scanf a string in a char pointer [duplicate]

Question

I want to scan a string and point a char pointer to this scanned string.

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

    char *string;
    scanf("%s",&string);
    printf("%s\n",string);

}

But gives a warning saying

warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char **'

How do I scan a string in a char * without warnings.

Edit : It worked for the below code with warnings and I drilled down it to the above one so to be very specific.

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

    int n = 2;

    char *strings[2];
    for(int i = 0; i < n; i++){
        scanf("%s",&strings[i]);
        printf("%s\n",&strings[i]);
    }

}

Answer 1

First of all, %s in scanf() expects a pointer to char, not a pointer to pointer to char.

Quoting C11, chapter §7.21.6.2 / p12, fscanf() (emphasis mine)

s
Matches a sequence of non-white-space characters.²⁸⁶⁾

If no l length modifier is present, the corresponding argument shall be a pointer to the initial element of a character array large enough to accept the sequence and a terminating null character, which will be added automatically. [...]

Change

scanf("%s",&string);

to

scanf("%s",string);

That said, you need to allocate memory to string before you actually try to scan into it. Otherwise, you'll end up using uninitialized pointer which invokes undefined behavior.