Pointers, Arrays, Strings and Malloc in C

Question

I'm currently learning about strings, pointers and arrays in C. I tried to write a program where an array holds three pointers to string addresses. It all seems to work but the program behaves strangely.

Here's the code:

char** getUserDetails()
    {

    char* host = "localhost";

    char* username = "root";

    char* password = "mypassword";

    // create array for holding pointers to strings

    char *userDetailsHolder[3];

    userDetailsHolder[0] = malloc(sizeof(char)*strlen(host));   
    strcpy(userDetailsHolder[0], host);

    userDetailsHolder[1] = malloc(sizeof(char)*strlen(username));
    strcpy(userDetailsHolder[1], username);

        userDetailsHolder[2] = malloc(sizeof(char)*strlen(password));
        strcpy(userDetailsHolder[2], password);

        return userDetailsHolder;
    }

    int main()
    {

    char** userDetails = getUserDetails();

    printf("Host: %s\nUsername: %s\nPassword: %s\n", userDetails[0], userDetails[1], userDetails[2]);

    printf("Host: %s\nUsername: %s\nPassword: %s\n", userDetails[0], userDetails[1], userDetails[2]);

            return 0;
    }

Output: The output indicates that something went terribly wrong

Host: localhost
Username: root
Password: mypassword
Host: root
Username: localhost
Password: Host: %s
Username: %s
Password: %s

The first printf seems to work, but the second has the wrong data in it. What did I do wrong?

Blagovest Buyukliev · Accepted Answer · 2011-04-08 12:57:49Z

8

The problem is that you are returning a pointer to an array which resides on the stack. userDetailsHolder is allocated on the stack and is not available after the function returns.

You could use malloc once more to allocate the array itself, and then it will be available after the function returns.

edited Apr 8, 2011 at 12:57

answered Apr 8, 2011 at 12:51

Blagovest Buyukliev

43.7k14 gold badges96 silver badges135 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Frank Vilea Over a year ago

Thanks that makes sense. I have never done that though. I tried to add the following code:

Frank Vilea Over a year ago

char** arrayOnStack = malloc(sizeof(userDetailsHolder)); arrayOnStack = userDetailsHolder; return arrayOnStack;

Blagovest Buyukliev Over a year ago

The new code would be char **userDetailsHolder = malloc(sizeof(char *) * 3)

Frank Vilea Over a year ago

Hey thanks, works perfectly. Is there some kind of rule you use to keep track of all the de-referencing? As a beginner I find it quite confusing to know at which level I'm at..

Blagovest Buyukliev Over a year ago

It's quite simple when you get it... Every asterisk in the declaration of a pointer corresponds to one dereferencing. For example, if you have declared an int *****a, the expression **a will yield a pointer of type int ***.

|

Kannan Goundan · Accepted Answer · 2011-04-08 12:55:11Z

3

Also, remember to allocate strlen(s)+1 bytes for strings. C strings are terminated with the zero byte and you need to make sure there's space for it.

answered Apr 8, 2011 at 12:55

Kannan Goundan

5,3023 gold badges29 silver badges35 bronze badges

1 Comment

Frank Vilea Over a year ago

Thanks, I already added it to the code: userDetailsHolder[0] = malloc(sizeof(char)*(strlen(host)+1));

Collectives™ on Stack Overflow

Pointers, Arrays, Strings and Malloc in C

2 Answers 2

6 Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

6 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related