In C89, I can't seem to make a character array from an existing one

Question

So my code breaks down at buffer[i] = envp[i].

I want to create a char**buffer of the environment variables, and make them lower case (didn't add the putc(tolower()) yet loop). But when I just try to make buffer[i] = envp[i], the compiler returns me this error:

error: assignment makes integer from pointer without a cast [-Wint-conversion] buffer[i] = envp[i]; ^

warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=] printf("%s\n", buffer[i]); ~^ ~~~~~~~~~ %d

int main(int argc, char *argv[], char * envp[])
{
    int i = 0;

    char *buffer = NULL;

    while(envp[i])
    {
        i++;
    }

    buffer = (char *)malloc(i * 8);


    for (i = 0; envp[i] != 0 ; i++)
    {
        buffer[i] = envp[i];
        printf("%s\n", envp[i]);
        printf("%s\n", buffer[i]);
    }

    return 0;

}

Please help, I've been breaking my head here :( . Thanks so much!!!

Answer 1

buffer is not the correct type to hold an array of strings. It should be defined as:

char **buffer;

Also, you should malloc as follows:

buffer = malloc(i * sizeof(*buffer));

Use sizeof(*buffer) explicitly instead of the magic number 8, as a pointer is not guaranteed to be 8 bytes. This is also prefered to sizeof(char *) as it does not depend on the type of buffer Also, don't cast the return value of malloc.

Since env is terminated by a NULL pointer, you'll want to do the same with buffer. You should also copy the strings with strdup instead of just copying the pointers so you can work on a separate copy:

char **buffer = malloc((i+1) * sizeof(*buffer));

for (i = 0; envp[i] != 0 ; i++)
{
    buffer[i] = strdup(envp[i]);
    printf("%s\n", envp[i]);
    printf("%s\n", buffer[i]);
}
buffer[i] = NULL;