Strings and character with printf

Question

I was confused with usage of  c and  s in the following C program  include  lt stdio h gt       void main         char name    quot siva quot       printf  quot  s n quot  name       printf  quot  c n quot   name      Output is siva s  Why we need to use pointer to display a character  c  and pointer is not needed for a string I am getting error when i use printf  quot  c n quot   name    Error i got is str c  In function    main     str c 9 2  warning  format     c    expects type    int     but argument 2 has type    char

User · Answer

c   is designed for a single character a char  so it print only one element Passing the char array as a pointer you are passing the address of the first element of the array that is a single char  and then will be printed     s  printf   c n   name       will print   i   and so on      Pointer is not needed for the  s because it can work directly with String of characters

User · Answer

The name of an array is the address of its first element  so name is a pointer to memory containing the string  siva    Also you don t need a pointer to display a character  you are just electing to use it directly from the array in this case   You could do this instead   char c    name  printf   c n   c

User · Answer

You re confusing the dereference operator   with pointer type annotation    Basically  in C   means different things in different places    In a type    means a pointer  int is an integer type  int  is a pointer to integer type As a prefix operator    means  dereference   name is a pointer   name is the result of dereferencing it  i e  getting the value that the pointer points to  Of course  as an infix operator    means  multiply

User · Answer

If you want to display a single character then you can also use name 0  instead of using pointer   It will serve your purpose but if you want to display full string using  c  you can try this    include lt stdio h gt  void main          char name    siva       int i      for i 0 i lt 4 i                  printf   c    name i

User · Answer

The thing is that the printf function needs a pointer as parameter  However a char is a variable that you have directly acces  A string is a pointer on the first char of the string  so you don t have to add the   because   is the identifier for the pointer of a variable

User · Answer

If you try this    include lt stdio h gt   void main      char name    siva    printf  name    p n   name    printf   amp name 0     p n    amp name 0     printf  name printed as   s is  s n  name    printf   name    c n   name    printf  name 0     c n   name 0        Output is   name   0xbff5391b    amp name 0    0xbff5391b name printed as  s is siva  name   s name 0    s   So  name  is actually a pointer to the array of characters in memory  If you try reading the first four bytes at 0xbff5391b  you will see  s    i    v  and  a   Location     Data                     0xbff5391b    0x73   s   --- gt  name 0  0xbff5391c    0x69   i   --- gt  name 1  0xbff5391d    0x76   v   --- gt  name 2  0xbff5391e    0x61   a   --- gt  name 3  0xbff5391f    0x00    0  --- gt  This is the NULL termination of the string   To print a character you need to pass the value of the character to printf  The value can be referenced as name 0  or  name  since for an array name    amp name 0     To print a string you need to pass a pointer to the string to printf  in this case  name  or   amp name 0

[c] Strings and character with printf

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