What does extern mean in a function
declaration?
Using extern in a function declaration we can make a function such that
it can used outside the file in which it is defined.
An extern variable, function definition, or declaration also makes the
described variable or function usable by the succeeding part of the
current source file. This declaration does not replace the definition.
The declaration is used to describe the variable that is externally
defined.
If a declaration for an identifier already exists at file scope, any
extern declaration of the same identifier found within a block refers to
that same object. If no other declaration for the identifier exists at
file scope, the identifier has external linkage.
What can I safely assume about the initial values of variables which
are not explicitly initialized?
It depends on complier which may assign any garbage value to a variable
if it is not initialized.
What is the difference between char a[] = “string”; and char *p =
“string”;?
In the first case 6 bytes are allocated to the variable a which is
fixed, where as in the second case if *p is assigned to some other value
the allocate memory can change.
What’s the auto keyword good for?
Answer1
Not much. It declares an object with automatic storage duration. Which
means the object will be destroyed at the end of the objects scope. All
variables in functions that are not declared as static and not
dynamically allocated have automatic storage duration by default.
For example
int main()
{
int a; //this is the same as writing “auto int a;”
}
Answer2
Local variables occur within a scope; they are “local” to a function.
They are often called automatic variables because they automatically
come into being when the scope is entered and automatically go away when
the scope closes. The keyword auto makes this explicit, but local
variables default to auto auto auto auto so it is never necessary to
declare something as an auto auto auto auto.
What is the difference between char a[] = “string”; and char *p =
“string”; ?
Answer1
a[] = “string”;
char *p = “string”;
The difference is this:
p is pointing to a constant string, you can never safely say
p[3]=’x';
however you can always say a[3]=’x';
char a[]=”string”; - character array initialization.
char *p=”string” ; - non-const pointer to a const-string.( this is
permitted only in the case of char pointer in C++ to preserve backward
compatibility with C.)
Answer2
a[] = “string”;
char *p = “string”;
a[] will have 7 bytes. However, p is only 4 bytes. P is pointing to an
adress is either BSS or the data section (depending on which compiler —
GNU for the former and CC for the latter).
Answer3
char a[] = “string”;
char *p = “string”;
for char a[]…….using the array notation 7 bytes of storage in the static
memory block are taken up, one for each character and one for the
terminating nul character.
But, in the pointer notation char *p………….the same 7 bytes required, plus
N bytes to store the pointer variable “p” (where N depends on the system
but is usually a minimum of 2 bytes and can be 4 or more)……
How do I declare an array of N pointers to functions returning
pointers to functions returning pointers to characters?
Answer1
If you want the code to be even slightly readable, you will use typedefs.
typedef char* (*functiontype_one)(void);
typedef functiontype_one (*functiontype_two)(void);
functiontype_two myarray[N]; //assuming N is a const integral
Answer2
char* (* (*a[N])())()
Here a is that array. And according to question no function will not
take any parameter value.
What does extern mean in a function declaration?
It tells the compiler that a variable or a function exists, even if the
compiler hasn’t yet seen it in the file currently being compiled. This
variable or function may be defined in another file or further down in
the current file.
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