Thread: Make function return string...

Yes, except the string will be destroyed when the function returns. Not to mention the return type 'char' won't work for arrays.

What you want is to return an address to a static variable in the function (note that means you'll only be able to return one string at any one time).

Code:

char * nameOfFunction(void)
{
    static char str[] = "Hello world";
    return str;
}

Or a better technique,

Code:

void nameOfFunction(char * buff, size_t n)
{
    if(buff == NULL)
        return;
    memset(buff, 0, n);
    strncpy(buff, "Hello world", n - 1);
    return;
}

/* ... */

char str[256];
nameOfFunction(str, sizeof(str));
printf("%s\n", str);

What is a static variabel? and whats is void?

I think you need a book, tutorial, reference or something at this point.

Why isnt there & in front of str, cause it seems logical to me that you passes the address into the pointer buff, and not the data inside str.

When an array is passed as an argument, it is converted to a pointer to its first element, thus the address of its first element is passed.

And why do you need to fill buff up with 0?

To ensure that there is a null character at the end of the string.

And finally why isnt there a * in front of buff in:

buff is the string, *buff is the first character of the string.

Originally Posted by laserlight

buff is the string, *buff is the first character of the string.

So you mean that when i use buff in memset function for example, the compiler already looks at it as a pointer to the mem block, but when u uses the * it just points to the first section in the array?

So inside buff we have the memory address of str...right?

No, buff is a pointer that (initially) contains the address of the first character of str.

And dont i have to tell the compiler that buff is a pointer and it shut write to the memory address inside? ... and why not?

By declaring buff as a parameter with a type of char*, you have told the compiler that buff is a pointer.

So you mean that when i use buff in memset function for example, the compiler already looks at it as a pointer to the mem block, but when u uses the * it just points to the first section in the array?

When I take one glance at this:

Code:

void nameOfFunction(char * buff, size_t n)

I can tell you that buff is a pointer to a char. Since buff is a pointer to a char, *buff is a char.