Thread: Multi-dimensjonal (char) array

Originally Posted by cnewbie1

However, I'm not sure if a two-dimensional array is a pointer to a pointer, or a pointer with pointers to arrays?

mstr is an array of 5 arrays of 10 char. It is neither a pointer to a pointer nor a pointer with pointers to arrays. However, it can be converted to a pointer to an array of 10 char. This is still different from a pointer to a pointer, and certainly different from a pointer to a char.

For example:

Code:

void foo(char (*y)[10]);

int main(int argc, char *argv[])
{
    char x[5][10];
    foo(x);
    return 0;
}

void foo(char (*y)[10])
{
    y[0][0] = '\0';
}

It's a question of how you envision it...

If you have: char Strings[10][50]

You have 10 buffers (or lines) of 50 characters each.

Strings points to the first line first character Strings[0][0].

Strings[1] is actually a pointer to the second buffer, first character.

Strings[3][30] is the character at the 4th buffer, 31st position.

If it looks like I'm missing by one... I am. Don't forget that C arrays always index from 0, not 1.

Originally Posted by CommonTater

Strings points to the first line first character Strings[0][0].

No, it does not.

Strings is not a pointer to char, and also cannot be implicitly converted to a pointer to char, so cannot point at Strings[0][0].

If you want evidence of that, you will find that the test

Code:

if (Strings == &(Strings[0][0])) {}

will not even compile.

Originally Posted by cnewbie1

However, doing the same thing with a 2-dimensional array doesn't work:

I suggest that you make use of a typedef, e.g.,

Code:

#include <stdio.h>

typedef char string_type[10];

string_type *foo(string_type *y);

int main(void)
{
    string_type x[5];
    string_type *r = foo(x);
    printf("%c\n", r[0][0]);
    return 0;
}

string_type *foo(string_type *y)
{
    y[0][0] = 'C';
    return y;
}

Originally Posted by CommonTater

Strings[1] is actually a pointer to the second buffer, first character.

You're thinking of an array of pointers - in this case, the entire array is one contiguous block of data.

Code:

int main( void )
{
    const int rows = 13, cols = 2;
    char data[ rows ][ cols ];
    int ltr = 'a';
    for( int rdx = 0; rdx < rows; ++rdx )
    {
        for( int cdx = 0; cdx < cols; ++cdx )
        {
            data[ rdx ][ cdx ] = ltr++;
        }
    }
/*
    Out of bounds?
*/    
    data[ 0 ][ 'j' - 'a' ] = '!';
    for( int rdx = 0; rdx < rows; ++rdx )
    {
        for( int cdx = 0; cdx < cols; ++cdx )
        {
            putchar( data[ rdx ][ cdx ] );
        }
        putchar( '\n' );
    }
}

Originally Posted by cnewbie1

scout: That works with a void method, as mentioned further up. But with a char method, receiving and returning a multi-dimensional array, I cannot get it to work that way. Using a typedef, however, worked beautifully.

mstring2 takes a multidimensional array and returns the same multidimensional array address (the mstr2 becomes a single array though). just play with it and hopefully it'll help. . yeah the typedef solution is sweet.

Code:

#include <stdio.h>
#include <stdlib.h>

char *mstring2(char *);

int main(void)
{
    char mbuf[5][10];
    char *mstr2 = mstring2(*mbuf);  // takes a multidimensional array and returns multidimensional array (address)
   
            printf("%p\n", mstr2);               // check if equal
            printf("%p\n", &mbuf[0][0]);         // 
      
    printf("this should be 49 = %d\n",mbuf[4][9]);
    printf("this should be 1 = %d\n",mbuf[0][1]);
    char k = 'k';
    char p = 'p';
    *(mstr2+(4*10)+9) = 3;                      
    *(mstr2+(0*10)+1) = k;
    *(mstr2+(1*10)+1) = p;
    printf("this should be 3 = %d\n",mstr2[49]);
    printf("this should be 3 = %d\n",mbuf[4][9]);
    printf("this should be k = %c\n",mbuf[0][1]);
    printf("this should be p = %c\n",mbuf[1][1]);

    getchar();
    return 0;
}

char *mstring2(char *k) {
    
    for (int i=0;i<50; i++)
    {
        *(k+i) = i;
    }
	return k;
}