CODEWITHSUNDEEP
cstringpointers
OldSchool
OldSchool

Reputation: 2183

confusion about character array initialisation

What is the difference between 

   char ch [ ] = "hello";

and

   char ch [ ] = { 'h','e','l','l','o','\0'};

and why we can only do 

   char *p = "hello";

But cant do

   char *p = {'h','e','l','l','o','\0'};

Upvotes: 3

Views: 96

Answers (5)

ThunderGr
ThunderGr

Reputation: 2367

The two array declarations are the same. As for the pointer declaration, the form char *p = {'h','e','l','l','o','\0'}; is not valid, simply because it was not included in the compiler design.

There is no theoretical reason, in my knowledge, why that declaration should not be valid, when char *p = "hello"; is.

Upvotes: 2

ajay
ajay

Reputation: 9680

Let's take it one by one. This following is initializing a char array ch using the string literal "hello".

char ch[] = "hello";

The following is initializing the array ch using an array initialization list. This is equivalent to the above statement.

char ch[] = {'h', 'e', 'l', 'l', 'o', '\0'};

The following is initializing a char pointer p to point to the memory where the string literal "hello" is stored. This is read-only memory. Attempting to modify its contents will not give compile error because string literal in C are not const qualified unlike in C++, but will cause undefined behaviour or even program crash.

char *p = "hello";
const char *p = "hello";  // better

The following last statement is plain wrong.

char *p = {'h','e','l','l','o','\0'};

p is a char pointer here, not an array and can't be initialized using array initialization list. I have highlighted the words array and pointer above to emphasize that array and pointer are different types. In some cases, an array is implicitly converted to a pointer to its first element like when an array is passed to a function or assigned to a pointer of the same type. This does not mean they are the same. They have different pointer arithmetic and different sizeof values.

Upvotes: 1

Ed Heal
Ed Heal

Reputation: 60037

Open upon a time they made a mistake about const

The bit "hello" should be a const char * const but they where lazy and just used char *. But to keep the faith alive they let that one slip.

Then they said. Ok. They can be equal.

Then they had char [] = { 'a', 'b', ...}; and all was good in the world

The the evil monster came and thrust upon them char *p = "hello". But the evil monster was in a good mood and said it should be const char *p = "hello" but I would be happy with that.

He went home and but the evil monster was not amused. He dictated over his realm char *p = {'h','e','l','l','o','\0'}; is the sign of a heretic.

Basically there was a cock-up. Just do it right from now and there is old code kicking around that needs to be satisfied.

Upvotes: 1

RahulKT
RahulKT

Reputation: 171

There is no difference in 

char ch [ ] = "hello";
char ch [ ] = { 'h','e','l','l','o','\0'};

But check below 

char *p = "hello";  //This is correct

char *p = {'h','e','l','l','o','\0'};  //This is wrong

If you want to make it correct you need to use 

char *p[]={'h','e','l','l','o','\0'}; //This works

Upvotes: -2

ouah
ouah

Reputation: 145899

char ch [ ] = "hello";
char ch [ ] = { 'h','e','l','l','o','\0'};

There is no difference. ch object will be exactly the same in both declarations.

On why we cannot do:

char *p = {'h','e','l','l','o','\0'};

An initializer list of more than one value can only be used for objects of aggregate type (structures or array). You can only initialize a char * with a pointer value.

Actually:

char ch [ ] = "hello";

and

char *p = "hello";

are not the same. The first initializes an array with the elements of a string literal and the second is a pointer to a string literal.

Upvotes: 3

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