Return an array of strings from a function

Question

I need to return a char** but when I try to do this, the compiler tells me that I want to return the address of a local variable. How can I do that? I know that I should allocate space for this variable but how? Here is my code, but the second printf doesn't appear and the function returns nothing:

char** parse_cmd(const char* cmdline) {
char** arguments = (char**)malloc(100);
int i;
int j=0, k=0;
printf("%s\n", cmdline);

for(i=0; i<100; i++) {
    arguments[i] = malloc(100);
}

for(i = 0; i < strlen(cmdline); i ++) {
    if(cmdline[i] != ' ') {
        arguments[j][k] = cmdline[i];
        k++;
    } else {
        arguments[j][k] = '\0';
        j++;
        k = 0;
    }
}

printf("%s\n", arguments[1]);

return arguments;
}

Casting the return value of malloc in C is considered bad practice by many (like me) for a variety of reasons. — Chris Lutz
– Chris Lutz, Commented May 30, 2011 at 5:11
Don't call strlen() each time around the loop; it gets expensive. However, that's not the main problem here... — Jonathan Leffler
– Jonathan Leffler, Commented May 30, 2011 at 5:53
What happens if you get 3 spaces in a row? How do you mark the end of the returned argument list? I see no null pointer, and no count. — Jonathan Leffler
– Jonathan Leffler, Commented May 30, 2011 at 6:48

Dov Grobgeld · Accepted Answer · 2011-05-30 05:03:51Z

4

You need to do multiple allocations. The first for the char** and then for each of the char*. E.g. something like

  char **args = (char**)malloc(100);
  int i;
  for (i=0; i<100; i++) 
    args[i] = malloc(100);

  // Rest of program

  return args;

answered May 30, 2011 at 5:03

Dov Grobgeld

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3 Comments

nyanev Over a year ago

Now works, but if my cmdline is / test /' arguments[0]` is /*** * is strange symbol.

sickgemini Over a year ago

@n_yanev - This is because you haven't terminated your string for each argument. In the else clause, you need to add a '\0' to the end before you iterate j. Like arguments[j][k] = '\0'

nyanev Over a year ago

You can see my edited code in my first post. But I have the same error.

Jonathan Leffler · Accepted Answer · 2011-05-30 15:08:15Z

2

Here's the code I assembled - and tested. It uses dynamic memory allocation for both the argv argument list and for each argument as it is assembled. The function release_cmd() releases the allocated space. The function cleanup() is internal and releases allocated space on a failure, before returning a null double-pointer. This simplifies the error handling. There's a minimal test harness in the prompt() function and main(). I haven't run in under valgrind, but the MacOS X implementation of malloc() quite often spots problems so I'm moderately confident there's no gross memory abuse - but there could be an off-by-one leak. I haven't tested the cleanup() code by faking an allocation failure.

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

void release_cmd(char **argv)
{
    for (size_t i = 0; argv[i] != 0; i++)
        free(argv[i]);
    free(argv);
}

static char **cleanup(size_t argc, char **argv)
{
    argv[argc] = 0;
    release_cmd(argv);
    return 0;
}

char **parse_cmd(const char* cmdline)
{
    size_t argc = 2;
    char **argv = malloc(argc * sizeof(char *));

    if (argv == 0)
        return 0;

    size_t j = 0;  // Index into argv
    size_t len = strlen(cmdline);

    for (size_t i = 0; i < len; i++)
    {
        while (isspace(cmdline[i]))
            i++;
        if (cmdline[i] == '\0')
            break;
        if (j > argc - 2)
        {
            size_t newc = (argc * 2);
            char **newv = realloc(argv, newc * sizeof(char *));
            if (newv == 0)
                return cleanup(argc, argv);
            argv = newv;
            argc = newc;
        }
        size_t argl = 2;    // Length of argument string
        argv[j] = malloc(argl);
        size_t k = 0;       // Index into argv[j]
        while (cmdline[i] != '\0' && !isspace(cmdline[i]))
        {
            if (k > argl - 2)
            {
                size_t newl = argl * 2;
                char  *news = realloc(argv[j], newl);
                if (news == 0)
                    return cleanup(argc, argv);
                argv[j] = news;
                argl    = newl;
            }
            argv[j][k++] = cmdline[i++];
        }
        argv[j][k] = '\0';
        argv[j] = realloc(argv[j], k+1);    // Shrink to fit!
        j++;
    }
    argv[j] = 0;
    argv = realloc(argv, (j+1)*sizeof(*argv));  // Shrink to fit!

    return argv;
}

static int prompt(const char *prompt, char *buffer, size_t bufsiz)
{
    printf("%s", prompt);
    return (fgets(buffer, bufsiz, stdin) != 0);
}

int main(void)
{
    char line[1024];

    while (prompt("cmd? ", line, sizeof(line)) != 0)
    {
        char **argv = parse_cmd(line);
        char **args = argv;
        while (*args)
            puts(*args++);
        release_cmd(argv);
    }
    putchar('\n');
    return 0;
}

edited May 30, 2011 at 15:08

answered May 30, 2011 at 6:45

Jonathan Leffler

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2 Comments

Jonathan Leffler Over a year ago

The bits I'm not quite happy with are the outer for loop, since the inner code goes tampering with the outer loop index i, and it might be cleaner to package the two realloc() chunks into separate functions. The loop could be replaced with just while (1) (you have to declare i and do not need len any more). I've now checked with valgrind and have fixed the argv = realloc(argv, (j+1)*sizeof(*argv)); in the code.

Jonathan Leffler Over a year ago

(Just for the record, the buggy code was argv = realloc(argv, j+1);, created by careless analogy with the similar realloc() for over-long character strings. The 'shrink to fit' realloc() operations may save space if the doubling algorithm allocated 128 bytes, say, as the string grew longer than 64 bytes.

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