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Question

I'm in the process of teaching myself C and I'm mistified as to what's causing the following issue: when I create an array in a method and return it as a pointer to the calling function, none of the content is correct. I've boiled down this problem to the following example:

char * makeArr(void){
	char stuff[4];
	stuff[0]='a';
	stuff[1]='b';
	stuff[2]='c';
	stuff[3]='d';
	printf("location of stuff:%p\n",stuff);
	int i;
    
    for(i = 0; i < 4; i++){
	   printf("%c\n",stuff[i]);
    }
    return stuff;
}
int main(void){
	char* myarr;
	myarr = makeArr();
	int i;
	printf("\n");
	printf("location of myarr:%p\n", myarr);
	for(i = 0; i < 4; i++){
		printf("%c\n",myarr[i]);
	}
}

The output returns the following:

location of stuff:0028FF08
a
b
c
d
location of myarr:0028FF08
Ä
ÿ
(
(a null character)

So I've verified that the locations between the two values are the same, however the values differ. I imagine that I'm missing some critical C caveat; I could speculate it's something to do with an array decaying into a pointer or a problem with the variable's scope, but and any light that could be shed on this would be much appreciated.

Answer

Your array stuff is defined locally to the function makeArr. You should not expect it to survive past the life of that function.

char * makeArr(void){
  char stuff[4];

Instead, try this:

char * makeArr(void){
  char *stuff=(char*)calloc(4, sizeof(char));

This dynamically creates an array which will survive until you free() it.