Print array with pointer in C

Question

I have a pointer to an array of char and I need to print the entire array (not one after the other but the entire array). I have the error: zsh: segmentation fault. I don't understand why. My code take a text from stdin and put the chars in an array. If the first array char_array isn't big enough I put them in the second one called bigger_array.

int main() {
    int c;
    int index = 0;
    size_t size = 100;                        
    char *char_array = malloc(size * sizeof(char)); 
    if (char_array != 0) {
        while ((c = getchar()) != EOF && index < size) {
            char_array[index] = c;
            index++;
        }
        size_t newSize = size * 100;
        char *bigger_array = realloc(char_array, newSize * sizeof(char)); 
        if (bigger_array != 0) {
            bigger_array = char_array;
            while ((c = getchar()) != EOF && index < newSize) {
                bigger_array[index] = c;
                index++;
            }
            size = newSize;
        }
    }
    printf(*char_array);
    return 0;
}

You are using bigger_array even when the first array had enough elements for the input. — MikeCAT
– MikeCAT, Commented Sep 26, 2020 at 13:24
Trying to print what is pointed at by char_array when char_array == 0 is bad. — MikeCAT
– MikeCAT, Commented Sep 26, 2020 at 13:26
When printing out a array your best bet is to use a loop, but you need to look over your code, what is it you want to check with char_array != 0, if you reallocate memory to a new char pointer why do you then have bigger_array = char_array?. Also you are not taking a file from stdin — forsb
– forsb, Commented Sep 26, 2020 at 13:28

MikeCAT · Accepted Answer · 2020-09-26 13:23:31Z

2

bigger_array = char_array; should be char_array = bigger_array;. Otherwise, the pointer to newly allocated buffer is overwritten to the pointer to old buffer and bad thing will happen.
printf(*char_array); is bad because it is passing a integer (one character) where pointer is required. It should be printf("%.*s", index, char_array);. Note that the length to print is specified and you don't need to add terminating null-character thanks to that.

answered Sep 26, 2020 at 13:23

MikeCAT

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Comments

chqrlie · Accepted Answer · 2020-09-26 14:14:14Z

There are multiple issues in your code:

you do not include <stdio.h> nor <stdlib.h>
you should free char_array and store big_array to char_array so char_array points to the bigger allocated array.
printf(*char_array) has undefined behavior for multiple reasons. You can print the characters with fwrite(char_array, 1, index, stdout) or with printf("%.*s\n", (int)index, char_array).
there is no need for separate loops, just reallocate the array on demand.

Here is a modified version:

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

int main() {
    int c;
    size_t index = 0;
    size_t size = 100;
    char *char_array = malloc(size);
    if (char_array != NULL) {
        while ((c = getchar()) != EOF) {
            if (index == size) {
                size_t new_size = size + size / 2 + size / 8; /* increase by the golden ratio */
                char *new_array = realloc(char_array, new_size);
                if (new_array == NULL) {
                    fprintf(stderr, "out of memory\n");
                    free(char_array);
                    return 1;
                }
                size = new_size;
                char_array = new_array;
            }
            char_array[index++] = c;
        }
        printf("%.*s\n", (int)index, char_array);
    }
    return 0;
}

Halfa · Accepted Answer · 2020-09-30 13:40:29Z

0

Considere this as a comment ^^

I advise you to use flags for warning with your complier. for exemple for gcc flags -W -Wextra -Wshadow are good to use and prevent lost of time.

answered Sep 30, 2020 at 13:40

Halfa

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