Array of dynamic length

Question

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

int main(void) {
    char *names = NULL;
    int capacity = 0;
    int size = 0;

    printf("Type 'end' if you want to stop inputting names\n");
    while (1) {
        char name[100];
        printf("Input:\n");
        fgets(name, sizeof(name), stdin);

        if (strncmp(name, "end", 3) == 0) {
            break;
        }

        if (size == capacity) {
            char *temp = realloc(names, sizeof(char) * (size + 1));

            if (!temp) {
                if (names) {
                    return 1;
                }
            }
            names = temp;
            capacity++;
        }

        names[size] = name;
        size++;
    }

    for (int i = 0; i < size; i++) {
        printf("OUTPUT :%c\n", names[i]);
    }

    if (names) {
        free(names);
    }
}

I am trying to create an array of dynamic length in C but I don't know what is wrong with my code? I think it is cause of how I take the user input and the problem occurs when the code names[size] = name is executed.

What if the user name is "endebbe"? :) Hint: the usual way to signal end of input is using the `EOF` signal. — pmg, Aug 16 '19 at 08:13
You are messed up between `char *` and `char`. Example `names[size] = name;` `names` is also `char *` and `name` is also `char *`. — kiran Biradar, Aug 16 '19 at 08:50
You should handle your dynamic array in functions ans not directly in `main`. — Jabberwocky, Aug 16 '19 at 08:59
Possible duplicate of [C dynamically growing array](https://stackoverflow.com/questions/3536153/c-dynamically-growing-array) — dandan78, Aug 17 '19 at 09:52

score 2 · Answer 1 · answered Aug 16 '19 at 10:07

You need to declare an array of pointers to strings (char**), not a pointer to a single character string (char*).

This means that when you want to add a new entry, not only do you need to make space for the new pointer in your array (type is char* not char) but you also have to separately allocate storage for the string itself and then set the pointer you just made space for to that allocated string.

score 0 · Answer 2 · answered Aug 17 '19 at 11:40

There are multiple problems in the code:

names should be defined as a pointer to an array of char *: char **names = NULL;
you must make copies of the strings read with fgets(), otherwise all entries in names will point to the same array name.
you actually do not need to distinguish capacity and size if you reallocate one extra slot at a time.
the format "OUTPUT :%c\n" is incorrect for a string.

Here is a corrected version:

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

int main(void) {
    char **names = NULL;
    int capacity = 0;
    int size = 0;

    printf("Type 'end' if you want to stop inputting names\n");
    for (;;) {
        char name[100];
        printf("Input:\n");
        if (!fgets(name, sizeof(name), stdin)) {
            break;
        }
        if (strncmp(name, "end", 3) == 0) {
            break;
        }
        if (size == capacity) {
            // reallocate array with geometric growth
            int new_capacity = capacity + (capacity / 2) + 4;
            char *temp = realloc(names, sizeof(*names) * new_capacity);
            if (!temp) {
                free(names);
                printf("out of memory\n");
                return 1;
            }
            names = temp;
            capacity = new_capacity;
        }
        name[strcspn(name, "\n")] = '\0';  // strip the trailing newline if any
        names[size++] = strdup(name);      // allocate a copy of the string
    }
    for (int i = 0; i < size; i++) {
        printf("OUTPUT: %s\n", names[i]);
    }
    free(names);
    return 0;
}

I tried to compile it but there is 1 error ":25:19: error: cannot initialize a variable of type 'char *' with an rvalue of type 'void *' char *temp = realloc(names, sizeof(*names) * new_capacity);" How do I solve it? — drazer, Aug 18 '19 at 14:26
@drazer: you should compile the code as **c**, not **c++**. the gcc option is `-x c`. Yo compile as c++, you need a cast: `char *temp = (char *)realloc(names, sizeof(*names) * new_capacity);` — chqrlie, Aug 18 '19 at 22:26

Array of dynamic length

2 Answers2