return 0;
}

which gives the compilation error:

gcc -o foo foo.c
foo.c: In function 'main':
foo.c:21: error: incompatible types in assignment

Compilation exited abnormally with code 1 at Sun Feb  1 20:05:37

You need to have a pointer, which is an lvalue, to which to assign the results.

This code, for example:

int main() {
    int a[10];
    int * ip;

    /* a = arrayGen(a,9);  */
    ip = a ; /* or &a[0] */
    ip = arrayGen(ip,9);

    return 0;
}

compiles fine:

gcc -o foo foo.c

Compilation finished at Sun Feb  1 20:09:28

Note that because of the identity at top, you can treat ip as an array if you like, as in this code:

int main() {
    int a[10];
    int * ip;
    int ix ;

    /* a = arrayGen(a,9);  */
    ip = a ; /* or &a[0] */
    ip = arrayGen(ip,9);

    for(ix=0; ix < 9; ix++)
        ip[ix] = 42 ;

    return 0;
}

---

Full example code

Just for completeness here's my full example:

int gen(int max){
    return 42;
}

int* arrayGen(int arrAddr[], int maxNum)
{
    int counter=0;
    while(arrAddr[counter] != '\0') {
        arrAddr[counter] = gen(maxNum);
        counter++;
    }
    return arrAddr;
}

int main() {
    int a[10];
    int * ip;
    int ix ;

    /* a = arrayGen(a,9);  */
    ip = a ; /* or &a[0] */
    ip = arrayGen(ip,9);

    for(ix=0; ix < 9; ix++)
        ip[ix] = 42 ;

    return 0;
}

Answer 6

Why even return arrAddr? Your passing a[10] by reference so the contents of the array will be modified. Unless you need another reference to the array then charlies suggestion is correct.

Answer 7

Try calling your parameter int* arrAddr, not int arrAddr[]. Although when I think about it, the parameters for the main method are similar yet that works. So not sure about the explanation part.

Edit: Hm all the resources I can find on the internet say it should work. I'm not sure, I've always passed arrays as pointers myself so never had this snag before, so I'm very interested in the solution.