I wrote a function that that takes some argument and a pointer argument . when calling the function , sometimes I need to pass along the pointer for use inside the function and sometimes I don't .
what are the consequences of calling a function with fewer arguments ? it compiles correctly and during runtime its still fine , but is this good programming ? is it better if I call the function with a dummy variable ?
Thanks and sorry for beginner question .
If you call a function with too few arguments and the compiler doesn't complain, then you're doing something wrong.
You can write a function declaration/definition that doesn't specify how many arguments it requires:
void func();
/* ... */
func();
func(arg1);
func(arg1, arg2);
All three of those calls will be accepted by the compiler, but at least two of them are incorrect.
That form of function declaration/definition has been obsolescent since the 1989 ANSI C standard.
Never use this form.
Functions declaration should always be written as prototypes, i.e., declarations that specify the number and type(s) of the parameters. As a special case, (void) denotes a function with no parameters.
void func(int arg);
/* ... */
func(); /* rejected by compiler */
func(arg1); /* accepted -- but only if arg1 is of type int or convertible to int */
func(arg1, arg2); /* rejected by compiler */
If you manage to write code that calls a function with an incorrect number of arguments and get it past the compiler, the behavior is undefined. It might appear to "work", but it could blow up in your face when, for example, you compile it with a different compiler, or with the same compiler and different options.
One complication: some functions are variadic, taking a variable number of arguments. The most common example of this is printf. For variadic functions, the required arguments are typically specified by the function's documentation -- and it's just as important to get the arguments right. The difference is that, for variadic functions, your compiler won't necessarily tell you that a call is incorrect.
The , ... syntax (in the function declaration) and the macros defined in <stdarg.h> are the only legitimate way to write and use C functions that take a variable number and type(s) of arguments.
One of the differences of C++ over plain C is that it incorporates name mangling, which lets you specify a single function with varying return types and parameters. Two functions that share the same name but have different parameters:
int foo();
int foo(int param1, char* param2);
can be done in C++ because it actually alters the name of functions behind-the scenes as it compiles it. When you want to pass in a different number of parameters, it's essentially the same as having two different function names that it calls:
int foo1();
int foo2(int param1, char* param2);
When you pass in fewer parameters than it expects, some compilers should at least throw warnings; others won't even compile the program at all.
Let's say that you pass in 2 parameters to a function that expects 3. When you try to reference that 3rd parameter within your function, what do you expect the value to be? Most of the time, it will be a garbage value, and definitely something your function does not expect.
I would suggest just passing in a dummy value to functions like that: a NULL for a pointer or a 0 or negative "uninitialized" kind of value for other types. You can at least test for those values in your function. Or just write a second function that takes a different number of parameters.
I would not do this. I would only call the function with the declared number of arguments.
It is extremely confusing to the reader of the code.
The reader of the code might even think the function has been overloaded and look
for the other definitions of the function with fewer arguments.
The omitted arguments can have any value at all - anything is considered legal behavior by the C compiler. The program may work perfectly in a test environment and then crash for any or no reason.
If I absolutely had to have a variable number of arguments, for a legitimate purpose, I would try to put the arguments in a list instead of using a variable number of arguments. If someone else required me to use variable arguments, I would use varargs instead of the default behavior, and then only in the case where all the arguments are the same type.
Related
Is it necessary to declare function before using it? Does the compiler give an error if we don't use a function declaration and call it directly. Please answer according to standard C.
If yes then what does it mean that argument of type are converted to int and float to double if arguments to function are not defined?
In ANSI C, you do not have to declare a function prototype; however, it is a best practice to use them. The only reason the standard allows you to not use them is for backward compatibility with very old code.
If you do not have a prototype, and you call a function, the compiler will infer a prototype from the parameters you pass to the function. If you declare the function later in the same compilation unit, you'll get a compile error if the function's signature is different from what the compiler guessed.
Worse, if the function is in another compilation unit, there's no way to get a compilation error, since without a a prototype there's no way to check. In that case, if the compiler gets it wrong, you could get undefined behavior if the function call pushes different types on the stack than the function expects.
Convention is to always declare a prototype in a header file that has the same name as the source file containing the function.
With prototypes, the compiler can verify you are calling the function correctly (using the right number and type of parameters).
Without prototypes, it's possible to have this:
// file1.c
void doit(double d)
{
....
}
int sum(int a, int b, int c)
{
return a + b + c;
}
and this:
// file2.c
// In C, this is just a declaration and not a prototype
void doit();
int sum();
int main(int argc, char *argv[])
{
char idea[] = "use prototypes!";
// without the prototype, the compiler will pass a char *
// to a function that expects a double
doit(idea);
// and here without a prototype the compiler allows you to
// call a function that is expecting three argument with just
// one argument (in the calling function, args b and c will be
// random junk)
return sum(argc);
}
In C, you need to define everything before you use it. I think the idea is that it can compile everything in one pass, since it always has all of the information it needs by the time it gets to it.
I'm not sure what you're asking with the second part. If the function isn't defined, it just won't work.
The short answer: in C89/90 it is not necessary, in C99 it is necessary.
In C89/90 the function does not have to be declared in order to be called. For undeclared function the compiler will make an assumption about function return type (assumes int) and about function parameter types (will derive them from the argument types at the point of the call).
In C99 the function has to be previously declared in order to be called. Note, that even in C99 there's no requirement to provide a prototype for the function, only a declaration is needed, i.e. a non-prototype declaration is OK. This means that the compiler no longer has to make any assumptions about the function return type (since the declaration always specifies it explicitly), but it still might have to make assumptions about the function parameters (if the declaration is not a prototype).
The C99 standard in 6.5.2.2 "Function call" describes what happens in a function call expression, including if no prototype has been seen by the compiler for the function:
If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument
Later, it describes what happens if the function call expression doesn't match what how the function is actually defined:
If the function is defined with a type that is not compatible with the type (of the expression) pointed to by the expression that denotes the called function, the behavior is undefined.
Clearly, to help ensure correctness, using function prototypes is the right thing to do. Enough such that C++ made requiring function prototypes one of the 'breaking' changes with c.
Is it possible to make declaration of variadic function so that it doesn't end with "..."?
Today I learned more about exec from unistd.h but through the day I've seen three (two actually) different declaration of execl:
1) int execl ( const char * path, const char * arg0, ..., (char*)NULL ); was shown to us in school and I imagined I would have to end the function call with a NULL value
2) int execl(const char *path, const char *arg, ... /* (char *) NULL */); is what I've found in the exec(3) man page. That would probably mean I still have to end it with a NULL value, but it is not enforced.
3) int execl(const char *path, const char *arg, ...); is what I found here. This one would probably normally put me to rest with the first one being a simplification for students, the second was a varning and this is the real thing (even though I would probably have normally higher regard for both options one and two.)
But then I found on the same site this declaration:
int execle(const char *path, const char *arg, ..., char * const envp[]);
Same question applies, I was unable to create variadic function not ending in ... with gcc telling me that it's expecting ')' before ',' token pointing to the comma after the three dots.
So finally, is it possible to make variadic functions ending with a NULL characters (execl) and if not, is it possible to make it end with predefined variable (execle)?
I tried to compile with gcc 6.3.1, I also tried --std=c11.
Is it possible to make declaration of variadic function so that it doesn't end with "..."?
Is it possible is a slippery question, but consider these facts:
the standard says that "If a function that accepts a variable number of arguments is defined without a parameter type list that ends with the ellipsis notation, the behavior is undefined" (C2011, 6.9.1/8)
Perhaps that answers the question already, but if you choose to mince words and focus on function declarations that are not definitions, then
a function definition is also a declaration
the C language standard requires all declarations of the same function to be "compatible" (else program behavior is undefined) (C2011 6.7/4)
two function declarations with mismatched parameter lists are not compatible (C2011, 6.2.7/3)
Thus, if you declare a variadic function that in fact is also defined, and that function's parameter list does not end with ..., then the program's behavior is undefined.
The documentation you've been reading for execle() and execl() is written to express and discuss those functions' expectations, but to the extent that it seems to present variadic function declarations in which the last element of the parameter list is not ..., those are not actually valid C function declarations.
So finally, is it possible to make variadic functions ending with a NULL characters (execl) and if not, is it possible to make it end with predefined variable (execle)?
It is not possible to describe such calling conventions via conforming C declarations. Variadic functions can have such expectations, and can enforce them at runtime, but they can be enforced at compile time only by a compiler that relies on special knowledge of the functions involved, or on C language extensions that allow such constraints to be described.
The declaration of a variadic function can only specify the required arguments, and the compiler can enforce their types. The variable-length part never has any type checking done. And the variable-length part is always at the end. The declaration for execle() is not meant as an actual C declaration, but just to describe to the programmer how he should construct the arguments.
It's not possible to enforce that the last argument to execl() is NULL. Variadic functions don't know how many arguments were supplied, they determine it from the values of the arguments. printf() assumes that it has enough arguments to fill in all the operators in the format string, and execl() iterates through the arguments until it finds NULL (execle() is similar, but it reads one additional argument to get envp). If you don't end with NULL, it will just keep going, reading garbage and causing undefined behavior.
The declaration you see is the one in the man pages of execl. The declaration for execle in glib is the following: int execle (const char *path, const char *arg, ...). The implementation assumes the last argument is a char**, and uses it for envp. I don't think you can enforce such a rule in C.
I read here that in C void foo() means a function foo taking an unspecified number of arguments of unspecified type.
Can anyone give me or point me to an example where a C function takes an unspecified number of arguments? What can this be applied to in C? I couldn't find anything on the web.
That's an old-style function declaration.
This declaration:
void foo();
declares that foo is a function returning void that takes an unspecified but fixed number and type(s) of arguments. It doesn't mean that calls with arbitrary arguments are valid; it means that the compiler can't diagnose incorrect calls with the wrong number or type of arguments.
Somewhere, perhaps in another translation unit (source file), there has to be a definition of the function, perhaps:
void foo(x, y)
long x;
double *y;
{
/* ... */
}
This means that any call to foo that doesn't pass two arguments of type long and double* is invalid, and has undefined behavior.
Prior to the 1989 ANSI C standard, these were the only kind of function declaration and definition available in the language, and the burden of writing correct function calls was entirely on the programmer. ANSI C added prototypes, function declarations that specify the types of a function's parameters, which allow compile-time checking of function calls. (This feature was borrowed from early C++.) The modern equivalent of the above would be:
void foo(long x, double *y);
/* ... */
void foo(long x, double *y) {
/* ... */
}
Old-style (non-prototype) declarations and definitions are still legal, but they're officially obsolescent, which means that, in principle, they could be removed from a future version of the language -- though since they're still around in the 2011 standard I don't know that that will ever actually happen.
There is no good reason to use old-style function declarations and definitions in modern C code. (I've seen arguments for using them in some corner cases, but I find them unconvincing.)
C also supports variadic functions like printf, which do take an arbitrary number of arguments, but that's a distinct feature. A variadic function must be declared with a prototype, which includes a trailing , .... (Calling a variadic function with no visible prototype isn't illegal, but it has undefined behavior.) The function itself uses macros defined in <stdarg.h> to process its parameters. As with old-style function declarations, there is no compile-time checking for arguments corresponding to the , ... (though some compilers may check some calls; for example gcc warns if the arguments in a printf call are inconsistent with the format string).
This literally means that you are not telling the compiler what arguments the function takes, this means that it will not protect you from calling it with any arbitrary set of arguments. You would need to say in the definition precisely what arguments are actually taken for the function to be implemented however.
You could for example use this if you were generating a header file to describe a foreign function in external code, however you did not know what the function's signature actually was, it would still then be callable using your header but if you provide the wrong arguments in the call results are undefined.
The declaration void foo(); is not necessarily the same thing as a function that takes a variable number of arguments.
All that it is is a function declaration that indicates nothing about the number of arguments the function takes (actually, this is not exactly true, see below about argument promotions). The declaration isn't a function prototype, which provides information about the number and types of the parameters (using an ellipsis to indicate variable length parameters).
The function definition (where the body of the function is provided) might take a fixed number of arguments, and might even have a prototype.
When such a function declaration is used, it is up to the programmer making a call to foo() to get the arguments that foo() expects (and their types) correct - the compiler has no information about the argument list foo() requires. It also restricts the type of parameters that the function definition can use because when the function is called without a prototype, therefore the default argument promotions will be applied to the arguments at the function call. So a function that's declared without a prototype can only expected to get promoted arguments.
See the following C99 standard sections for some details:
6.7.5.3/14 "Function declarators (including prototypes)
...
The empty list in a function declarator that is not part of a
definition of that function specifies that no information about the
number or types of the parameters is supplied.
...
6.5.2.2 Function calls
...
If the expression that denotes the called function has a type that
does not include a prototype, the integer promotions are performed on
each argument, and arguments that have type float are promoted to
double. These are called the default argument promotions. If the
number of arguments does not equal the number of parameters, the
behavior is undefined. If the function is defined with a type that
includes a prototype, and either the prototype ends with an ellipsis
(, ...) or the types of the arguments after promotion are not
compatible with the types of the parameters, the behavior is
undefined. If the function is defined with a type that does not
include a prototype, and the types of the arguments after promotion
are not compatible with those of the parameters after promotion, the
behavior is undefined, except for the following cases:
one promoted type is a signed integer type, the other promoted type is the corresponding unsigned integer type, and the value is
representable in both types;
both types are pointers to qualified or unqualified versions of a character type or void.
The C function calling standard allows for a function to be called with zero or more arguments and the number of arguments may or may not match the function interface.
The way this works is that it is up to the caller to adjust the stack after the called function returns rather than the called function adjusting the stack unlike other standards such as Pascal which require the called function to manage the stack adjustment properly.
Because the caller knows what arguments and their types have been pushed onto the stack before the called function is called and the called function does not, it is up to the caller to clear the pushed arguments from the stack after the called function returns.
With updated C standards, the function call interface description has become more complex in order to allow the compiler to detect and report interface problems that the original K&R C standard allowed to go undetected by the compiler.
The standard now is that you specify variable argument lists using elipsis notation of three periods or dots after the last known and specified argument in the called functions interface specification or declaration.
So you would see something like the following for some of the Standard C Library I/O functions:
int sprintf (char *buffer, char *format, ...);
This indicates that the function sprintf requires that the first argument be a char pointer to a buffer, the second argument be a char pointer to a format string, and there may be other additional arguments. In this case any additional arguments would be what are needed to be inserted for the print format specifiers in the format string. If the format string is just a text string with no format specifies (something like %d for an integer for instance) then there would be no other arguments.
The newer C Standards specify a set of functions/macros for the use with variable argument lists, the varg functions. With these functions/macros the called function can step through the variable part of an argument list and process the arguments. These functions look something like the following:
int jFunc (int jj, char *form, ...)
{
va_list myArgs;
int argOne;
va_start (myArgs, form);
argOne = va_arg (myArgs, int);
va_end (myArgs);
return 0;
}
The problem that we have with variable argument lists is that C does not have a way of communicating the variable argument or even how many arguments. So the designer of the function has to provide a mechanism. In the case of the C Standard Library I/O functions this is done with the format that indicates the number of arguments following the format string by specifying format specifiers for each argument. And since there is nothing that does a consistency check, you can end up with a format string that specifies more or less than the actual arguments resulting in either garbage output or less output than expected.
Since modern C compilers have some degree of backwards compatibility for old C source code, that means that you can use some of the older constructs and the compiler will allow it though hopefully with a warning.
The new function interface specifications are designed to reduce the chances of using a function incorrectly. So the new standards recommend that you use the function interface declaration so that the compiler can help you by detecting interface problems and incorrect variable usage in the function call.
However if you want to be a risk taker you do not have to use this safety net so if you like you can just define a function with an empty argument list and wing it.
You might also find an answer I put into this question about currying in C that uses variable argument lists along with a way to determine how many arguments are provided.
As mentioned in many other answers, a function taking an unspecified number of arguments of unspecified type just means it's caller doesn't know about its definition. It's completely different from variadic functions where you must define with ... and then use va_list in stdarg.h to obtain the arguments
It's actually common in the past and there are many standard functions using that feature, for example open()/openat()/_open() with an optional last parameter
fd = open(filename, O_RDONLY);
fd = open(filename, O_CREAT | O_WRONLY, 0777);
See open() function parameters
main() is also a function that can receive varying number of arguments
int main();
int main(int argc, char **argv);
int main(int argc, char **argv, char **envp);
int main(int argc, char **argv, char **envp, char **apple);
In fact the Windows x64 calling convention is designed to support such unspecified number of arguments so it's a bit unfortunate for most modern usecases
So today I figured (for the first time admittedly) that int foo() is in fact different from int foo(void) in that the first one allows any number of inputs and the second one allows zero.
Does int foo() simply ignore any given inputs? If so, what's the point of allowing this form of function? If not, how can you access them and how is this different from having a variable argument list (e.g. something like int foo (int argc, ...))?
The key to understanding this is understanding the call stack. What happens when you call a function in C (with the standard x86 ABI — other platforms may vary) is that the caller pushes all of the arguments, in reverse order, onto the stack before calling the callee. Then, the callee can read as many of those arguments as it wants. If you use foo(void), obviously you won't read any. If you use foo(int), you'll be able to read one word into the stack below your current frame.
Using foo() (with no args specified) means that the compiler won't care to check the arguments you pass to foo. Anything goes; there's no contract. This can be useful if foo has two different implementations that take different types as arguments, and I want to be able to switch them out programmatically. (It's also sometimes useful when writing functions in assembler, where you can handle the stack and arguments yourself.) But remember that the compiler is not going to do any type-checking for you, so you have to be very careful.
This is different from foo(int, ...), since in that case, the function actually has a way to access all of the arguments (using varargs). There's no reason ever to actually define a function with foo(), since it's basically equivalent to foo(void). foo() is only useful for declarations.
First, take int foo(). This form is a function declaration that does not provide a prototype - that is, it doesn't specify the number and types of its arguments. It is an archaic form of function declaration - it's how functions were originally declared in pre-ANSI C. Such a function however does have a fixed number and type of arguments - it's just that the declaration doesn't tell the compiler what it is, so you are responsible for getting it right where you call the function. You declare such a function with arguments like so:
int foo(a, b)
int a;
char *b;
{
return a + strlen(b);
}
This form of function declaration is now obsolete.
int foo(void) is a modern declaration that provides a prototype; it declares a function that takes no arguments. int foo(int argc, ...) also provides a prototype; it declares a function that has one fixed integer argument and a variable argument list.
In standard conform C, you have to use int foo(void) if the function does not accept any parameters.
I guess it is compiler dependant what happens, when you pass arguments to a function with empty braces. But I don't think there is a way to access these parameters.
As for main, the only standard conform (pure c) ways to write them are either int main(void) or int main(int argc, char **argv) (or char *argv[] which is the same).
Well, in C++ the two forms are equivalent and they both declare a function that takes no arguments. If you try to call the function and pass in an argument, the compile will give an error.
On the other hand, C and Objective-C both treat the two forms differently. In these languages the first form declares a function that takes an unknown number of arguments, whereas the second form declares a function that takes no arguments at all. So, in C the following is valid code:
int foo() {
return 5;
}
int main() {
return foo(1, 2, 3);
}
The compiler doesn't complain, and the code runs fine (a C++ compiler would give an error on this same code).
Generally what you want in C and Objective-C is to use the second form and include the explicit void to indicate that the function should take no arguments. However, it's more common in C++ to use the first form because it's equivalent and shorter.
Is declaring an header file essential? This code:
main()
{
int i=100;
printf("%d\n",i);
}
seems to work, the output that I get is 100. Even without using stdio.h header file. How is this possible?
You don't have to include the header file. Its purpose is to let the compiler know all the information about stdio, but it's by no means necessary if your compiler is smart (or lazy).
You should include it because it's a good habit to get into - if you don't, then the compiler has no real way to know if you're breaking the rules, such as with:
int main (void) {
puts (7); // should be a string.
return 0;
}
which compiles without issue but rightly dumps core when running. Changing it to:
#include <stdio.h>
int main (void) {
puts (7);
return 0;
}
will result in the compiler warning you with something like:
qq.c:3: warning: passing argument 1 of ‘puts’ makes pointer
from integer without a cast
A decent compiler may warn you about this, such as gcc knowing about what printf is supposed to look like, even without the header:
qq.c:7: warning: incompatible implicit declaration of
built-in function ‘printf’
How is this possible? In short: three pieces of luck.
This is possible because some compilers will make assumptions about undeclared functions. Specifically, parameters are assumed to be int, and the return type also int. Since an int is often the same size as a char* (depending on the architecture), you can get away with passing ints and strings, as the correct size parameter will get pushed onto the stack.
In your example, since printf was not declared, it was assumed to take two int parameters, and you passed a char* and an int which is "compatible" in terms of the invocation. So the compiler shrugged and generated some code that should have been about right. (It really should have warned you about an undeclared function.)
So the first piece of luck was that the compiler's assumption was compatible with the real function.
Then at the linker stage, because printf is part of the C Standard Library, the compiler/linker will automatically include this in the link stage. Since the printf symbol was indeed in the C stdlib, the linker resolved the symbol and all was well. The linking was the second piece of luck, as a function anywhere other than the standard library will need its library linked in also.
Finally, at runtime we see your third piece of luck. The compiler made a blind assumption, the symbol happened to be linked in by default. But - at runtime you could have easily passed data in such a way as to crash your app. Fortunately the parameters matched up, and the right thing ended up occurring. This will certainly not always be the case, and I daresay the above would have probably failed on a 64-bit system.
So - to answer the original question, it really is essential to include header files, because if it works, it is only through blind luck!
As paxidiablo said its not necessary but this is only true for functions and variables but if your header file provides some types or macros (#define) that you use then you must include the header file to use them because they are needed before linking happens i.e during pre-processing or compiling
This is possible because when C compiler sees an undeclared function call (printf() in your case) it assumes that it has
int printf(...)
signature and tries to call it casting all the arguments to int type. Since "int" and "void *" types often have same size it works most of the time. But it is not wise to rely on such behavior.
C supprots three types of function argument forms:
Known fixed arguments: this is when you declare function with arguments: foo(int x, double y).
Unknown fixed arguments: this is when you declare it with empty parentheses: foo() (not be confused with foo(void): it is the first form without arguments), or not declare it at all.
Variable arguments: this is when you declare it with ellipsis: foo(int x, ...).
When you see standard function working then function definition (which is in form 1 or 3) is compatible with form 2 (using same calling convention). Many old std. library functions are so (as desugned to be), because they are there form early versions of C, where was no function declarations and they all was in form 2. Other function may be unintentionally be compatible with form 2, if they have arguments as declared in argument promotion rules for this form. But some may not be so.
But form 2 need programmer to pass arguments of same types everywhere, because compiler not able to check arguments with prototype and have to determine calling convention osing actual passed arguments.
For example, on MC68000 machine first two integer arguments for fixed arg functions (for both forms 1 and 2) will be passed in registers D0 and D1, first two pointers in A0 and A1, all others passed through stack. So, for example function fwrite(const void * ptr, size_t size, size_t count, FILE * stream); will get arguments as: ptr in A0, size in D0, count in D1 and stream in A1 (and return a result in D0). When you included stdio.h it will be so whatever you pass to it.
When you do not include stdio.h another thing happens. As you call fwrite with fwrite(data, sizeof(*data), 5, myfile) compiler looks on argruments and see that function is called as fwrite(*, int, int, *). So what it do? It pass first pointer in A0, first int in D0, second int in D1 and second pointer in A1, so it what we need.
But when you try to call it as fwrite(data, sizeof(*data), 5.0, myfile), with count is of double type, compiler will try to pass count through stack, as it is not integer. But function require is in D1. Shit happens: D1 contain some garbage and not count, so further behaviour is unpredictable. But than you use prototype defined in stdio.h all will be ok: compiler automatically convert this argument to int and pass it as needed. It is not abstract example as double in arument may be just result of computation involving floating point numbers and you may just miss this assuming result is int.
Another example is variable argument function (form 3) like printf(char *fmt, ...). For it calling convention require last named argument (fmt here) to be passed through stack regardess of its type. So, then you call printf("%d", 10) it will put pointer to "%d" and number 10 on stack and call function as need.
But when you do not include stdio.h comiler will not know that printf is vararg function and will suppose that printf("%d", 10) is calling to function with fixed arguments of type pointer and int. So MC68000 will place pointer to A0 and int to D0 instead of stack and result is again unpredictable.
There may be luck that arguments was previously on stack and occasionally read there and you get correct result... this time... but another time is will fail. Another luck is that compiler takes care if not declared function may be vararg (and somehow makes call compatible with both forms). Or all arguments in all forms are just passed through stack on your machine, so fixed, unknown and vararg forms are just called identically.
So: do not do this even you feel lucky and it works. Unknown fixed argument form is there just for compatibility with old code and is strictly discouraged to use.
Also note: C++ will not allow this at all, as it require function to be declared with known arguments.