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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)?
Related
This question already has answers here:
Why pass by const reference instead of by value?
(8 answers)
Which is faster? Pass by reference vs pass by value C++
(1 answer)
Performance cost of passing by value vs. by reference or by pointer?
(6 answers)
Closed 5 months ago.
When I pass a std::vector into a C++ function, like so:
void myFunc(std::vector<int> myVec)
{
// do something
}
Does the vector myVec get copied or is in reality only a pointer passed? I'm asking from optimization perspective; should I always explicitly pass a pointer to the vector or is there a difference in speed? Assume that I only want to read the data in the function, not manipulate the vector itself.
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How does dereferencing of a function pointer happen?
(5 answers)
Calling a function through a function pointer - dereference the pointer or not? What's the difference?
(2 answers)
Closed 12 months ago.
what's the difference between the second and third line:
int (*myFunction)() = fooFunction;
(*fooFunction)();
fooFunction();
Why do both do the same thing? Is it just how C++'s syntax is? I thought that a function pointer needs to be dereferenced in order to call it. How does the third line work, it isn't dereferenced.
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.
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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:
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)