Why don't I need to dereference a character pointer in C before printing it? -

why code work? expect need dereference ptr, printf("%s\n", *ptr); before print out, segmentation fault if try way.

#include <stdio.h>  int main(int argc, char *argv[]) {         char name[] = "jordan";         char *ptr = name;         printf("%s\n", ptr); } 

hope guys give me insight.

when print string need starting address of string.

printf("%s\n", ptr);                 ^ address %s    

it prints chars till \0 nul encounter.

whereas print chat int .. need value variable:

printf("%c\n", *ptr);                ^ * %c print first char 

where in scanf() string need give address:

scanf("%s", ptr);             ^ string address 

also int scanf() char

scanf("%c", ptr);             ^ read @ first location char address  

note: scanf() need address %c store scanned value in memory.

be careful ptr points constant string can't use in scanf.

why segmentation fault following code ?

    printf("%s\n", *ptr); 

when this, because of %s printf interprets *ptr address, not address , if treat address points location read protected program(process) causes segmentation fault.

your ptr via name points constant string in memory ("jordan") in below diagram:

name 2002 ┌─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐ │ 'j' │ 'o' │ 'r' │ 'd' │ 'a' │ 'n' │'\0' │ ........ └─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘   ^   |  ptr = name    ==> ptr = 2002      *ptr = 'j' 

in printf("%s\n", *ptr); *ptr = 'j' , ascii value of char 'j' 74 74 address not under process control , trying read memory location , memory violation , segmentation fault occurs.

if compiles code containing printf("%s\n", *ptr); proper option -wall gcc warning below:

warning: format ‘%s’ expects argument of type ‘char *’, argument 2 has type ‘int’

says %s need (expects ) address of type char* putting value

notice:

printf("%s\n",   *ptr);         ^          ^ argument-2         argument-1