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I'm new at C and I'm trying to do an exercise which asks to insert some strings and then store them. First it requests a multidimensional array where we have for every row of the array a string, and then as an array of pointers. Here's the code for the first part. I don't know how to store into an array some strings that are not already written.

For the second one I have no idea since I've never done exercises with pointers before.

#include <stdio.h>

int main(){
int n; //number of strings
int x; //number of characters per string

    printf("How many strings do you want to insert?");
    scanf("%d", &n);

    if ((n >= 1) && (n <= 20)){
        printf("How many characters per string?");
        scanf("%d", &x);
        char str[x];

            if (x <= 10){
                for(int i = 0; i < n; i++){
                    printf("Insert a string:");
                    scanf("%s", str);
                    for(int j = 0; j < x; j++){
                        char arr[j];
                        arr[j] = str[x];
                        printf("%s", arr);
                    }
        }
            }
            else {
                printf("Error:the number of characters must be < 10");
            }
    }
    else {
        printf("Error: the number must be < 20");
    }

return 0;
}
Schiele
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  • Sorry, but I'm afraid you'll need to go back to your favorite book/tutorial and read a bit about how to declare and use arrays. – Hulk Apr 29 '20 at 15:12
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    Yes, do some simpler things with pointers before in order to get familiar with the concept of pointers. – Jabberwocky Apr 29 '20 at 15:13
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    Hints: in C, `char str[5];` declares an array of 5 characters. `char str[x];` would be what is called a Variable Length Array, this is only supported by some versions of the C standard and probably not what you should use as a beginner (even if it might happen to work). You should be careful not to create an array with size zero, accessing that would be Undefined Behavior (you do this at `char arr[j];`). – Hulk Apr 29 '20 at 15:21
  • I agree. I noticed a couple of mistakes and I was writing an answer listing them. Then I came to the part in which you copy the string to your array... but there's not any 2-D array.. – Roberto Caboni Apr 29 '20 at 15:21

1 Answers1

2

... requests a multidimensional array where we have for every row of the array a string, and then as an array of pointers.

After getting the qualified number of strings, allocate an array of pointers to char.

if ((n >= 1) && (n <= 20)){
  char **string_list = calloc(n, sizeof *string_list);
  assert(string_list);  // or other error checking

(Notice no type in = calloc(n, sizeof *string_list);. Easier to code right, review and maintain.)

Read the strings in a working temp buffer. As "How many characters per string?" likely means the number of characters not including the null character, our str[] needs a +1 in size.

        // char str[x]; // too small
        char str[x+1];

Yet we know x <= 10 and can use a fixed buffer size and limit input length

      for(int i = 0; i < n; i++){
        char str[10+1];
        printf("Insert a string:");
        scanf("%10s", str);  // Notice the 10 - a width limit

        // TBD check if scanf() returned 1 and if str is longer than x

Now allocate a copy of the str

        string_list[j] = strdup(str);
        assert(string_list[j]);  // or other error checking
      }

Later, when done with string_list[], clean-up and free allocations.

for (int i=0; i<n; i++) {
  free(string_list[i]);
}
free(string_list);

What is weak about this:

It uses scanf() rather than fgets() and then parses, has minimal error checking, does not take in strings with spaces, does not handle over-long input, strdup() is not standard -yet, etc.

So the above is a baby step. Better code would handle the weak issues.

chux - Reinstate Monica
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