This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
In a C function declaration, what does “…” as the last parameter do?
What does this mean ,...); it is written at the end of a function in a code i am debuging.
like this void abc( int a, int b, ...);
It means the function can take any number of extra arguments. For example, consider printf; the first argument is the format string, and then there can be any number of arguments after that for all of the modifiers. This would be represented by using ... after the first argument when defining the function.
That specifies a variable number of arguments which can be accessed using the macros in the cstdarg header.
Related
This question already has answers here:
Why does argument matching in C++ ignore array sizes?
(1 answer)
In a function declaration, what does passing a fixed size array signify? [duplicate]
(3 answers)
Why can one specify the size of an array in a function parameter?
(3 answers)
Closed 3 years ago.
In C++, arrays are supposed to be passed to functions by reference. Hence, in what follows, function foo should implicitly use array inp by reference.
void foo(double inp[10]) {}
void foo1(double (&inp)[10]) {}
My question is, since both functions supposedly have the same interpretation of the input variable, why can we call foo in what follows, but we cannot call foo1?
int main()
{
double ary[20];
foo(ary); // compiles without any problem.
foo1(ary); // compiler error: invalid initialization of reference of type ‘double (&)[10]’ from expression of type ‘double [20]’
return 0;
}
since both functions supposedly have the same interpretation of the input variable
But they don't. A function argument of the type double inp[10] is automatically adjusted to a pointer double*. The 10 plays no part in providing type information here. Since all arrays decay to pointers, that will allow you to pass an array of any size.
A reference to an array does not get adjusted however. The type information is still there, and it must be a reference to an array of exactly ten doubles. The only way to pass a reference to an array of any size is to have a separate function for it, which you may write a function template to accomplish
template<std::size_t N>
void foo2(double (&inp)[N]) {}
This question already has answers here:
Calling a function through its address in memory in c / c++
(6 answers)
Closed 5 years ago.
I have a function at a known memory address(for example: 0x11111111). The function returns an int, and takes a uint32_t pointer as its only argument.
How would I call this function using c++? I have seen a few examples of calling a function by its address, but I can't seem to find one that takes a pointer as its argument
EDIT:I seen that. Doesn't address how to call the function that takes a pointer as an argument
If you’re sure that there’s a function there, you could call it by casting the address to a function pointer of the appropriate type, then calling it. Here’s C code to do this:
typedef int (*FunctionType)(uint32_t*);
FunctionType function = (FunctionType)0x11111111;
function(arg);
This can easily be modified to support any number of function arguments and any return type you’d like. Just tweak the argument types list of the FunctionType typedef.
Or, in one line (gulp):
(((int (*)(uint32_t *)) 0x11111111)(arg);
This question already has answers here:
Is f(void) deprecated in modern C and C++? [duplicate]
(6 answers)
Closed 6 years ago.
I have seen in C++ program, during function declaration if there is no parameter for the function void is declared as parameter like this:
int F1(void)
How is it different than:
int F1()
There is no difference. Using void is just a more explicit way to declare the same thing. Personally, I never use that syntax and rarely see anyone else use it either.
It's the same thing in C++, and is a holdover from C.
Here's an excerpt from the C++ 2003 standard (C.1.6):
Change: In C++, a function declared with an empty parameter list takes no arguments.
In C, an empty parameter list means that the number and type of the function arguments are unknown"
Example:
int f(); // means int f(void) in C++
// intf(unknown) in C
Rationale: This is to avoid erroneous function calls (i.e. function calls with the wrong number or type of arguments).
Effect on original feature: Change to semantics of well-defined feature. This feature was marked as “obsolescent” in C.
Both of them are exactly the same, leaving the argument field empty as () is the one I prefer, some prefer writing (void) just so that someone editing the code may be ensured that no arguments are required. Makes no difference though, just a readability thing.
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:
Question about a function definition (three dots in parameters..)
I'm new to c++, what is the meaning of the ... (triple dot) inside the method or function
sample
void strArr::set(const char *value, ...) {
// some code here
}
It's a way to express a variable number of arguments, used to be called va_args or variadic function.
Take a look here to learn about them in C++, they do exist in other programming languages though, since it's just a generic concept.
That's a variadic function, it can take any number of parameters (more than 1 in this case).
Same principle as printf and others:
printf("%d", 1);
printf("%d%d%d", 1,1,1);
printf("%d%d%d%d%d%d", 1,1,1,1,1,1);
These aren't overloads, but the same function taking a variable number of arguments.