How to free memory of a char pointer inside a struct pointer

Question

I'm able to assign values with strdup and print the values with this: (*test + index)->email but I have no idea how to free the memory allocated to the variable email. I thought about freeing test+index but I guess this would cause a memory leak, right? Taking into account that the struct has allocated memory and each of the pointer inside it have memory allocated with strdup.

Edit:

The code is roughly like this:

struct random {
    char *email;
} Random;

void function(Random **struct) {
    char *temp = calloc(100, sizeof(char));
    *struct = calloc(5, sizeof(Random));
    for (int i = 0; i < 5; i++) {
        scanf("%s", temp);
        (*struct + i)->email = strdup(temp); //This works
    }
    free((*struct + 3)->email); //Gives segmentation fault
}

int main() {
    Random *struct;

    function(&struct)
}

Did you omit any code here? Because scanf into uninitialized temp is going to be undefined behavior. temp needs to be malloced, or made into an array, not just a char*. Also, you use *struct + i here, which makes no sense, given there is no i variable, nor a randomSize variable. — ShadowRanger
– ShadowRanger, Commented Sep 19, 2019 at 2:56
@ShadowRanger Yes, I tried to simplify the example as much as possible. I forgot the malloc part. temp is initialized with calloc. — Calpurnius Kazik
– Calpurnius Kazik, Commented Sep 19, 2019 at 2:58
why are you adding i ? you are going to have to put in enough code to repro the issue — Grady Player
– Grady Player, Commented Sep 19, 2019 at 2:59
You're going to need to provide an actual minimal reproducible example (make sure it compiles, runs, and has the same error as your full code), we can't just guess at what other mistakes you're making when you chop up the code this badly. — ShadowRanger
– ShadowRanger, Commented Sep 19, 2019 at 2:59
C11 Standard - 6.4.1 Keywords, your use of struct as a variable name is not allowed. — David C. Rankin
– David C. Rankin, Commented Sep 19, 2019 at 3:30

chqrlie · Accepted Answer · 2019-09-20 07:36:02Z

The posted code does not compile:

you cannot use struct as the name of a variable. struct is a keyword.
Random is a global variable, not a type.

It is idiomatic and much simpler in C to return the result instead of passing its address as an argument.

Following these remarks, and adding basic checks, the code should be simplified as:

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

typedef struct Random {
    char *email;
} Random;

Random *function(void) {
    char *temp = calloc(100, sizeof(char));
    if (temp == NULL)
        return NULL;
    Random *s = calloc(5, sizeof(Random));
    if (s != NULL) {
        for (int i = 0; i < 5; i++) {
            if (scanf("%99s", temp) != 1)
                *temp = '\0';
            (s + i)->email = strdup(temp); //This works
        }
        free((s + 3)->email); //Gives segmentation fault
    }
    free(temp);
    return s;
}

int main() {
    Random *s = function();
    // ...
}

This code, semantically equivalent to your posted fragment, does not have undefined behavior where you indicate, your actual code must be doing something else.

Jason Nguyen · Accepted Answer · 2019-12-29 04:44:56Z

1

If I'm not mistaken, wouldn't it this?

free(*(test+index)->email);
free(text+index);

edited Dec 29, 2019 at 4:44

answered Sep 19, 2019 at 2:21

Jason Nguyen

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2 Comments

Calpurnius Kazik Over a year ago

I get a segumentation fault when I try that. Maybe the edit might clarify things? I'm not sure.

Jonathan Leffler Over a year ago

It would be free((*(test + index))->email), I think. Or, better, free(test[index]->email); and free(text[index]);`.

user3629249 · Accepted Answer · 2019-09-20 04:04:31Z

0

please read/understand: precedence of C operators

Then note that the de-reference operator * has a lower precedence than the + operator,

Therefore, The posted code needs to be modified to use something like:

(*mystruct)+i = ...

etc. Otherwise, the + will be executed before the *

answered Sep 20, 2019 at 4:04

user3629249

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1 Comment

chqrlie Over a year ago

unary * evaluates before binary +, which is usually referred to as having higher precedence, not lower. *s + i is equivalent to (*s) + i, the parentheses do not change anything.

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