why am i unable to compile the program containing the code
char name[10];
name= "Rajesh";
While i am able to compile a program with
char name[10]="Rajesh";
7 Answers
That's because your code snippet is not performing declaration, but assignment:
char name[10]; // Declaration
name= "Rajesh"; // Assignment.
And arrays are not directly assignable in C.
The name name actually resolves to the address of its first element (&name[0]), which is not an lvalue, and as such cannot be the target of an assignment.
String Variable Declarations and Assignments
String variables can be declared just like other arrays:
char phrase[14];
String arrays can be initialised or partially initialised at the same time as being declared, using a list of values enclosed in "{}" braces (the same is true of arrays of other data types). For example, the statement
char phrase[14] = {'E','n','t','e','r',' ','a','g','e',':',' ','\0'};
both declares the array "phrase" and initialises it to the state. The statement
char phrase[14] = "Enter age: ";
is equivalent. If the "14" is omitted, an array will be created just large enough to contain both the value ""Enter age: "" and the sentinel character "'\0'", so that the two statements
char phrase[] = {'E','n','t','e','r',' ','a','g','e',':',' ','\0'};
char phrase[] = "Enter age: ";
are equivalent both to each other and to the statement
char phrase[12] = "Enter age: ";
However, it is important to remember that string variables are arrays, so we cannot just make assignments and comparisons using the operators "=" and "==". We cannot, for example, simply write
phrase = "You typed: "; //Wrong way
Instead, we can use a special set of functions for string assignment and comparison.
Edited :
And other way is to do that, using pointer : -
Declare variable
char const *phrase; /* a pointer to type character */
And initialize variable as where you want, as
phrase = "Test string";
7 Comments
NealB
+1 excellent answer, however, I think a lot of coufusion could be avoided if we just didn't use the word "string" with respect to arrays of characters in C. C simply does not have a "string" data type - just an array of characters, a convention with respect to how these arrays are constructed (ie. ends with a null) and a bunch of runtime routines to maniplate arrays following this convention (eg strcpy).
xanatos
I was tempted to write the pointer example, but then I thought I was opening a can of worms. The
pA now is "pointing" to a string literal that mustn't be modified (and that sadly sometimes can be modified). Without even thinking about the implicit conversion char[] => char* and "what are the difference between "Enter age: " and "Test string"?
Useless
BTW, I think you need to declare
char const *pA; for it to work - otherwise you're losing constness from the string literal.xanatos
@Useless Sadly for historical reasons (retrocompatibility with old code),
*pA can be a char*Useless
Although it may SEGV if you try writing to your allegedly non-const
char* ;-) |
You cannot assign values to string arrays by using assignment.
In C, You can only initialize arrays not assign them, a array of characters is no exception for this rule.
You will need to use string copying functions like strcpy or strncpy and so on.
However you can encapsulate a string in a struct and simulate this:
typedef struct Yourstring Yourstring;
struct Yourstring
{
char a[24];
};
Yourstring a = { "abcd" };
Yourstring b = a;
Yourstring c = { 0 };
c = b;
3 Comments
xanatos
I think you should explain him better. explain him the difference between
char name[10] and char* and the differences that there are.
Samuele Mattiuzzo
i'll add this as a comment: your first choice is a mix between C (first statement) and C++ (second statement). this link should be useful to have some comparisons cs.smu.ca/~porter/csc/ref/c_cpp_strings.html
ktf
Be careful - copying structs in this way is not possible with all C compiler implementations.
char name[10];
In this first example, you're declaring name to be an array of ten characters. The symbol name is now interpreted as the starting address of this array, but while you can write into the array, you can't move the symbol name.
So, this:
name= "Rajesh";
would mean pointing name away from the array you declared and at the location of the string literal "Rajesh" which is stored elsewhere in memory. You just can't do this.
What you can do is either:
strcpy(name, "Rajesh");
which copies your string literal from it's immutable location in your executable, into the char array you declared, or:
char const *pointer_to_name = "Rajesh";
which doesn't copy anything, but merely stores the address of your immutable string literal into a variable where you can use it, or your second example:
char name[10]="Rajesh";
which declares name to be an array of 10 characters and initialises it.
1 Comment
Useless
I really feel I should warn you to be careful of
strcpy as well, but the post was long enough. However: be careful of strcpy. It doesn't check the destination is big enough for the string, and you can easily cause problems if it isn't.char name[10];
name= "Rajesh";
Here name is an array of characters.
The simple name is basically pointer to first element of array and it cannot be assigned with some value like it's done in above statement.
My point is
char name[10]; name= "Rajesh";
Explanation: This is not the correct declaration of array. Strings are nothing but collection of characters terminated with '\0' operator. So, the array index (which in this case is 'name', basically points to address of 1st character in array i.e name holds the address of character 'R' in 'Rajesh'.
If you want to initialize like as mentioned above, the better approach could have been:
char name[10]; *name= "Rajesh";
Now, the above declaration won't throw any error, but still it will throw warning, like:
assignment makes integer from pointer without a cast [-Wint-conversion] *name = "Rajesh"
2 Comments
chqrlie
The simple
name is basically pointer to first element of array NO, this is a confusing analogy: name is the name of the array. Arrays cannot be assigned to with =. If name was a pointer, Any valid pointer value could be assigned to it.chqrlie
I'm sorry, but your rephrased answer is still confusing:
name does not point to address of first character, it does not hold the address of character R in Rajesh, it is the address of that character. name is not a pointer at all, it is the array itself. Furthermore the syntax *name = "Rajesh" may be more readable as name[0] = "Rajesh", which makes it obvious that one is trying to store a string into a character. It is a shame C compilers accept this nonsense with just a warning, it should be an error.AS char name[10]="Rajesh" is definition compiler understands what are you trying to do and corrects your mistake. In c++ strings written in "" are constant and some compilers put them to stringpools to save space. name="...." means that you are trying to assign a constant to a non-constant pointer which is not allowed.
you should use strcpy to copy a string into an array.
2 Comments
char name[10] ="Rajesh";
This one is an array initialization. The compiler knows of it. It's a one-shot trick. You can use it only when you define your variable. It would be equivalent to:
char name[10] = { 'R', 'a', 'j', 'e', 's', 'h', '\0' };
The other one is illegal, because you can't use array initialization outside of array definition.
Comments
I dont remember where exactly i read it but C standard says, you can assign a string value for an array at the defination but not after the defination.
char a[10]="rajesh" its a defination hence works
char a[10];a="rajesh"; fails its not a defination
rather you need to use strcpy(a,"rajesh") to assing a value for a string, if its not a defination
