CODEWITHSUNDEEP
c
Debabratta Jena
Debabratta Jena

Reputation: 431

Expression syntax error in the following code

char name[25]; 
name[] = "abcd";

The code above gives me expression syntax error.

char name[25]; 
name = "abcd";

The code above gives me an Lvalue Required error.

But following code gives no error:

char name[25] = "abcd";

Can anyone please explain?

Upvotes: 0

Views: 451

Answers (2)

Vlad from Moscow
Vlad from Moscow

Reputation: 311068

If to place semicolons as it is required then this construction

char name[25]; name[]="abcd";

may be rewritten for visibility like

char name[25]; 
name[]="abcd";

So iy is seen that in the second line there us absent a type specifier that the line would be a valid declaration

char name[25]; 
char name[]="abcd";

In this line where again there is absent a type specifier for the second identifier

char name[25]; name ="abcd"

and we will rewrite like

char name[25]; 
char name ="abcd";

then name has the type char but is initialized by a string literal instead of one character. So it is evident that name shall be an array or a pointer to char, For example

char name[25]; 
char name[] ="abcd";

or

char name[25]; 
char *name ="abcd";

or for example

char name[25]; 
char name[26] ="abcd";

Of course the names of identifiers shall be different. Otherwise the compiler again will issue an error due to redefinition of the same identifier name in the same scope.

Upvotes: 3

Amadan
Amadan

Reputation: 198436

char name[25] (a declaration of an array) does this:

  • Reserves 25 bytes of memory
  • Declares name as an array of characters (which is almost but not exactly like char *) pointing to that memory.

char name[25] = "abcd" (a declaration of an array with an initialiser) does this:

  • Reserves 25 bytes of memory, filled with "abcd\0"
  • Declares name as an array of characters (which is almost but not exactly like char *) pointing to that memory.

(The case of name[] = "abcd" is not a syntax supported by C.)

In both cases, one of the critical differences between pointers and arrays is that the target of a pointer can change; the target of an array cannot. I.e. you can never assign anything to name declared as char[] above, but you can assign to name declared as char *, as follows. char *name; name = "abcd" (a declaration of a pointer, assignment of a literal character array to a pointer) does this:

  • Reserves 5 bytes of memory, filled by "abcd\0"
  • Declares name as a pointer to character (which is almost but not exactly like char[]) pointing to undefined target
  • Assigns the address of the memory occupied by "abcd\0" to the variable name.

Upvotes: 3

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