Question

I’m trying to return an array of char*‘s to a function. I’ve simplified my code to a test case that clones a char array that instead of containing chars holds pointers to those chars.

/*
 * code.c
 */
#include <stdio.h>

char* makePointerCopy(char cIn[]);

int main() {
    char cTest[] = {'c', 't', 's', 't'};
    char* cPTest[] = makePointerCopy(cTest);
    printf("%p %c", cPTest, *cPTest);
    fflush(stdout);
    return 0;
}

char* makePointerCopy(char cIn[]) {
    char* cOut[sizeof(cIn)/sizeof(cIn[0])];
    int iCntr;

    for (iCntr = 0; iCntr < sizeof(cIn)/sizeof(cIn[0]); iCntr++)
        cOut[iCntr] = cIn + iCntr;

    return cOut;
}

A couple of warnings aside, this is what the compiler has to say about this code snippet:

invalid initializer (at char* cPTest[] = makePointerCopy(cTest);)

Why does this happen?

Answer 1

Because makePointerCopy returns a char*, not a char*[].

You should be able to change that line to:

char* cPTest = makePointerCopy(cTest);

More specifically, the reason that you get THAT error message, rather than something about types, is that array initializers are required to be compile-time constants.

From http://bytes.com/topic/c/answers/215573-invalid-initializer

Even if the declaration is not at file
scope, it would be illegal in both C90
and C99. C90 demands compile-time
constant initializers for automatic
and register arrays. And both C90 and
C99 require a character array to be
initialized with a) a string literal,
or b) a brace-enclosed initializer
list.

Still, the type mismatch is the actual problem here.