What is the difference between char s and char s

Question

In C  one can use a string literal in a declaration like this   char s      hello     or like this   char  s    hello     So what is the difference  I want to know what actually happens in terms of storage duration  both at compile and run time

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char  s1    Hello world      Points to fixed character string which is not allowed to modify char s2      Hello world      As good as fixed array of characters in string so allowed to modify     s1 0     J      Illegal s2 0     J      Legal

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C99 N1256 draft  There are two different uses of character string literals    Initialize char     char c      abc           This is  more magic   and described at 6 7 8 14  Initialization       An array of character type may be initialized by a character string literal  optionally   enclosed in braces  Successive characters of the character string literal  including the   terminating null character if there is room or if the array is of unknown size  initialize the   elements of the array    So this is just a shortcut for   char c       a    b    c     0      Like any other regular array  c can be modified  Everywhere else  it generates an    unnamed array of char What is the type of string literals in C and C    with static storage that gives UB if modified   So when you write   char  c    abc     This is similar to        unnamed is magic because modifying it gives UB     static char   unnamed      abc   char  c     unnamed    Note the implicit cast from char   to char    which is always legal   Then if you modify c 0   you also modify   unnamed  which is UB   This is documented at 6 4 5  String literals       5 In translation phase 7  a byte or code of value zero is appended to each multibyte   character sequence that results from a string literal or literals  The multibyte character   sequence is then used to initialize an array of static storage duration and length just   sufficient to contain the sequence  For character string literals  the array elements have   type char  and are initialized with the individual bytes of the multibyte character   sequence            6 It is unspecified whether these arrays are distinct provided their elements have the   appropriate values  If the program attempts to modify such an array  the behavior is   undefined     6 7 8 32  Initialization  gives a direct example      EXAMPLE 8  The declaration  char s      abc   t 3     abc         defines  plain  char array objects s and t whose elements are initialized with character string literals       This declaration is identical to  char s        a    b    c     0     t        a    b    c           The contents of the arrays are modifiable  On the other hand  the declaration  char  p    abc         defines p with type  pointer to char  and initializes it to point to an object with type  array of char  with length 4 whose elements are initialized with a character string literal  If an attempt is made to use p to modify the contents of the array  the behavior is undefined    GCC 4 8 x86-64 ELF implementation  Program    include  lt stdio h gt   int main void        char  s    abc       printf   s n   s       return 0      Compile and decompile   gcc -ggdb -std c99 -c main c objdump -Sr main o   Output contains    char  s    abc   8   48 c7 45 f8 00 00 00    movq    0x0 -0x8  rbp  f   00          c  R X86 64 32S  rodata   Conclusion  GCC stores char  it in  rodata section  not in  text   Note however that the default linker script puts  rodata and  text in the same segment  which has execute but no write permission  This can be observed with   readelf -l a out   which contains    Section to Segment mapping    Segment Sections       02      text  rodata   If we do the same for char      char s      abc     we obtain   17    c7 45 f0 61 62 63 00    movl    0x636261 -0x10  rbp    so it gets stored in the stack  relative to  rbp

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char s      hello     declares s to be an array of char which is long enough to hold the initializer  5   1 chars  and initializes the array by copying the members of the given string literal into the array   char  s    hello     declares s to be a pointer to one or more  in this case more  chars and points it directly at a fixed  read-only  location containing the literal  hello

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Just to add  you also get different values for their sizes   printf  sizeof s      zu n   sizeof s       6 printf  sizeof  s     zu n   sizeof s       4 or 8   As mentioned above  for an array   0  will be allocated as the final element

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As an addition  consider that  as for read-only purposes the use of both is identical  you can access a char by indexing either with    or    lt var gt     lt index gt   format   printf   c   x 1          Prints r   And   printf   c     x   1      Prints r   Obviously  if you attempt to do    x   1     a     You will probably get a Segmentation Fault  as you are trying to access read-only memory

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The difference here is that   char  s    Hello world     will place  Hello world  in the read-only parts of the memory  and making s a pointer to that makes any writing operation on this memory illegal    While doing   char s      Hello world     puts the literal string in read-only memory and copies the string to newly allocated memory on the stack  Thus making  s 0     J     legal

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An example to the difference  printf  quot hello quot    2     llo char a      quot hello quot    2    error  In the first case pointer arithmetics are working  arrays passed to a function decay to pointers

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Given the declarations  char  s0    hello world   char s1      hello world     assume the following hypothetical memory map                        0x01  0x02  0x03  0x04         0x00008000   h     e     l     l          0x00008004   o           w     o          0x00008008   r     l     d    0x00             s0      0x00010000  0x00  0x00  0x80  0x00 s1      0x00010004   h     e     l     l          0x00010008   o           w     o          0x0001000C   r     l     d    0x00   The string literal  hello world  is a 12-element array of char  const char in C    with static storage duration  meaning that the memory for it is allocated when the program starts up and remains allocated until the program terminates   Attempting to modify the contents of a string literal invokes undefined behavior     The line  char  s0    hello world     defines s0 as a pointer to char with auto storage duration  meaning the variable s0 only exists for the scope in which it is declared  and copies the address of the string literal  0x00008000 in this example  to it  Note that since s0 points to a string literal  it should not be used as an argument to any function that would try to modify it  e g   strtok    strcat    strcpy    etc       The line  char s1      hello world     defines s1 as a 12-element array of char  length is taken from the string literal  with auto storage duration and copies the contents of the literal to the array   As you can see from the memory map  we have two copies of the string  hello world   the difference is that you can modify the string contained in s1     s0 and s1 are interchangeable in most contexts  here are the exceptions   sizeof s0    sizeof  char   sizeof s1    12  type of  amp s0    char    type of  amp s1    char     12     pointer to a 12-element array of char   You can reassign the variable s0 to point to a different string literal or to another variable   You cannot reassign the variable s1 to point to a different array

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This declaration   char s      hello     Creates one object - a char array of size 6  called s  initialised with the values  h    e    l    l    o     0    Where this array is allocated in memory  and how long it lives for  depends on where the declaration appears   If the declaration is within a function  it will live until the end of the block that it is declared in  and almost certainly be allocated on the stack  if it s outside a function  it will probably be stored within an  initialised data segment  that is loaded from the executable file into writeable memory when the program is run   On the other hand  this declaration   char  s   hello     Creates two objects    a read-only array of 6 chars containing the values  h    e    l    l    o     0   which has no name and has static storage duration  meaning that it lives for the entire life of the program   and a variable of type pointer-to-char  called s  which is initialised with the location of the first character in that unnamed  read-only array    The unnamed read-only array is typically located in the  text  segment of the program  which means it is loaded from disk into read-only memory  along with the code itself   The location of the s pointer variable in memory depends on where the declaration appears  just like in the first example

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char s      Hello world     Here  s is an array of characters  which can be overwritten if we wish   char  s    hello     A string literal is used to create these character blocks somewhere in the memory which this pointer s is pointing to  We can here reassign the object it is pointing to by changing that  but as long as it points to a string literal the block of characters to which it points can t be changed

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In the case of   char  x    fred     x is an lvalue -- it can be assigned to  But in the case of   char x      fred     x is not an lvalue  it is an rvalue -- you cannot assign to it

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First off  in function arguments  they are exactly equivalent   void foo char  x   void foo char x        exactly the same in all respects   In other contexts  char   allocates a pointer  while char    allocates an array  Where does the string go in the former case  you ask  The compiler secretly allocates a static anonymous array to hold the string literal  So   char  x    Foo      is approximately equivalent to  static const char   secret anonymous array      Foo   char  x    char      secret anonymous array    Note that you must not ever attempt to modify the contents of this anonymous array via this pointer  the effects are undefined  often meaning a crash    x 1     O      BAD  DON T DO THIS    Using the array syntax directly allocates it into new memory  Thus modification is safe   char x      Foo   x 1     O      No problem    However the array only lives as long as its contaning scope  so if you do this in a function  don t return or leak a pointer to this array - make a copy instead with strdup   or similar  If the array is allocated in global scope  of course  no problem

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In the light of comments here it should be obvious that   char   s    hello    Is a bad idea  and should be used in very narrow scope    This might be a good opportunity to point out that  const correctness  is a  good thing   Whenever and wherever You can  use the  const  keyword to protect your code  from  relaxed  callers or programmers  which are usually most  relaxed  when pointers come into play   Enough melodrama  here is what one can achieve when adorning pointers with  const    Note  One has to read pointer declarations right-to-left   Here are the 3 different ways to protect yourself when playing with pointers    const DBJ  p means  p points to a DBJ that is const         that is  the DBJ object can t be changed via p   DBJ  const p means  p is a const pointer to a DBJ         that is  you can change the DBJ object via p  but you can t change the pointer p itself   const DBJ  const p means  p is a const pointer to a const DBJ         that is  you can t change the pointer p itself  nor can you change the DBJ object via p   The errors related to attempted const-ant mutations are caught at compile time  There is no runtime space or speed penalty for const    Assumption is you are using C   compiler  of course     --DBJ

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char  str    Hello     The above sets str to point to the literal value  Hello  which is hard-coded in the program s binary image  which is flagged as read-only in memory  means any change in this String literal is illegal and that would throw segmentation faults   char str      Hello     copies the string to newly allocated memory on the stack  Thus making any change in it is allowed and legal   means str 0     M     will change the str to  Mello    For more details  please go through the similar question   Why do I get a segmentation fault when writing to a string initialized with  quot char  s quot  but not  quot char s   quot

[c] What is the difference between char s[] and char *s?

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