Why this code output nothing ?? Help plz

Question

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

int main(){
char *str ="12345";
printf("%s",*str);
return 0;
}
// output : nothing why ?

Answer 1

Mik_RDC the %s format specifier needs a pointer. Remove the dereference and it will work.

Change:
printf("%s",*str);

To:
printf("%s",str);

Answer 2

Mik_RDC I'll give a short backgrounder that might clarify:

The character string "12345" is stored at some location in memory. The location has a unique numeric address.

If we store the address in a variable, p, then p is a pointer to the data. p has its own memory location just like an integer variable. If you print the contents of p as a numeric value, it shows the memory address of "12345". This is the pointer value that %s needs to locate the string.

For other purposes use *p to access the data that p refers to. This is called "dereferencing the pointer". For instance, in printf("%c", *p), the specifier %c prints a single character. It expects a character value, not an address like %s does, so *p is appropriate. It prints the first character of the string, which is '1'.

Answer 3

Big thanks bro Brian

Answer 4

Brian * (asterix) is'nt use to access pointer value ?