Thread: returning a string array from a function

Hello,

I have a short example of a C library that I have coded below along with the header. I am calling this function from a C++ client app.

Some of the warning I get are as follows. This normally gives me a segmentation stack run time error. So I guess some about memory not been allocated being assigned to null address.

warning: return from incompatible pointer type
warning: function returns address of local variable

I have tried many different ways to do this. As you can see from my commenting out pieces of code.

However, I can return pointer without any problems. However, my code needs to return a string array. i.e. char msgTest[] = "Hello how are you";

Can any one offer any advice on this?

Many thanks for any suggestions.

Code:

//C library code
//Header cHeader.h
#ifdef __cplusplus 
extern "C"
{
#endif
	char* returnMessage(char *message);
#ifdef __cplusplus
}
#endif

//Implementation cDriver.c
char* returnMessage(char *message)
{
	printf("Message: %s\n", message);
	
	char *msg = NULL;
// 	//msg = message;
// 	printf("msg = message: %s\n", msg);
// 	//strcpy(msg, message);
// 	printf("strcpy(msg, message): %s\n", msg);
	
	char msgTest[] = "Hello how are you";
	strcpy(msg, msgTest);
	return &msgTest;
}

And the C++ code for calling the above function.

Code:

int main(int argc, char** argv)
{
	char *ptrTest = NULL;
	char test[] = "Hello";
	ptrTest = test;
	char *ptrRtn = NULL;

	ptrRtn = returnMessage(ptrTest);
	cout << "PtrRtn: " << ptrRtn << endl;

        return 0;
}

Just remove the &:

Code:

return msgTest;

Hello,

Code:

return msgTest;

I get the following warning: function returns address of local variable.

I am not sure, but as this is a local variable the memory way not been available when the function returns? Just a guess.

Thanks,

Hello,

thanks for the reply,

i have used malloc, which worked. Howerver, At which point do I free the memory? As I can't do it after I have returned, or after I have copied the contents?

Code:

char* returnMessage(char *message)
{
	printf("Message: %s\n", message);
	
	char *msg = (char *) malloc(sizeof(char));
	
	char msgTest[] = "Hello how are you";
	
	strncpy(msg, msgTest, sizeof(msgTest));
	
        //when to free the memory that has been allocated.
	return msg;
}

I have noticed I get a stack dump.

Many thanks,

Hello,

Thanks for your help. I have used the code below:

Code:

char* returnMessage(char *message)
{
	printf("Message: %s\n", message); //Just testing the correct message is passed in, that is all.
	
	char msgTest[] = "Hello how are you";

	char *msg = malloc(strlen(msgTest)); //Allocated the length of the string.

	strncpy(msg, msgTest, strlen(msgTest)); //Used strlen instead of sizeof
	
	return msg;
}

Is the reason I am using strncpy in a bad way, because I previously used sizeof?

Just a final question about freeing memory that has been allocated. As the function returnMessage allocated memory, how can I free that in the calling function, as it will not have scope?

Many thanks,

Best to do it like this:

Code:

char* ReturnMessage(char* out, size_t outSize)
{
  char msgTest[] = "Hello how are you";
  strncpy(out, msgTest, outSize);
  return out;
}

int main(int argc, char* argv[])
{
  char message[1024];
  ReturnMessage(message, sizeof(message));
  return 0;
}

Hello,

3) Sizeof evaluates the size of a type, but won't work on a pointer because the size where it points to is unknown. Why? I'll let you think on that.

My best answer, as the pointer will only point to the first character in an array.

When freeing memory I have freed the memory in the calling function. As below.

Code:

int main(int argc, char** argv)
{
	char *ptrTest = NULL;
	char test[] = "Hello";
	ptrTest = test;
	char *ptrRtn = NULL;

	ptrRtn = returnMessage(ptrTest);
	cout << "PtrRtn: " << ptrRtn << endl;
        
        free(ptrRtn); //Freeing the memory that was allocated in the returnMessage function.
        return 0;
}

Any corrections please advise,

Thanks,

sizeof does not evaluate the expression, so it doesn't matter what pointers point to in a sizeof expression.