How do I return a char array from a function?

I've tried the f开发者_JAVA技巧ollowing:

char[10] testfunc()
{
    char[10] str;

    return str;
}

---

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
}

---

You have to realize that char[10] is similar to a char* (see comment by @DarkDust). You are in fact returning a pointer. Now the pointer points to a variable (str) which is destroyed as soon as you exit the function, so the pointer points to... nothing!

Usually in C, you explicitly allocate memory in this case, which won't be destroyed when the function ends:

char* testfunc()
{
    char* str = malloc(10 * sizeof(char));
    return str;
}

Be aware though! The memory pointed at by str is now never destroyed. If you don't take care of this, you get something that is known as a 'memory leak'. Be sure to free() the memory after you are done with it:

foo = testfunc();
// Do something with your foo
free(foo); 

---

A char array is returned by char*, but the function you wrote does not work because you are returning an automatic variable that disappears when the function exits.

Use something like this:

char *testfunc() {
    char* arr = malloc(100);
    strcpy(arr,"xxxx");
    return arr;
}

This is of course if you are returning an array in the C sense, not an std:: or boost:: or something else.

As noted in the comment section: remember to free the memory from the caller.

---

As you're using C++ you could use std::string.

---

With Boost:

boost::array<char, 10> testfunc()
{
    boost::array<char, 10> str;

    return str;
}

A normal char[10] (or any other array) can't be returned from a function.

---

When you create local variables inside a function that are created on the stack, they most likely get overwritten in memory when exiting the function.

So code like this in most C++ implementations will not work:

char[] populateChar()
{
    char* ch = "wonet return me";
    return ch;
}

A fix is to create the variable that want to be populated outside the function or where you want to use it, and then pass it as a parameter and manipulate the function, example:

void populateChar(char* ch){
    strcpy(ch, "fill me, Will. This will stay", size); // This will work as long as it won't overflow it.
}

int main(){
    char ch[100]; // Reserve memory on the stack outside the function
    populateChar(ch); // Populate the array
}

A C++11 solution using std::move(ch) to cast lvalues to rvalues:

void populateChar(char* && fillme){
    fillme = new char[20];
    strcpy(fillme, "this worked for me");
}

int main(){
    char* ch;
    populateChar(std::move(ch));
    return 0;
}

Or this option in C++11:

char* populateChar(){
    char* ch = "test char";
    // Will change from lvalue to r value
    return std::move(ch);
}

int main(){
    char* ch = populateChar();
    return 0;
}

---

With c++17 you can use next code:

auto testfunc() {
    union {
        char str[14];
    } res = { .str = "Hello, World!" };
    return res;
}

and then use you string as

const auto str = testfunc();
std::cout << str.str;