This question already has answers here:
What does `*&` in a function declaration mean?
(8 answers)
Closed 8 years ago.
I am Converting a C++ code into C program. I come across a function that takes void draw(char *&a). how to write this function *& in C program. I tried using same *&a but it is showing , (...etc expected. Please help.
Thanks
There is no pass by reference in C. Therefore you need to change the function declaration to
void draw(char **a)
Related
This question already has answers here:
Passing references to pointers in C++
(10 answers)
Closed 4 years ago.
I have two function signatures in C++
void printArray(int* arrayPtr);
void printArray(int*& arrayPtr);
I understand the 1st function. It says the function takes in a arrayPtr argument which is of type that's a pointer pointing to an integer.
Both function signature works, but I have a hard time understanding the 2nd signature(*&) and what benefits it offers?
It's exactly the same as type versus type&; the first is a value and the second is a reference. The fact that type is a pointer doesn't change that.
This question already has answers here:
Why do function pointer definitions work with any number of ampersands '&' or asterisks '*'?
(5 answers)
How do function pointers in C work?
(12 answers)
Closed 4 years ago.
I'd like to know what is the difference between these two (seemingly correct) ways of passing a function to another function:
void fun(int p(int x));
void fun(int (*p)(int x));
Also, I know that in the second case we're passing a function pointer, but is the first approach also a function pointer (but without the star)?
This question already has answers here:
what does this mean char (*(*a[4])())[5]?
(4 answers)
Closed 7 years ago.
Can somebody explain what does it mean?
void f(int (*)[7]) {}
There is a site devoted to dealing with C gibberish and converting it to English:
cdecl.org Try it) It is convenient and sometimes even entertaining.
It states that void f(int (*)[7]) means: declare f as function (pointer to array 7 of int) returning void.
This is a function definition, which can accept a pointer to an array of int with 7 size.
Declare an 7-size array, int a[7];, and you can pass the address of it into f, like f(&a);
Check the live: http://cpp.sh/8ztz
It's definition of function that takes one parameter. That parameter is an unnamed pointer to array of ints.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
what’s the meaning of *&
what * &pSth mean?
is this a pointer or a ref?
Why/When we need that?
--code--
ClassName::GetSth(TypeName* &pSth)
{
//some code
}
It means "reference to pointer to Typename".
TypeName* &pSth is a reference to TypeName pointer.
Equivalent syntax in C is TypeName** pSth
This question already has an answer here:
Closed 10 years ago.
Possible Duplicate:
What does “void *(*)(void *)” mean in c++?
What does the type void(*)(void *) mean?
I came across this type in the example code for the book "Mastering Algorithms with C"
void list_init(List *list, void (*destroy)(void *data))
{
...
...
}
It's a function pointer.
void (*destroy)(void *data)
destroy is a pointer to a function which returns void and takes a void* as an argument.
cdecl.org is a useful tool for discerning complex C declarations. Also, take a look at the spiral rule.
In this specific case, its a pointer to which any function can be cast to void(*)(void *) and the function parameter void * can be any type.