Problem with returning an array to a function in C

Question

My problem is that I can't return the processed string to the function. The function's job is that it must accept a string, change its letters, and return the processed string to the function.


char *dna_strand(const char *dna)
{
    int a;
   for (a = 1; *(dna+a) != '\0' ; a++) {
}
 char array[a];
    for (int i = 0; i < a; ++i) {
        if ('A' == *(dna + i)) {
            array[i] = 'T';
        } else if ('T' == *(dna + i)){ 
            array[i] = 'A';
        } else{
            array[i] = *(dna + i);
        }
    }
      return array;
}

Error : [1]: https://i.sstatic.net/uwvCh.png

Please post error message as text instead of image. Format the code so it's not misleading. Use strlen() + 1 instead of the first loop. — user9706
– user9706, Commented Apr 25, 2022 at 9:20
VLA is an C99 extension and shall be avoided. Better use alloca instead of array[i] or in your case, malloc then return the allocated string, but then you must free it. — Ilian Zapryanov
– Ilian Zapryanov, Commented Apr 25, 2022 at 9:23
@IlianZapryanov alloca would not work as op wants to return the value. — user9706
– user9706, Commented Apr 25, 2022 at 9:24
you are promising a string but an array is not null terminated — Özgür Güzeldereli
– Özgür Güzeldereli, Commented Apr 25, 2022 at 9:32
The function doesn't return an array. It returns a pointer to an array. Then the array is destroyed. Then the function which calls this function tries to use the array. — Stack Exchange Broke The Law
– Stack Exchange Broke The Law, Commented Apr 25, 2022 at 10:14

score 2 · Accepted Answer · 2022-04-25 09:50:00Z

2

The local variable char array[a]; is out of scope for caller so you cannot use that to return a value. Instead allocate memory on the heap with malloc() or as herestrdup():

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

char *dna_strand(const char *dna) {
    char *array = strdup(dna);
    if (!array) return array;
    for (int i = 0; array[i]; i++) {
        if (dna[i] == 'A') array[i] = 'T';
        else if (dna[i] == 'T') array[i] = 'A';
    }
    return array;
}

int main() {
    char *s = dna_strand("ATTCG");
    printf("%s\n", s);
    free(s);
    return 0;
}

If you want to be fancy use strpbrk() to find the next letter we need to map instead of stepping one letter at a time:

char *dna_strand(const char *dna) {
    char *array = strdup(dna);
    for (char *p = array; p; p = strpbrk(p, "AT")) {
        if (*p == 'A') *p++ = 'T';
        else *p++ = 'A';
    }
    return array;
}

edited Apr 25, 2022 at 9:50

answered Apr 25, 2022 at 9:31

user9706

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1 Comment

Eric Postpischil Over a year ago

Scope is not the applicable concept here. Objects may be used in code outside the scope of their identifiers. The appropriate concepts are storage duration and lifetime; objects may not be used after their lifetimes end.

Vlad from Moscow · Accepted Answer · 2022-04-25 09:27:23Z

1

You may not return a pointer to a local array with automatic storage duration because after exiting the function the array will not be alive and the returned pointer will be invalid.

You need to allocate an array dynamically.

Pay attention to that this loop

for (a = 1; *(dna+a) != '\0' ; a++) {

can invoke undefined behavior when the user will pass to the function an empty string.

You should write at least like

int a = 0;
while ( *(dna+a) != '\0' ) ++a;
++a;

or

int a = 0;
do ; while ( *( dna + a++ ) != '\0' );

After that you can allocate dynamically a character array

char *array = malloc( a );

edited Apr 25, 2022 at 9:27

answered Apr 25, 2022 at 9:21

Vlad from Moscow

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Comments

Darth-CodeX · Accepted Answer · 2022-04-25 10:17:17Z

0

You cannot return a pointer to a local variable's address, because it will be destroyed when it will goes out from the function's scope.

Since, a is not a constant value, you cannot initialize array as static. Hence, the only option is left is using heap-allocation.

a can be initialize like this:

char *array = calloc(a + 1, sizeof(char));

Note: calloc() is defined in stdlib.h header file.

edited Apr 25, 2022 at 10:17

answered Apr 25, 2022 at 9:26

Darth-CodeX

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6 Comments

user9706 Over a year ago

sizeof(char) is defined as 1

Jabberwocky Over a year ago

Wording is not accurate: You cannot return a local variable -> You cannot return a pointer to a local variable

Eric Postpischil Over a year ago

@AllanWind: sizeof (char) is indeed 1 by definition although it is not defined to be 1. (The fact that sizeof (char) evaluates to 1 is a logical conclusion of the definitions of sizeof and char, but it is not the explicit definition of sizeof (char).) However, it is not good advice to suggest it not be used, as your comment perhaps implies. If char is explicitly called out where the code relies in it, this helps both to communicate to readers what the code is doing and to make it easier to update for future changes…

Eric Postpischil Over a year ago

… For example, if somebody wishes to migrate character-handling code to wide-character-handling code, the fact the original other explicitly used char in sizeof and other places makes it easier for the migrator to find them and replace them with wchar_t.

user9706 Over a year ago

@EricPostpischil if you make such a change, you need to port your code. As op used the type the would have to change sizeof(char) to sizeof(new_type). To your point, they should use the variable name instead (*array).

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