The following code cannot be compiled by VC++ and clang.
int f()
{
return 0;
}
int main()
{
// error : called object type 'int' is not a function or function pointer
int f = f();
}
It is necessary in some cases. For example, I have a function to calculate the character count of a string, which is named count, however, another function parameter is also expressively named as count.
size_t count(char* sz)
{
return strlen(sz);
}
bool check_count(char* sz, size_t count)
{
return count == count(sz); // ???
}
How to resolve this issue?
In C++ you can define a namespace for your objects, in your example you could do:
namespace MyFunctions {
int f()
{
return 0;
}
}
int main()
{
int f = MyFunctions::f();
}
The answer is simple. This is not supported. C, as many other languages, cannot support absolutely every scenario. It is unreasonable to put out such a goal. Nobody ever tried to achieve this.
In your particular case, you should rename your parameter. Function always has limited scope. It is always recompiled as a whole. Names of params in prototypes in the header files may have different names. Renaming param in the body of the function will work in 99.9999% of cases.
Related
I have two functions with a little different functionality, so I can't make them as template functions.
int func64(__int64 a) {
return (int) a/2;
}
int func32(int a) {
return a--;
}
Depending on variable b64, I would like to call func64 or func32. I don't want check if b64 is true many times in my code, so I use pointers to functions.
void do_func(bool b64) {
typedef int (*pfunc32)(int);
typedef int (*pfunc64)(__int64);
pfunc32 call_func;
if (b64)
call_func = func64; //error C2440: '=' : cannot convert from 'int (__cdecl *)(__int64)' to 'pfunc32'
else
call_func = func32;
//...
call_func(6);
}
How can I avoid this error and cast call_func to pfunc32 or pfunc64?
The language requires all functions called through the same function pointer to have the same prototype.
Depending on what you want to achieve, you could use the pointer/cast aproach already mentioned (which satisfies this requirement at the loss of type safety) or pass a union instead:
union u32_64
{
__int64 i64;
int i32;
};
int func64(union u32_64 a) {
return (int) a.i64/2;
}
int func32(union u32_64 a) {
return --a.i32;
}
void do_func(bool b64) {
typedef int (*pfunc)(union u32_64);
pfunc call_func;
if (b64)
call_func = func64;
else
call_func = func32;
//...
union u32_64 u = { .i64 = 6 };
call_func(u);
}
Pass a void pointer and cast it in the function body.
Of course this means less compiler control if you use the wrong type; if you call func64 and pass an int to it the program will compile and produce wrong results without giving you any tip of what is going wrong.
int func64(void *a) {
__int64 b = *((__int64*) a);
return (int) b/2;
}
int func32(void *a) {
int b = *((int *) a)
return b-1;
}
I need to call func32() or func64() depending on flag b64
So do that:
void do_func(bool b64) {
if (b64)
func64(6);
else
func32(6);
}
Well, first of all, please note that function func32 is returning the input argument as is.
This is because with return a--, you are returning the value of a before decrementing it.
Perhaps you meant to return a-1 instead?
In any case, you can simply declare this function as int func32(__int64 a).
This way, it will have the same prototype as function func64, but will work exactly as before.
BTW, calling a function through a pointer might be more "expensive" than a simple branch operation, so depending on your application, you might be better off with a simple if/else conditional statement...
Make a wrapper for func64:
int func64_as_32(int a) {
return func64(a);
}
Now you can assign either func32 or func64_as_32 to call_func since they have the same signature. The value you pass in, 6, has type int so passing it to func64_as_32 has the same effect as passing it directly to func64.
If you have call sites where you pass in a value of type __int64 then you'd do it the other way around, wrap func32.
As bool in C++ converts to int ( true => 1, false => 0 ) you can use b64 as array index. So take SJuan76's advice, convert your functions prototype to int f(void*) and put them into array int (*array_fun[2])(void* x); . You can call these functions then like that :
int p = 6;
array_fun[b64](&p);
What's the difference between:
void function();
int main()
{......}
void function()
{......}
vs
void function()
{.......}
int main();
It seems odd to declare a function before main then define it after main when you could just declare and define it before main. Is it for aesthetic purposes? My teacher writes functions like the first example.
It's just for code organization purposes ("aesthetics", I guess). Without forward declarations you'd need to write every function before it's used, but you may want to write the bodies of a function in a different order for organizational purposes.
Using forward declarations also allows you to give a list of the functions defined in a file at the very top, without having to dig down through the implementations.
Forward declarations would also be necessary in the case of mutually recursive functions. Consider this (silly) example:
bool is_odd(int); // neccesary
bool is_even(int x) {
if (x == 0) {
return true;
} else {
return is_odd(x - 1);
}
}
bool is_odd(int x) {
return is_even(x - 1);
}
Note I mean to make this answer supplementary, others have already given good answers to this questions.
Note that knowing how to forward declare things in C++ becomes very important. Once you begin using header files it basically becomes mandatory. Header files will allow you to build a prototype of functions and classes/structs then define them in a corresponding .cpp file. This is a very important organizational feature of C++.
// MyClass.h
class MyClass
{
public:
MyClass(int x);
void printInt();
private:
int myInt;
};
// MyClass.cpp
MyClass::MyClass(int x)
{
MyClass::myInt = x;
}
void MyClass::printInt()
{
std::cout << MyClass::myInt;
}
Doing things this way makes it so you're not completely bound to make a huge hodgepodge of code. Especially if you're writing real programs that will have a considerably large amount of source code.
So while in the question you asked forward declaring is really just more of a preference, later on it really won't be a choice.
Your first example is in the Top-Down style, the second in Bottom-Up style.
It's largely aesthetic, in that some people prefer one over the other.
For example, someone might prefer to get a high-level overview of the program before getting the details (top-down), while another might person might prefer to see the details first (bottom-up).
A tangible benefit of the declarations in the top-down approach is that you can more easily re-organize the function definitions without having to worry about ordering.
Look at this example:
int max(int num1, int num2)
{
// local variable declaration
int result;
if (num1 > num2)
result = num1;
else
result = num2;
return result;
}
where, int max(int num1, int num2) is behaving as a function.
Now, here we have a program
#include <iostream>
using namespace std;
// function declaration
int max(int num1, int num2);
int main ()
{
// local variable declaration:
int a = 100;
int b = 200;
int ret;
// calling a function to get max value.
ret = max(a, b);
cout << "Max value is : " << ret << endl;
return 0;
}
// function returning the max between two numbers
int max(int num1, int num2)
{
// local variable declaration
int result;
if (num1 > num2)
result = num1;
else
result = num2;
return result;
}
I kept max() function along with main() function and compiled the source code. While running final executable, it would produce the following result:
Max value is : 200
Calling a Function:
While creating a C++ function, you give a definition of what the function has to do. To use a function, you will have to call or invoke that function.
When a program calls a function, program control is transferred to the called function. A called function performs defined task and when its return statement is executed or when its function-ending closing brace is reached, it returns program control back to the main program.
To call a function, you simply need to pass the required parameters along with function name, and if function returns a value, then you can store returned value.
When I define a function of a class, I call another function of the same class within it. But when I do not type the class name it gives segmentation fault. Check below.
Header file
class DomainSolver
{
int fnc1 (UserDefinedType & var);
int fnc2 (UserDefinedType & var);
};
C file
int DomainSolver::fnc2 (UserDefinedType & var)
{
return 0;
}
int DomainSolver::fnc1 (UserDefinedType & var)
{
// fnc2 (var); // does not work
DomainSolver::fnc2(var); // works
return 0;
}
Wild guess… since the code you presented does not have any issues…
The function being called is declared virtual in a base class, so even if the virtual keyword is not present in the declaration here it is virtual.
The function being called does not access any member of the object.
You are calling the function on an invalid pointer or reference (for example through a null pointer or on an object that has already been deleted.
If all those guesses are right, the use of the qualification inhibits the dynamic dispatch mechanism, avoiding the dereference of an invalid pointer to the vtable. The code is still wrong (due to the third point above), but it seems to work.
The solution is not to call a member function through an invalid pointer or reference.
Although as pointed out by Zac's reply, the functions as you present them are not properly formed, there shouldn't be a difference between calling the scoped version; if you are getting a segfault in one case and not the other it's possibly because of code elsewhere.
Here is an example that works just fine:
dsolver.hh
#ifndef DSOLVER_HH
#define DSOLVER_HH
class DomainSolver
{
public:
int fnc1 (int& var);
int fnc2 (int& var);
};
#endif
dsolver.cc
#include <iostream>
#include "dsolver.hh"
int DomainSolver::fnc1 (int& var)
{
std::cout << "fnc1\n";
fnc2( var );
var = -1;
return var;
}
int DomainSolver::fnc2 (int& var)
{
std::cout << "fnc2\n";
var = 100;
return var;
}
main.cc
#include <iostream>
#include "dsolver.hh"
int main()
{
DomainSolver my_dsolver;
int my_int = 5;
my_dsolver.fnc1(my_int);
return 0;
}
Assuming this is close to your actual code, you have undefined behavior in fnc1:
int DomainSolver::fnc1 (UserDefinedType & var)
{
// fnc2 (var); // does not work
DomainSolver::fnc2(var); // works
// missing return!!!
}
You declare it to return an int, but then never return anything (in either case). Both cases are UB, so anything they do is technically "valid", since your code is not.
This code should be:
int DomainSolver::fnc1 (UserDefinedType & var)
{
return fnc2 (var);
}
As a side note: This is a good example of why you should never ignore the warnings given by the compiler (as you should have received a warning with both versions).
EDIT
With your latest edit adding a return value to fnc1, you'll need to show more of your actual code as there is not enough there to properly diagnose the problem (with the return being there, there is nothing wrong with your shown code).
I'm trying to implement a minheap in C++. However the following code keeps eliciting errors such as :
heap.cpp:24:4: error: cannot convert 'complex int' to 'int' in assignment
l=2i;
^
heap.cpp:25:4: error: cannot convert 'complex int' to 'int' in assignment
r=2i+1;
^
heap.cpp: In member function 'int Heap::main()':
heap.cpp:47:16: error: no matching function for call to 'Heap::heapify(int [11], int&)'
heapify(a,i);
^
heap.cpp:47:16: note: candidate is:
heap.cpp:21:5: note: int Heap::heapify(int)
int heapify(int i) //i is the parent index, a[] is the heap array
^
heap.cpp:21:5: note: candidate expects 1 argument, 2 provided
make: * [heap] Error 1
#include <iostream>
using namespace std;
#define HEAPSIZE 10
class Heap
{
int a[HEAPSIZE+1];
Heap()
{
for (j=1;j<(HEAPISZE+1);j++)
{
cin>>a[j];
cout<<"\n";
}
}
int heapify(int i) //i is the parent index, a[] is the heap array
{
int l,r,smallest,temp;
l=2i;
r=2i+1;
if (l<11 && a[l]<a[i])
smallest=l;
else
smallest=i;
if (r<11 && a[r]<a[smallest])
smallest=r;
if (smallest != i)
{
temp = a[smallest];
a[smallest] = a[i];
a[i]=temp;
heapify(smallest);
}
}
int main()
{
int i;
for (i=1;i<=HEAPSIZE;i++)
{
heapify(a,i);
}
}
}
Ultimately, the problem with this code is that it was written by someone who skipped chapters 1, 2 and 3 of "C++ for Beginners". Lets start with some basics.
#include <iostream>
using namespace std;
#define HEAPSIZE 10
Here, we have included the C++ header for I/O (input output). A fine start. Then, we have issued a directive that says "Put everything that is in namespace std into the global namespace". This saves you some typing, but means that all of the thousands of things that were carefully compartmentalized into std:: can now conflict with names you want to use in your code. This is A Bad Thing(TM). Try to avoid doing it.
Then we went ahead and used a C-ism, a #define. There are times when you'll still need to do this in C++, but it's better to avoid it. We'll come back to this.
The next problem, at least in the code you posted, is a misunderstanding of the C++ class.
The 'C' language that C++ is based on has the concept of a struct for describing a collection of data items.
struct
{
int id;
char name[64];
double wage;
};
It's important to notice the syntax - the trailing ';'. This is because you can describe a struct and declare variables of it's type at the same time.
struct { int id; char name[64]; } earner, manager, ceo;
This declares a struct, which has no type name, and variables earner, manager and ceo of that type. The semicolon tells the compiler when we're done with this statement. Learning when you need a semicolon after a '}' takes a little while; usually you don't, but in struct/class definition you do.
C++ added lots of things to C, but one common misunderstanding is that struct and class are somehow radically different.
C++ originally extended the struct concept by allowing you to describe functions in the context of the struct and by allowing you to describe members/functions as private, protected or public, and allowing inheritance.
When you declare a struct, it defaults to public. A class is nothing more than a struct which starts out `private.
struct
{
int id;
char name[64];
double wage;
};
class
{
public:
int id;
char name[64];
double wage;
};
The resulting definitions are both identical.
Your code does not have an access specifier, so everything in your Heap class is private. The first and most problematic issue this causes is: Nobody can call ANY of your functions, because they are private, they can only be called from other class members. That includes the constructor.
class Foo { Foo () {} };
int main()
{
Foo f;
return 0;
}
The above code will fail to compile, because main is not a member of Foo and thus cannot call anything private.
This brings us to another problem. In your code, as posted, main is a member of Foo. The entry point of a C++ program is main, not Foo::main or std::main or Foo::bar::herp::main. Just, good old int main(int argc, const char* argv[]) or int main().
In C, with structs, because C doesn't have member functions, you would never be in a case where you were using struct-members directly without prefixing that with a pointer or member reference, e.g. foo.id or ptr->wage. In C++, in a member function, member variables can be referenced just like local function variables or parameters. This can lead to some confusion:
class Foo
{
int a, b;
public:
void Set(int a, int b)
{
a = a; // Erh,
b = b; // wat???
}
};
There are many ways to work around this, but one of the most common is to prefix member variables with m_.
Your code runs afoul of this, apparently the original in C passed the array to heapify, and the array was in a local variable a. When you made a into a member, leaving the variable name exactly the same allowed you not to miss the fact that you no-longer need to pass it to the object (and indeed, your heapify member function no-longer takes an array as a pointer, leading to one of your compile errors).
The next problem we encounter, not directly part of your problem yet, is your function Heap(). Firstly, it is private - you used class and haven't said public yet. But secondly, you have missed the significance of this function.
In C++ every struct/class has an implied function of the same name as the definition. For class Heap that would be Heap(). This is the 'default constructor'. This is the function that will be executed any time someone creates an instance of Heap without any parameters.
That means it's going to be invoked when the compiler creates a short-term temporary Heap, or when you create a vector of Heap()s and allocate a new temporary.
These functions have one purpose: To prepare the storage the object occupies for usage. You should try and avoid as much other work as possible until later. Using std::cin to populate members in a constructor is one of the most awful things you can do.
We now have a basis to begin to write the outer-shell of the code in a fashion that will work.
The last change is the replacement of "HEAPSIZE" with a class enum. This is part of encapsulation. You could leave HEAPSIZE as a #define but you should expose it within your class so that external code doesn't have to rely on it but can instead say things like Heap::Size or heapInstance.size() etc.
#include <iostream>
#include <cstdint> // for size_t etc
#include <array> // C++11 encapsulation for arrays.
struct Heap // Because we want to start 'public' not 'private'.
{
enum { Size = 10 };
private:
std::array<int, Size> m_array; // meaningful names ftw.
public:
Heap() // default constructor, do as little as possible.
: m_array() // says 'call m_array()s default ctor'
{}
// Function to load values from an istream into this heap.
void read(std::istream& in)
{
for (size_t i = 0; i < Size; ++i)
{
in >> m_array[i];
}
return in;
}
void write(std::ostream& out)
{
for (size_t i = 0; i < Size; ++i)
{
if (i > 0)
out << ','; // separator
out << m_array[i];
}
}
int heapify(size_t index)
{
// implement your code here.
}
}; // <-- important.
int main(int argc, const char* argv[])
{
Heap myHeap; // << constructed but not populated.
myHeap.load(std::cin); // read from cin
for (size_t i = 1; i < myHeap.Size; ++i)
{
myHeap.heapify(i);
}
myHead.write(std::cout);
return 0;
}
Lastly, we run into a simple, fundamental problem with your code. C++ does not have implicit multiplication. 2i is the number 2 with a suffix. It is not the same as 2 * i.
int l = 2 * i;
There is also a peculiarity with your code that suggests you are mixing between 0-based and 1-based implementation. Pick one and stick with it.
--- EDIT ---
Technically, this:
myHeap.load(std::cin); // read from cin
for (size_t i = 1; i < myHeap.Size; ++i)
{
myHeap.heapify(i);
}
is poor encapsulation. I wrote it this way to draw on the original code layout, but I want to point out that one reason for separating construction and initialization is that it allows initialization to be assured that everything is ready to go.
So, it would be more correct to move the heapify calls into the load function. After all, what better time to heapify than as we add new values, keeping the list in order the entire time.
for (size_t i = 0; i < Size; ++i)
{
in >> m_array[i];
heapify(i);
}
Now you've simplified your classes api, and users don't have to be aware of the internal machinery.
Heap myHeap;
myHeap.load(std::cin);
myHeap.write(std::cout);
Could someone please tell me if this is possible in C or C++?
void fun_a();
//int fun_b();
...
main(){
...
fun_a();
...
int fun_b(){
...
}
...
}
or something similar, as e.g. a class inside a function?
thanks for your replies,
Wow, I'm surprised nobody has said yes! Free functions cannot be nested, but functors and classes in general can.
void fun_a();
//int fun_b();
...
main(){
...
fun_a();
...
struct { int operator()() {
...
} } fun_b;
int q = fun_b();
...
}
You can give the functor a constructor and pass references to local variables to connect it to the local scope. Otherwise, it can access other local types and static variables. Local classes can't be arguments to templates, though.
C++ does not support nested functions, however you can use something like boost::lambda.
C — Yes for gcc as an extension.
C++ — No.
you can't create a function inside another function in C++.
You can however create a local class functor:
int foo()
{
class bar
{
public:
int operator()()
{
return 42;
}
};
bar b;
return b();
}
in C++0x you can create a lambda expression:
int foo()
{
auto bar = []()->int{return 42;};
return bar();
}
No but in C++0x you can http://en.wikipedia.org/wiki/C%2B%2B0x#Lambda_functions_and_expressions which may take another few years to fully support. The standard is not complete at the time of this writing.
-edit-
Yes
If you can use MSVC 2010. I ran the code below with success
void test()
{
[]() { cout << "Hello function\n"; }();
auto fn = [](int x) -> int { cout << "Hello function (" << x << " :))\n"; return x+1; };
auto v = fn(2);
fn(v);
}
output
Hello function
Hello function (2 :))
Hello function (3 :))
(I wrote >> c:\dev\loc\uniqueName.txt in the project working arguments section and copy pasted this result)
The term you're looking for is nested function. Neither standard C nor C++ allow nested functions, but GNU C allows it as an extension. Here is a good wikipedia article on the subject.
Clang/Apple are working on 'blocks', anonymous functions in C! :-D
^ ( void ) { printf("hello world\n"); }
info here and spec here, and ars technica has a bit on it
No, and there's at least one reason why it would complicate matters to allow it. Nested functions are typically expected to have access to the enclosing scope. This makes it so the "stack" can no longer be represented with a stack data structure. Instead a full tree is needed.
Consider the following code that does actually compile in gcc as KennyTM suggests.
#include <stdio.h>
typedef double (*retdouble)();
retdouble wrapper(double a) {
double square() { return a * a; }
return square;
}
int use_stack_frame(double b) {
return (int)b;
}
int main(int argc, char** argv) {
retdouble square = wrapper(3);
printf("expect 9 actual %f\n", square());
printf("expect 3 actual %d\n", use_stack_frame(3));
printf("expect 16 actual %f\n", wrapper(4)());
printf("expect 9 actual %f\n", square());
return 0;
}
I've placed what most people would expect to be printed, but in fact, this gets printed:
expect 9 actual 9.000000
expect 3 actual 3
expect 16 actual 16.000000
expect 9 actual 16.000000
Notice that the last line calls the "square" function, but the "a" value it accesses was modified during the wrapper(4) call. This is because a separate "stack" frame is not created for every invocation of "wrapper".
Note that these kinds of nested functions are actually quite common in other languages that support them like lisp and python (and even recent versions of Matlab). They lead to some very powerful functional programming capabilities, but they preclude the use of a stack for holding local scope frames.
void foo()
{
class local_to_foo
{
public: static void another_foo()
{ printf("whatevs"); }
};
local_to_foo::another_foo();
}
Or lambda's in C++0x.
You can nest a local class within a function, in which case the class will only be accessible to that function. You could then write your nested function as a member of the local class:
#include <iostream>
int f()
{
class G
{
public:
int operator()()
{
return 1;
}
} g;
return g();
}
int main()
{
std::cout << f() << std::endl;
}
Keep in mind, though, that you can't pass a function defined in a local class to an STL algorithm, such as sort().
int f()
{
class G
{
public:
bool operator()(int i, int j)
{
return false;
}
} g;
std::vector<int> v;
std::sort(v.begin(), v.end(), g); // Fails to compile
}
The error that you would get from gcc is "test.cpp:18: error: no matching function for call to `sort(__gnu_cxx::__normal_iterator > >, __gnu_cxx::__normal_iterator > >, f()::G&)'
"
It is not possible to declare a function within a function. You may, however, declare a function within a namespace or within a class in C++.
Not in standard C, but gcc and clang support them as an extension. See the gcc online manual.
Though C and C++ both prohibit nested functions, a few compilers support them anyway (e.g., if memory serves, gcc can, at least with the right flags). A nested functor is a lot more portable though.
No nested functions in C/C++, unfortunately.
As other answers have mentioned, standard C and C++ do not permit you to define nested functions. (Some compilers might allow it as an extension, but I can't say I've seen it used).
You can declare another function inside a function so that it can be called, but the definition of that function must exist outside the current function:
#include <stdlib.h>
#include <stdio.h>
int main( int argc, char* argv[])
{
int foo(int x);
/*
int bar(int x) { // this can't be done
return x;
}
*/
int a = 3;
printf( "%d\n", foo(a));
return 0;
}
int foo( int x)
{
return x+1;
}
A function declaration without an explicit 'linkage specifier' has an extern linkage. So while the declaration of the name foo in function main() is scoped to main(), it will link to the foo() function that is defined later in the file (or in a another file if that's where foo() is defined).