is there a way for Realloc in functions correctly? [duplicate]

Question

this function get an memory allocated input(with size : 1 bytes) from user and put it in heap data. it works correctly in function but when this function ends and come back to main, input will be corrupted. it happens when re-allocate function change the address of strings in heap memory. can anyone help me to fix it?

int main() {
    char* input = (char*)malloc(sizeof(char));
    example(input);
    printf("%s", input);
}

   
void example(char* input) {
    int i = 0;
    char* a = "qweqeqweqweqweqweqwasdfsdfsdgasdgg";
    for (int i = 0; i < 20; i++) {
        input = (char*)realloc(input, sizeof(char) * i + 1);
        input[i] = a[i];
    }
}

example is only getting a copy of the pointer input. input in main will not change when you change input in example. If you run your program in a debugger, you can see that. Side note: if realloc fails the old pointer will still be valid, but when you write p = realloc(p,...) you lose the old value and thus have a memory leak. — Werner Henze
– Werner Henze, Commented Jan 30, 2021 at 19:43
The duplicate question has two good answers. The accepted answer gives one of two ways of solving the problem; the second answer outlines both ways of solving the problem (and is arguably the better answer, therefore). Basically, you either have the called function return the reallocated (or just allocated) pointer, or you pass a pointer to the pointer that will be (re)allocated. — Jonathan Leffler
– Jonathan Leffler, Commented Jan 30, 2021 at 20:01

dbush · Accepted Answer · 2021-01-30 21:25:10Z

2

The realloc function can return a pointer that was different from what was passed in. If that happens, the copy of input you have in the main function is invalid.

You can fix this by having example return the modified pointer and assigning to back to input in the main function. Also:

Don't cast the return value of realloc
If realloc fails the original pointer is still valid, so assign to a temporary and check the return value
You need to add a terminating null byte to the string and allocate additional space for it.

So your function would become this:

char *example(char* input) {
    int i = 0;

    char* a = "qweqeqweqweqweqweqwasdfsdfsdgasdgg";
    for (int i = 0; i < 20; i++) {
        char *tmp = realloc(input, i + 1 + 1);
        if (!tmp) {
            perror("relloc failed");
            free(input);
            exit(1);
        }
        input = tmp;
        input[i] = a[i];
    }
    intput[i] = 0;
    return input;
}

And you call it like this:

input = example(input);

You should also move the definition of example to before main, that way it can be called correctly.

edited Jan 30, 2021 at 21:25

answered Jan 30, 2021 at 19:43

dbush

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4 Comments

Werner Henze Over a year ago

Another solution would be to pass a pointer to input, so void example(char** input). As in the question, your code has a memory leak.

0___________ Over a year ago

OPs code has more issues

chux Over a year ago

Same issue with sizeof(char) * i + 1. Recommend sizeof(char) * (i + 1) if code wants to scale by the size.

chux Over a year ago

As OP is looking for a string, this answer is incomplete as it lack a null character assignment.

0___________ · Accepted Answer · 2021-01-30 21:22:40Z

1

The way you call realloc is wrong. You should use a temporary variable to save the pointer because if realloc fails you will loose the reference the original memory block.

char *example(char* input);

int main(void) 
{
    char *input = malloc(sizeof(*input));
    char *tmp;
    if((tmp == example(input))) input = tmp;
    printf("%s", input);
}

   
char *example(char* input) {
    int i = 0;
    char* a = "qweqeqweqweqweqweqwasdfsdfsdgasdgg";
    for (int i = 0; i < 20; i++) 
    {
        char *tmp = realloc(input, sizeof(*input) * (i + 2));
        if(tmp)
        {
            input = tmp;
            input[i] = a[i];
            input[i + 1] = 0;
        }
        else
        {
            /* handle allocation error */
        }
    }
    return tmp;
}

You can also use pointer to pointer but you need to save the original pointer to avoid potential memory leak:

int main(void) 
{
    char *input = malloc(sizeof(*input));
    char *tmp = input;

    example(&tmp);
    if(tmp) input = tmp;
    printf("%s", input);
}

   
void example(char** input) {
    int i = 0;
    char* a = "qweqeqweqweqweqweqwasdfsdfsdgasdgg";
    for (int i = 0; i < 20; i++) 
    {
        *input = realloc(*input, sizeof(*input) * (i + 2));
        if(*input)
        {
            *input = tmp;
            (*input)[i] = a[i];
            (*input)[i + 1] = 0;
        }
        else
        {
            /* handle allocation error */
        }
    }
}

Another problem in your code: you do not null terminate your string. You need to read warnings. You have more issues in your code - for example, you call function without prototype.

edited Jan 30, 2021 at 21:22

answered Jan 30, 2021 at 19:56

0___________

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2 Comments

Alireza Arbabi Over a year ago

can you help me to write the code with double pointer? your code doesn't work correctly and give runtime error in realloc

0___________ Over a year ago

@AlirezaArbabi copy paste error

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