Returning an array using C

You can't return arrays from functions in C. You also can't (shouldn't) do this:

char *returnArray(char array []){ char returned [10]; //methods to pull values from array, interpret them, and then create new array return &(returned[0]); //is this correct?} 

returned is created with automatic storage duration and references to it will become invalid once it leaves its declaring scope, i.e., when the function returns.

You will need to dynamically allocate the memory inside of the function or fill a preallocated buffer provided by the caller.

Option 1:

dynamically allocate the memory inside of the function (caller responsible for deallocating ret)

char *foo(int count) {    char *ret = malloc(count);    if(!ret)        return NULL;    for(int i = 0; i < count; ++i)         ret[i] = i;    return ret;}

Call it like so:

int main() {    char *p = foo(10);    if(p) {        // do stuff with p        free(p);    }    return 0;}

Option 2:

fill a preallocated buffer provided by the caller (caller allocates buf and passes to the function)

void foo(char *buf, int count) {    for(int i = 0; i < count; ++i)        buf[i] = i;}

And call it like so:

int main() {    char arr[10] = {0};    foo(arr, 10);    // No need to deallocate because we allocated     // arr with automatic storage duration.    // If we had dynamically allocated it    // (i.e. malloc or some variant) then we     // would need to call free(arr)}

C's treatment of arrays is very different from Java's, and you'll have to adjust your thinking accordingly. Arrays in C are not first-class objects (that is, an array expression does not retain it's "array-ness" in most contexts). In C, an expression of type "N-element array of T" will be implicitly converted ("decay") to an expression of type "pointer to T", except when the array expression is an operand of the sizeof or unary & operators, or if the array expression is a string literal being used to initialize another array in a declaration.

Among other things, this means that you cannot pass an array expression to a function and have it received as an array type; the function actually receives a pointer type:

void foo(char *a, size_t asize){  // do something with a}int bar(void){  char str[6] = "Hello";  foo(str, sizeof str);}

In the call to foo, the expression str is converted from type char [6] to char *, which is why the first parameter of foo is declared char *a instead of char a[6]. In sizeof str, since the array expression is an operand of the sizeof operator, it's not converted to a pointer type, so you get the number of bytes in the array (6).

If you're really interested, you can read Dennis Ritchie's The Development of the C Language to understand where this treatment comes from.

The upshot is that functions cannot return array types, which is fine since array expressions cannot be the target of an assignment, either.

The safest method is for the caller to define the array, and pass its address and size to the function that's supposed to write to it:

void returnArray(const char *srcArray, size_t srcSize, char *dstArray, char dstSize){  ...  dstArray[i] = some_value_derived_from(srcArray[i]);  ...}int main(void){  char src[] = "This is a test";  char dst[sizeof src];  ...  returnArray(src, sizeof src, dst, sizeof dst);  ...}

Another method is for the function to allocate the array dynamically and return the pointer and size:

char *returnArray(const char *srcArray, size_t srcSize, size_t *dstSize){  char *dstArray = malloc(srcSize);  if (dstArray)  {    *dstSize = srcSize;    ...  }  return dstArray;}int main(void){  char src[] = "This is a test";  char *dst;  size_t dstSize;  dst = returnArray(src, sizeof src, &dstSize);  ...  free(dst);  ...}

In this case, the caller is responsible for deallocating the array with the free library function.

Note that dst in the above code is a simple pointer to char, not a pointer to an array of char. C's pointer and array semantics are such that you can apply the subscript operator [] to either an expression of array type or pointer type; both src[i] and dst[i] will access the i'th element of the array (even though only src has array type).

You can declare a pointer to an N-element array of T and do something similar:

char (*returnArray(const char *srcArr, size_t srcSize))[SOME_SIZE]{  char (*dstArr)[SOME_SIZE] = malloc(sizeof *dstArr);  if (dstArr)  {    ...    (*dstArr)[i] = ...;    ...  }  return dstArr;}int main(void){  char src[] = "This is a test";  char (*dst)[SOME_SIZE];  ...  dst = returnArray(src, sizeof src);  ...  printf("%c", (*dst)[j]);  ...}

Several drawbacks with the above. First of all, older versions of C expect SOME_SIZE to be a compile-time constant, meaning that function will only ever work with one array size. Secondly, you have to dereference the pointer before applying the subscript, which clutters the code. Pointers to arrays work better when you're dealing with multi-dimensional arrays.

I am not saying that this is the best solution or a preferred solution to the given problem. However, it may be useful to remember that functions can return structs. Although functions cannot return arrays, arrays can be wrapped in structs and the function can return the struct thereby carrying the array with it. This works for fixed length arrays.

    #include <stdio.h>    #include <stdlib.h>    #include <string.h>    typedef    struct     {        char v[10];    } CHAR_ARRAY;    CHAR_ARRAY returnArray(CHAR_ARRAY array_in, int size)    {        CHAR_ARRAY returned;        /*        . . . methods to pull values from array, interpret them, and then create new array        */        for (int i = 0;  i < size; i++ )            returned.v[i] = array_in.v[i] + 1;        return returned; // Works!    }     int main(int argc, char * argv[])    {        CHAR_ARRAY array = {1,0,0,0,0,1,1};        char arrayCount = 7;        CHAR_ARRAY returnedArray = returnArray(array, arrayCount);         for (int i = 0; i < arrayCount; i++)            printf("%d, ", returnedArray.v[i]);  //is this correctly formatted?        getchar();        return 0;    }

I invite comments on the strengths and weaknesses of this technique. I have not bothered to do so.