Create Dynamic Array of Strings in C

Question

I know there are multiple questions about creating dynamic arrays in C but they didn't really help me so let me ask in a different way.

My program needs to read in a variable number of command line arguments, each of variable length. This function takes the argv array passed into main and should return an array of char* containing only those arguments that are environment settings. (The program needs to replicate the UNIX env command.) Here is what I have now:

char** getEnvVariables(char* args[], int start) {
    int i = start;
    char** env_vars = (char **) malloc(1); 
    while (args[i]) {
        printf("size of env_vars: %d\n", sizeof(env_vars));
        if (isEnvironmentSetting(args[i])) {
            printf("arg: %s\n", args[i]);
            printf("size of arg: %d\n", sizeof(args[i]));
            printf("new size of env_vars: %d\n", (sizeof(env_vars) + sizeof(args[i])));

            env_vars = realloc(env_vars, (sizeof(env_vars) + sizeof(args[i])));
            memcpy(env_vars, args[i], sizeof(args[i]));
            i++;
        }
        else
            break;
    }

    return env_vars;
}

My idea was to create the array with malloc() and then use realloc() to allocate the space needed for each char* and memcpy() to add the new char* to the array. But the array isn't actually growing. At each iteration of the loop, it's size is 8 bytes. I'm still very new to C and the hands-on memory management so any help is appreciated.

sizeof(...) is a compile-time construct (except in the case of VLAs, which you don't need to worry about here), so sizeof(env_vars) is the size of the pointer env_vars, not the size of the block of allocated memory that env_vars points to. There is no portable way to determine the latter; you just have to keep track yourself of the value that you pass to malloc/realloc. — ruakh
– ruakh, Commented Sep 21, 2012 at 3:35
sizeof is also not strlen, so sizeof(args[i]) is also wrong. — Jim Balter
– Jim Balter, Commented Sep 21, 2012 at 4:12

Scooter · Accepted Answer · 2012-09-21 05:42:02Z

You cannot copy the input C-strings to the output, unless you want to return a (char pointer) that points to all the strings concatenated together. To return an array of (char pointer) (or (char pointer pointer) ) you need to either malloc a new string and store the address of that in env_vars, or have env_vars store the address of args[i]. Here is an implementation of both approaches:

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

bool isEnvironmentSetting(const char * string)
{
   (void) string;
   return true;
}

char** getEnvVariables1(char* args[], int start) {
    int i = start;
    int num_args = 0;
    char** env_vars = NULL;
    char *string = NULL;
    printf("size of env_vars: %ld\n", num_args * sizeof(env_vars));
    while (args[i]) {
        if (isEnvironmentSetting(args[i])) {
            printf("arg: %s\n", args[i]);
            printf("size of arg: %ld\n", strlen(args[i]));
            num_args++;
            printf("new size of env_vars: %ld\n", num_args * sizeof(env_vars));
            env_vars = realloc( env_vars, sizeof(env_vars) * num_args );
            string = malloc(strlen(args[i]) + 1);
            strcpy(string,args[i]);
            env_vars[num_args - 1] = string;
            i++;
        }
        else
            break;
    }
    env_vars = realloc( env_vars, sizeof(env_vars) * (num_args + 1) );
    env_vars[num_args] = NULL;    
    return env_vars;
}

char** getEnvVariables2(char* args[], int start) {
    int i = start;
    int num_args = 0;
    char** env_vars = NULL;
    printf("size of env_vars: %ld\n", num_args * sizeof(env_vars));
    while (args[i]) {
        if (isEnvironmentSetting(args[i])) {
            printf("arg: %s\n", args[i]);
            printf("size of arg: %ld\n", strlen(args[i]));
            num_args++;
            printf("new size of env_vars: %ld\n", num_args * sizeof(env_vars));
            env_vars = realloc( env_vars, sizeof(env_vars) * num_args );
            env_vars[num_args - 1] = args[i];
            i++;
        }
        else
            break;
    }
    env_vars = realloc( env_vars, sizeof(env_vars) * (num_args + 1) );
    env_vars[num_args] = NULL;
    return env_vars;
}

int main(int argc, char *argv[])
{
   (void) argc;

   char **envVars = getEnvVariables1(argv,1);
   int i = 0;
   while (envVars[i] != NULL)
   {
      printf("env var: %s\n",envVars[i]);
      i++;
   }

   envVars = getEnvVariables2(argv,1);
   i = 0;
   while (envVars[i] != NULL)
   {
      printf("env var: %s\n",envVars[i]);
      i++;
   }

}

Thanks, this worked great. I went with your second approach since it was more memory efficient to copy just the pointers instead of allocating memory for a copy of the string.

Danil Onishchenko · Accepted Answer · 2012-09-21 03:42:54Z

1

You can get environment variables specifing main function this way:

int main(int argc, char *argv[], char *envp[])

envp variable likes argv and contains environment variables.

answered Sep 21, 2012 at 3:42

Danil Onishchenko

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1 Comment

Jim Balter Over a year ago

That's not what the OP is asking for. Read the spec of the env command.

qrdl · Accepted Answer · 2012-09-21 06:21:06Z

1

sizeof(env_vars) will always return the size of pointer, not the size of allocated memory the env_vars points to. You can use sizeof for such purpose only for statically-allocated arrays.

So in your case you need to maintain a separate variable that holds the array size.

answered Sep 21, 2012 at 6:21

qrdl

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