Pass character array by value and return a new character array from the function?

Question

I apologise if I'm completely misunderstanding C++ at the moment, so my question might be quite simple to solve. I'm trying to pass a character array into a function by value, create a new array of the same size and fill it with certain elements, and return that array from the function. This is the code I have so far:

char *breedMutation(char genome []){ size_t genes = sizeof(genome); char * mutation = new char[genes]; for (size_t a = 0 ;a < genes; a++) { mutation[a] = 'b'; } return mutation; }

The for loop is what updates the new array; right now, it's just dummy code, but hopefully the idea of the function is clear. When I call this function in main, however, I get an error of initializer fails to determine size of ‘mutation’. This is the code I have in main:

int main() { char target [] = "Das weisse leid"; //dummy message char mutation [] = breedMutation(target); return 0; }

I need to learn more about pointers and character arrays, which I realise, but I'm trying to learn by example as well.

EDIT: This code, which I'm trying to modify for character arrays, is the basis for breedMutation.

int *f(size_t s){ int *ret=new int[s]; for (size_t a=0;a<s;a++) ret[a]=a; return ret; }

The way you're tackling this problem is very c-ish, so it's not really a c++ question. In c++, one would typically work with std::strings and/or specially designed classes rather than plain char-arrays. — leftaroundabout
– leftaroundabout, Commented Nov 26, 2011 at 23:09

AusCBloke · Accepted Answer · 2011-11-26 23:11:21Z

Your error is because you can't declare mutation as a char[] and assign it the value of the char* being returned by breedMutation. If you want to do that, mutation should be declared as a char* and then deleted once you're done with it to avoid memory leaks in a real application.

Your breedMutation function, apart from dynamically allocating an array and returning it, is nothing like f. f simply creates an array of size s and fills each index in the array incrementally starting at 0. breedMutation would just fill the array with 'b' if you didn't have a logic error.

That error is that sizeof(genome); will return the size of a char*, which is generally 4 or 8 bytes on a common machine. You'll need to pass the size in as f does since arrays are demoted to pointers when passed to a function. However, with that snippet I don't see why you'd need to pass a char genome[] at all.

Also, in C++ you're better off using a container such as an std::vector or even std::array as opposed to dynamically allocated arrays (ones where you use new to create them) so that you don't have to worry about freeing them or keeping track of their size. In this case, std::string would be a good idea since it looks like you're trying to work with strings.

If you explain what exactly you're trying to do it might help us tell you how to go about your problem.

Thanks for the help; as I said, the for loop just contains dummy code at the moment, which fills everything with the letter b. In the actual application, it uses an algorithm for potentially mutating each allele. std::string looks like my best gamble at the moment.

hmjd · Accepted Answer · 2011-11-26 22:57:49Z

The line:

size_t genes = sizeof(genome);

will return the sizeof(char*) and not the number of elements in the genome array. You will need to pass the number of elements to the breedMutation() function:

breedMutation(target, strlen(target));

or find some other way of providing that information to the function.

Hope that helps.

EDIT: assuming it is the number of the elements in genome that you actually want.

Loki Astari · Accepted Answer · 2011-11-26 23:08:28Z

Array are very limited.
Prefer to use std::vector (or std::string)

std::string breedMutation(std::string const& genome) { std::string mutation; return mutation; } int main() { std::string target = "Das weisse leid"; //dummy message std::string mutation = breedMutation(target); }

Godwin · Accepted Answer · 2011-11-26 22:52:05Z

Try replacing the second line of main() with:

char* mutation = breedMutation(target);

Also, don't forget to delete your mutation variable at the end.

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