C Pointer issues with array

Question

I'm really new to programming with C and have encountered the following problem:

When reading about pointers at https://www.tutorialspoint.com/cprogramming/c_pointer_to_an_array.htm, They have the following example:

#include <stdio.h>

int main () {

   /* an array with 5 elements */
   double balance[5] = {1000.0, 2.0, 3.4, 17.0, 50.0};
   double *p;
   int i;

   p = balance;
 
   /* output each array element's value */
   printf( "Array values using pointer\n");
    
   for ( i = 0; i < 5; i++ ) {
      printf("*(p + %d) : %f\n",  i, *(p + i) );
   }

   printf( "Array values using balance as address\n");
    
   for ( i = 0; i < 5; i++ ) {
      printf("*(balance + %d) : %f\n",  i, *(balance + i) );
   }
 
   return 0;
}

This example works fine, but when I try to replicate it in my code with a similar situation I'm getting an error.

For example, I create and fill an array with values from a text file. This code works fine and outputs each string in the array as I would expect when finished

#include <stdio.h>
#include <string.h>

#define MAXCHAR 10000

int main() {
    
    FILE *fp;
    char* filename = "../MagicProg/Files/MagicProg_csv_ikoria.csv";
    char str[MAXCHAR];

    char c;
    
    char seps[] = "|";
    char *token;

    int total_line_count = 0;

    fp = fopen(filename, "r");
    
    if (fp == NULL)
    {
        printf("Could not open file %s", filename);
        return 1;
    }
    
    for (c = getc(fp); c != EOF; c = getc(fp)) 
    {
        if (c == '\n') 
        {
            total_line_count += 1; 
        }
    }   

    fclose(fp);
    fp = fopen(filename, "r");

    // declare the arrays
    char csv_card_names[total_line_count][30];
    
    int curr_line_index = 0;
    int line_token_count = 0;

    while (fgets(str, MAXCHAR, fp) != NULL)
    {
 
        line_token_count = 0;
        token = strtok( str, seps );
            
        while ( token != NULL )
        {
            if ( curr_line_index > 0 ) 
            {
                if ( line_token_count == 0 ) 
                {
                
                    strcpy(csv_card_names[curr_line_index - 1], token);
                }
            }
            
            line_token_count += 1;
            token = strtok( NULL, seps );
        }

        curr_line_index += 1;
    }
     
    for ( int i = 0; i < total_line_count; i++ ) {
        printf("LINE=\n");
        printf("%s\n", csv_card_names[i]);
        
    }
        
    fclose(fp);
    return 0;
    
    
}

However, In trying to understand arrays I'm trying to re-write the last block of the code where I output the array elements using pointers. From the example above, I try creating a pointer and assigning the array to it just before looping through total line count at the end

...

char *p[30];
p = csv_card_names;
 
for ( int i = 0; i < total_line_count; i++ ) {
    printf("LINE=\n");
    printf("%s\n", csv_card_names[i]);       
}

But Geany is giving me the error "assignment to expression with array type". I'm not sure how what I'm doing is different than the example, where they assign:

p = balance;

The array (i think) has values of strings equalling 30 chars, so I'm trying to create a point capable of referencing those but am getting this error. any help is appreciated thanks

`char *p[30]` -- is an *array-of-pointers* `char [30]` (e.g. 30 pointers to char) While `char (*p)[30]` -- is a *pointer-to-array* of `char[30]` (e.g. a pointer to a character array of 30 chars) As to why the parens are needed, see [C Operator Precedence](https://en.cppreference.com/w/c/language/operator_precedence) — David C. Rankin, Jun 28 '20 at 23:54
A few links that provide basic discussions of pointers may help. [Difference between char *pp and (char*) p?](https://stackoverflow.com/a/60519053/3422102) and [Pointer to pointer of structs indexing out of bounds(?)...](https://stackoverflow.com/a/60639540/3422102) — David C. Rankin, Jun 29 '20 at 00:00
Tip: Instead of `*(x + y)` just use `x[y]`. Not only is it less typing but it makes way more sense to anyone familiar with C. Remember you can do `x[y][z]` and even deeper nested structures, which just will not fly as computing those offsets is really not easy. — tadman, Jun 29 '20 at 00:55

score 3 · Answer 1 · answered Jun 28 '20 at 23:43

3

char *p[30] is not a pointer - it is an array of pointers. As any other array it cannot be assigned.

you need to declare it as pointer to array of 30 chars.

char (*p)[30] = ....

answered Jun 28 '20 at 23:43

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It's worth noting that `char*` is often better than `char (*p)[30]` because the length is largely irrelevant when dealing with pointers and needs to be specified in a more concrete way, like as an additional function argument. – tadman Jun 29 '20 at 00:39

C Pointer issues with array

1 Answers1