C、char** 奇怪的输出

Question

#include    <stdio.h>

char**      StringArray ( int n_size )
{
    char*   astr_allocate[ n_size ];
    char**  pstr_string_array = astr_allocate;
    int n_count;

    for ( n_count = 0; n_count < n_size; n_count++ )
        *(pstr_string_array + n_count) = " ";

    *(pstr_string_array + n_size) = "\0";

    return  pstr_string_array;
}


char*       String      ( int n_size )
{
    char    ach_allocate[ n_size ];
    char*   str_string = ach_allocate;
    int n_count;

    for ( n_count = 0; n_count < n_size; n_count++ )
        *(str_string + n_count) = ' ';

    *(str_string + n_size) = '\0';

    return  str_string;
}


void main ()
{

    int n_size      = 5;
    int n_count     ;
    char*   pch_string  = String ( n_size );
    char**  pstr_string = StringArray ( n_size );

    for ( n_count = 0; n_count < n_size; n_count++ )
        printf  ( "%c", *(pch_string + n_count) );

    for ( n_count = 0; n_count < n_size; n_count++ )
        printf  ( "%s", *(pstr_string + n_count) );

    printf  ( "\n\n" );
}

This produces wonderful outputs of "???"(Literal question marks) and random stuff like that. I am just trying to understand pointers and string type stuff more, if someone could help out that would be great thankyou!

additionally: Been writing and compiling this in a linux terminal and nano, if that changes anything

Answer 1

I would recommend that you study arrays, pointers and strings more.

• Since this program is a program for a regular hosted system (Linux), you must declare main as int main (void) .
• C does not have a pre-made string class, C strings are arrays of characters. As such they need to actually be allocated somewhere. You do try to allocate them as local variables inside a function, but you cannot return a pointer to a local variable, as that variable will cease to be valid once you leave the function. The memory which was previously reserved for your array may now be used for completely unrelated things at any time. See this .
• Arrays in C are zero-indexed. Meaning that given an array int arr[2]; the items have index [0] and [1]. Therefore you can't access it as arr[2] = ... or *(arr + 2) = ... , because that points past the end of the array.
• Given a pointer to character, which points at an array of characters (C string), you can't assign data to it by doing things like *(pstr_string_array + n_count) = " "; because that will only change where the pointer points at! You copy nothing. Instead, you must use strcpy(pointer, " ", 2) .
• A string literal " " is the same as an array of characters char arr [2] = {' ', '\\0'} . So code like "\\0" doesn't make any sense.
• Pointers to pointers are not arrays, nor do they point to arrays, nor are they compatible with arrays. There is a lot of incorrect books and tutorials out there teaching blatantly incorrect uses of pointer-to-pointer. There's actually very few cases where you need to use it, 2D arrays is not one of them, so just forget you ever saw pointer-to-pointer for now.
• Please note that *(pointer + n) is 100% equivalent to pointer[n] . The latter is easier to read, so use that form when possible.