can't access some elements in malloc generated 2D array

Question

I'm trying to make a 2D array in C with malloc for "hash map" project but as usual. something happend !

basically I have an issue of accessing the 3rd element of the 1st array arr[0][3], it's super weird (at least to me) ,so let me show you the code

1- create 2D array:

int main(void){
    char*** table;
    table = malloc(sizeof(char*) * 10);
    //create array inside table
    table[0] = malloc(sizeof(char) * 20);
    
    return 0;
}

2- assign some strings to 1st array table[0]:

    table[0][1] = "test1";
    table[0][2] = "test2";
    table[0][3] = "test3";
    table[0][4] = "test4";
    ...
    table[0][n] = "testn";

3- now the magic happens:

   // 10 => n >= 0;
   printf("%s", table[0][n]);

returns -> malloc(): corrupted top size Aborted (core dumped)

in that moment I tried everything a noob would do , so somehow I figured out that the 3rd string test3 is the problem. so if I remove the table[0][3] line, all works great !!

    table[0][1] = "test1";
    table[0][2] = "test2";
    //table[0][3] = "test3";
    table[0][4] = "test4";
    ...
    table[0][n] = "testn";
    
    // 10 => n >= 0;
    printf("%s", table[0][n]);

returns => "testn";

EDIT for Vlad From Moscow that works:

   for(int i=0; i<10; i++{
       table[0][n] = "test";
       //here works
       printf("%s", table[0][n];
   }
   //here also works fine
   printf("%s", table[0][3];

If char*** table; is correct, and you want an array of arrays of strings (a "3d" array), then your allocations are wrong. For example malloc(sizeof(char) * 20) will allocate enough space for 20 characters, not pointers to char. It's similar to char array[20]. You make the same problem with the other allocation as well, but there it works because a pointer is a pointer, no matter if it's char * or char **. — Some programmer dude
– Some programmer dude, Commented Oct 11, 2022 at 18:50
@Someprogrammerdude that's exactly what I want, like in python 2d_arr=[["test"], ["test1","test2"]] — at_Root
– at_Root, Commented Oct 11, 2022 at 18:55

Vlad from Moscow · Accepted Answer · 2022-10-11 19:09:53Z

2

You declared the pointer table like

char*** table;

So the expression table[0] used in this statement

table[0] = malloc(sizeof(char) * 20);

has the type char ** and you allocated memory for 20 characters.

If sizeof( char * ) is equal to 8 then the allocated memory can accommodate only two pointers of the type char *. If sizeof( char * ) is equal to 4 then the allocated memory can accommodate only 5 pointers of the type char *.

You need to do something like the following

char*** table;
table = malloc(sizeof(char**) * 10);
//create array inside table
for ( size_t i = 0; i < 10; i++ )
{
    table[i] = malloc(sizeof(char *) * 20);
}

These memory allocations will simulate a two dimensional array of the type char * table[10][20].

On the other hand, you could at once allocate memory for a two-dimensional array like

char * ( *table )[20] = malloc( sizeof( char *[10][20] ) );

edited Oct 11, 2022 at 19:09

answered Oct 11, 2022 at 18:54

Vlad from Moscow

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9 Comments

at_Root Over a year ago

yup it worked but why? why in my code only the 3rd string is returning malloc(): corrupted top size Aborted (core dumped) ,that's interesting !

Vlad from Moscow Over a year ago

@at_Root In which case does the error occur? In your code in the question?

at_Root Over a year ago

3 when I call printf to any of the allocated strings, directly or even by storing it in variable then print it or even store the len of the string the try to printf it

Vlad from Moscow Over a year ago

@at_Root If you mean your code in question then as I wrote the allocated memory can accommodate only two pointers of the type char * if sizeof( char * ) is equal to 8.

Vlad from Moscow Over a year ago

@at_Root Again the new code has undefined behavior. Undefined behavior means that anything can occur.

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