double pointers array to function initialization

Question

I'm trying to test the function below:

uint32_t atoi_len(uint8_t **buf, uint8_t len) {
    uint32_t value = 0;

do {
    if ((**buf < '0') || (**buf > '9')) return 0; 
    value *= 10;
    value += *((*buf)++) - '0';
} while (--len);

return value;
}

I have redefined uint32_t and uint8_t as int and char, just to make it work on my pc (Codeblocks).

What I'm trying to do is define a 2 dimensional array and pass this to this function as first argument (uint8_t **), but nothing works.

uint8_t buffer[10][10] = {"test", "test2", "stuff", "something"};

but i can't call:

atoi(buffer, len);

not even:

atoi_len(&buffer[0][0], len);
atoi_len(&buffer[0], len);

How can be initialized and accessed by a pointer? How do I pass the pointer to the function?

How can I build a 2 dimensional array, and pass the pointer to this to the function? I'm able to do it in one dimension easily:

char array[30] = "testof the test";
char* str = array; //or char* str = &array[0];

I would like to have this simple method to create a multidimensional array and assign a pointer to it, so that I can access array by pointer and even pass this pointer to a function.

Thank you very much, S

Thank you. I have problems declaring an array like char array[10][10] = {"df","dsf","det","ewr"}; but then I want a pointer char** buf; to point to array. char** buf = &array[0] OR &array doesn't work. Can you suggest something? thanks — rusty81, Feb 09 '19 at 23:22
How much of [this](https://stackoverflow.com/questions/8767166/passing-a-2d-array-to-a-c-function)' accepted answer also applies to C? — Phil M, Feb 09 '19 at 23:30
Are you trying to call `atoi` or `atoi_len`? `atoi` is a library function that takes only one argument. It looks like you intended to call `atoi_len` instead. — Tom Karzes, Feb 09 '19 at 23:41
Why do you want to pass a 2D-array? The code you show does only operate on the the 1st string. — alk, Feb 10 '19 at 09:23

score 1 · Answer 1 · answered Feb 10 '19 at 12:45

char array[][10] can be passed to a function as char (*buf)[10].
There seemed to be not much point of passing a two dimensional array to a function that only processed the first element, so this takes the argument to indicate the element to process.

#include <stdio.h>
#include <inttypes.h>

uint32_t atoi_len( char (*buf)[10], size_t each) {
    uint32_t value = 0;
    size_t len = 0;

    printf ( "buf[%zu] is %-12s", each, buf[each]);

    while ( buf[each][len]) {//not the zero terminator
        if ((buf[each][len] < '0') || (buf[each][len] > '9')) {
            return 0;
        }
        value *= 10;
        value += buf[each][len] - '0';
        len++;
    }

    return value;
}

int main( void) {
    char array[][10] = { "33", "-43,6,1", "8", "98", "12.3"};

    printf ( "result is %"PRIu32"\n", atoi_len ( array, 0));
    printf ( "result is %"PRIu32"\n", atoi_len ( array, 1));//has - and ,
    printf ( "result is %"PRIu32"\n", atoi_len ( array, 2));
    printf ( "result is %"PRIu32"\n", atoi_len ( array, 3));
    printf ( "result is %"PRIu32"\n", atoi_len ( array, 4));//has .
    return 0;
}

score 0 · Answer 2 · answered Feb 10 '19 at 00:50

I have solved for the moment with the following code:

Declaring a pointer to array

char *array[] = {"-43,6,-45,1","8","98"};

Passing this to the function in either of the two following ways

atoi_len(array, len);
atoi_len(&(*array), len);

I'm now starting to undestand, I've followed adresses in memory of every variable using Debug and Watch (CodeBlocks).

The last piece is to know if there is a simple/elegant way to do this. Can't use malloc because on embedded hardware i'm not allow to :).

Thanks to everyone.

double pointers array to function initialization

2 Answers2