struct containing arrays.

Firstly sorry and apologize for this silly question :devil:

lets say I have this example(it's spontaneously) :

Code:

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#include<stdio.h>
#include<string.h>
struct student {
        char name[15];}jay;
int main()
{
      jay.name="David";
      return 0;
}

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I know there're errors in this code, and that's why I came for :)
Well, to put the value "david" into the struct's name , I must access it and I accessed the struct by using operator "."(see in line 5), and now why I must write

Code:

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jay.name

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, and not a

Code:

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jay.name[15]

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, I'm saying that because the variable here not name alone it's name[15], so I must define that while trying to accessing... , and that's all..thanks.

You cannot assign to a char array that way. You need to use strcpy().

Quote:

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Originally Posted by Elkvis

You cannot assign to a char array that way. You need to use strcpy().

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So you're trying to say if I could assign, then I could use the statement I've talked about?

Quote:

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Originally Posted by RyanC

So you're trying to say if I could assign, then I could use the statement I've talked about?

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Try modifying the program so the variable in the struct is an integer instead of a character array. Then assign a value to it, and print the result. I think that should answer your question.

Quote:

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Originally Posted by Matticus

Try modifying the program so the variable in the struct is an integer instead of a character array. Then assign a value to it, and print the result. I think that should answer your question.

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You didn't get me, I know I must write

Code:

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strcpy(jay.name, "David")

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but why we cant write like jay.name[15]="David" ? isn't
the variable in the struct called name[15]? or we just forgetting always from the object "[length]" and name itself is the variable!

Quote:

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Originally Posted by RyanC

but why we cant write like jay.name[15]="David" ? isn't
the variable in the struct called name[15]?

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No, the member is named name, not name[15].

There are 3 different thing.

1. You've defined struct called student
2. Inside the struct you have a "member variable" and your asking compiler to allocate 15 bytes.
3. And to access that 15 bytes of memory you use a name and you named it as "name"

When you want to copy something to that memory you need to specify where to copy rhs too. And the way you do that is specifying the name in this case. Does that make sense?

Quote:

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Originally Posted by RyanC

but why we cant write like jay.name[15]="David" ? isn't
the variable in the struct called name[15]? or we just forgetting always from the object "[length]" and name itself is the variable!

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It wouldn't work the way you would want it to without a special case for strings written into the standard for the assignment operator (I think). The type on the right hand side of the assignment is a pointer to const char, which would need to be deep copied into the array on the left, or else it would just be undefined behavior. You can do a shallow copy of a string literal like this:

Code:

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const char* a = "Hello";
const char* b = a;

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Basically just make sure you use strcpy or similar function to make a copy of a string literal when assigning it to non-const memory and you should be fine.

Quote:

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Originally Posted by Alpo

It wouldn't work the way you would want it to without a special case for strings written into the standard for the assignment operator (I think). The type on the right hand side of the assignment is a pointer to const char, which would need to be deep copied into the array on the left, or else it would just be undefined behavior.

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The problem is that an array is not a modifiable lvalue, but the left hand operand of an assignment operator is required to be a modifiable lvalue.

Quote:

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Originally Posted by Alpo

The type on the right hand side of the assignment is a pointer to const char

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Not quite: the type of the string literal is actually char[6] that is not an lvalue, so it is effectively equivalent to const char[6], and since arrays are converted to pointers to their first elements, it effectively amounts to const char*, yet strictly speaking it isn't.