How to Return a Char Array from a Function

How do I return a char array from a function?

Best as an out parameter:

void testfunc(char* outStr){
  char str[10];
  for(int i=0; i < 10; ++i){
    outStr[i] = str[i];
  }
}

Called with

int main(){
  char myStr[10];
  testfunc(myStr);
  // myStr is now filled
}

How can I return a character array from a function in C?

You've got several options:

1) Allocate your array on the heap using malloc(), and return a pointer to it. You'll also need to keep track of the length yourself:

void give_me_some_chars(char **arr, size_t *arr_len)
{
    /* This function knows the array will be of length 2 */
    char *result = malloc(2);

    if (result) {
        result[0] = 'c';
        result[1] = 'a';
    }

    /* Set output parameters */
    *arr = result;
    *arr_len = 2;
}

void test(void)
{
    char *ar;
    size_t ar_len;
    int i;

    give_me_some_chars(&ar, &ar_len);

    if (ar) {
        printf("Array:\n");
        for (i=0; i<ar_len; i++) {
            printf(" [%d] = %c\n", i, ar[i]);
        }
        free(ar);
    }
}

2) Allocate space for the array on the stack of the caller, and let the called function populate it:

#define ARRAY_LEN(x)    (sizeof(x) / sizeof(x[0]))

/* Returns the number of items populated, or -1 if not enough space */
int give_me_some_chars(char *arr, int arr_len)
{
    if (arr_len < 2)
        return -1;

    arr[0] = 'c';
    arr[1] = 'a';

    return 2;
}

void test(void)
{
    char ar[2];
    int num_items;

    num_items = give_me_some_chars(ar, ARRAY_LEN(ar));

    printf("Array:\n");
    for (i=0; i<num_items; i++) {
        printf(" [%d] = %c\n", i, ar[i]);
    }
}

DO NOT TRY TO DO THIS

char* bad_bad_bad_bad(void)
{
    char result[2];      /* This is allocated on the stack of this function
                            and is no longer valid after this function returns */

    result[0] = 'c';
    result[1] = 'a';

    return result;    /* BAD! */
}

void test(void)
{
    char *arr = bad_bad_bad_bad();

    /* arr is an invalid pointer! */
}

How to return a char array created in function?

The simplest way would be to return a std::string, and if you needed access to the internal char array use std::string::c_str().

#include <iostream>
#include <string>

using namespace std;

string myGoodFunction(){
    char charArray[] = "Some string\n";
    return string(charArray);
}

int main(int argc, char** argv) {  
    cout << myGoodFunction();
    return 0;
}

If you need to return something other than a char array, remember that pointers can be used as iterators. This allows you to encapsulate an array in a vector or a similar structure:

vector<int> returnInts() {
    int someNums[] = { 1, 2, 3, 4 };
    return vector<int>(someNums, someNums + 4);
}

Return char[]/string from a function

Notice you're not dynamically allocating the variable, which pretty much means the data inside str, in your function, will be lost by the end of the function.

You should have:

char * createStr() {

    char char1= 'm';
    char char2= 'y';

    char *str = malloc(3);
    str[0] = char1;
    str[1] = char2;
    str[2] = '\0';

    return str;

}

Then, when you call the function, the type of the variable that will receive the data must match that of the function return. So, you should have:

char *returned_str = createStr();

It worths mentioning that the returned value must be freed to prevent memory leaks.

char *returned_str = createStr();

//doSomething
...

free(returned_str);

How can I return an char array from a function using malloc

Your array is local, we need to declare it dynamically. Please try below sample code.

char **arr = NULL;
arr = (char**)malloc(words*sizeof(char*))
/* then use it */
return arr;

How to return private char array from member function?

private or public is not matter. your problem is return char instead of char*.

I suggest you to use string instead of char array.

#include <iostream>
using namespace std;
class TEST
{
    char n[10];
public:
    char* getname()
    {
        cout<<"what's your name?:";
        cin.getline(n,10);
        return n;
    }
};
int main()
{   
    char* name;
    TEST obj;
    name = obj.getname();
    cout<<"Name :"<<name;
}

Returning char array in C function

 char *resultPtr = &result;
 return resultPtr;

You can't do this. result doesn't exist when function ends, you can't return result.

Modify your function like this:

void encryptDecrypt(char *text, char *result)
{
  for (int i = 0; i < LENGTH - 1; i++)
  { 
    result[i] = (char)(text[i] ^ KEY[i]);
  }
}

create an array at the caller site, and pass it as result. e.g.

char result[LENGTH] = {0}; 
encryptDecrypt(plainText, result);

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NOTE: Actually using %s for printing encrypted data is not best idea, because for example as a result of XOR you may get a null byte in between the text, which will be considered as a null terminator for your string, and printf won't show rest of the string. Consider something like this for printing cipher text.

Your format specifier for strlen is also wrong, use %zu insted of %i, or you will trigger undefined behaviour.

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