Check if the first character is '\0'. You should also probably check if your pointer is NULL.

char *c = "";
if ((c != NULL) && (c[0] == '\0')) {
   printf("c is empty\n");
}

You could put both of those checks in a function to make it convenient and easy to reuse.

Edit: In the if statement can be read like this, "If c is not zero and the first character of character array 'c' is not '\0' or zero, then...".

The && simply combines the two conditions. It is basically like saying this:

if (c != NULL) { /* AND (or &&) */
    if (c[0] == '\0') {
        printf("c is empty\n");
    }
}

You may want to get a good C programming book if that is not clear to you. I could recommend a book called "The C Programming Language".

The shortest version equivalent to the above would be:

if (c && !c[0]) {
  printf("c is empty\n");
}

Since C-style strings are always terminated with the null character (\0), you can check whether the string is empty by writing

do {
   ...
} while (url[0] != '\0');

Alternatively, you could use the strcmp function, which is overkill but might be easier to read:

do {
   ...
} while (strcmp(url, ""));

Note that strcmp returns a nonzero value if the strings are different and 0 if they're the same, so this loop continues to loop until the string is empty.

Hope this helps!

Given this code:

char text[50];
if(strlen(text) == 0) {}

Followed by a question about this code:

 memset(text, 0, sizeof(text));
 if(strlen(text) == 0) {}

I smell confusion. Specifically, in this case:

char text[50];
if(strlen(text) == 0) {}

... the contents of text[] will be uninitialized and undefined. Thus, strlen(text) will return an undefined result.

The easiest/fastest way to ensure that a C string is initialized to the empty string is to simply set the first byte to 0.

char text[50];
text[0] = 0;

From then, both strlen(text) and the very-fast-but-not-as-straightforward (text[0] == 0) tests will both detect the empty string.

The easy way to do it would be like this:

if (msg[0])
    printf("message: %s", msg);

If, at some later date, msg is a pointer, you would first want to assure it's not a NULL pointer.

if (msg && msg[0])
    printf("message: %s", msg);

A test program to demonstrate:

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

char *msg;

void test() {
    if (msg && msg[0])
        printf("string is %s\n", msg);
}

int main()
{
    msg = NULL;
    test();
    msg = calloc(10, 1);
    test();
    msg[0] = 'm';
    msg[1] = 'e';
    test();
    free(msg);
}

On my machine the output is:

string is me

Pointers aren't "empty", they either point at valid data or invalid data. A controlled way to make them point at invalid data is to assign them to NULL. Thus the correct way to write the program would be:

char *name = NULL;
int age;
store_stuff(&name, &age);

...

void store_stuff(char **name, int *age) {
  if(*name == NULL)
  {
    // handle error
  }
  ...
}