Problems while dynamically allocating memory for a read input function

Question

I am having trouble with a function that should read a string from the user. I am always getting (null) as the output.

Is this even a "right" approach for that kind of problem?

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

int getString(char *input);

int main(void)
{
        char *arr = NULL;
        printf("please enter string: ");
        getString(arr);
        printf("%s", arr);
        return 0;
}

int getString(char *input)
{
        int i;
        char c;
        char *tmp;
        input = malloc(sizeof(char));
        for (i = 0; (c = getchar()) != EOF && c != '\n'; ++i) {
                tmp = realloc(input, (i + 2) * sizeof(char));
                if (tmp == NULL) {
                        free(input);
                        printf("allocation error");
                        return -1;
                }
                input = tmp;
                input[i] = c;
        }
        input[i] = '\0';
        return 0;
}

Marco Bonelli · Accepted Answer · 2020-03-31 19:41:06Z

1

If you want to dynamically allocate the string you need to pass a pointer to char*, not just a char *. This way, the function can modify the real char * pointer and the caller will see the result. In your current code, the input variable only exists inside the function and does not affect the variable used by the caller, therefore your arr stays unchanged (NULL).

Something like this:

int getString(char **input)
{
        int i;
        char c;
        char *tmp, *cur = NULL;

        // No initial malloc() needed here.
        // Let realloc() do the job passing NULL the first time.

        for (i = 0; (c = getchar()) != EOF && c != '\n'; ++i) {
                tmp = realloc(cur, (i + 2) * sizeof(char));
                if (tmp == NULL) {
                        free(cur);
                        printf("allocation error");
                        return -1;
                }
                cur = tmp;
                cur[i] = c;
        }

        cur[i] = '\0';
        *input = cur;
        return 0;
}

And then pass the parameter like this:

getString(&arr);

answered Mar 31, 2020 at 19:41

Marco Bonelli

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2 Comments

sneeed Over a year ago

Don't I have to put an free(cur); before the return 0; to prevent a memory leak?

Marco Bonelli Over a year ago

@sneeed absolutely not, you give the memory to the caller (*input = cur). If you free before returning everything becomes invalid.After the function return it's the caller's job to use the data and free it when it's done.

Eraklon · Accepted Answer · 2020-03-31 19:40:26Z

0

You should return the input pointer, since it is local to your function and get deallocated when the program leaves from the function, so in main the arr is still NULL.

int* getString(char *input);

int main(void)
{
        //...
        arr = getString(arr);
        //...
}

int* getString(char *input)
{
        //...
        return input;
}

answered Mar 31, 2020 at 19:40

Eraklon

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1 Comment

sneeed Over a year ago

Tanks for your explanation!

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