C - Why no output?

Question

Does *str point to nothing?

char *str = "12345";
printf("%s",*str);

Answer 1

Yes even I encountered this question in the c challenges. the reason is simple %s cannot be used with a char*. This would help getting the complete string.
Printf ("%s",str);

Answer 2

* is used when you want to access single element,

printf("%c",*str); // prints 1
printf("%c",*(str+2)); //prints 3

Answer 3

I'm guessing it's due to the use of %s format specifier - which was meant for string, not a char. *s dereference the <str> and returns the first character in <str> ('1'). Expecting more details from others.

Answer 4

Paolo De Nictolis
Because you use the wrong format specifier ,the %s is basically used for the string not for character......

Answer 5

remove * before str,
printf("%s",str);

Answer 6

Adding to what everyone have said, the s format specifier requires the pointer to the initial character as its argument, and then the function will dereference that pointer on its own. You pass *str as its argument, *str has the value of 49 (the character '1'), so the function dereference that value which causes segmentation fault, or no output in SL playground

Answer 7

Because it already pointer. Try char *p2 = "54321";
printf ("%s",p2);

return 0;

Answer 8

%s expects char * by writing * before str u are passing char str which in case is 1. If u wanna see string just remove *. If u wanna see output from *str change %s to %c which is for char s