Strange function definition [duplicate] - c

I'm relatively new to C. I've come across a form of function syntax I've never seen before, where the parameter types are defined after that parameter list. Can someone explain to me how it is different than the typical C function syntax?
Example:
int main (argc, argv)
int argc;
char *argv[];
{
return(0);
}

That's the old-style syntax for parameter lists, which is still supported. In K&R C you could also leave off the type declarations and they would default to int. i.e.
main(argc, argv)
char *argv[];
{
return 0;
}
would be the same function.

What's also interesting is the calling convention difference of functions with, and functions without a prototype. Consider an old style definition:
void f(a)
float a; {
/* ... */
}
In this case, the calling convention is that all arguments are promoted before being passed to the function (for example, a float argument is first promoted to double, before being passed). So if f receives a double but the parameter has type float (which is perfectly valid) the compiler has to emit code that converts the double to a float prior to executing the function's body.
If you include a prototype, the compiler does not do such automatic promotions anymore and any data passed is converted to the types of the parameters of the prototype as if by assignment. So the following is not legal and results in undefined behavior:
void f(float a);
void f(a)
float a; {
}
In this case, the function's definition would convert the submitted parameter from double (the promoted form) to float because the definition is old-style. But the parameter was submitted as a float, because the function has a prototype. For example, clang gives
main.c:3:9: warning: promoted type 'double' of K&R function parameter is not compatible with the parameter type 'float' declared in a previous prototype [-Wknr-promoted-parameter]
Your options of solving the contradictions are the two following:
// option 1
void f(double a);
void f(a)
float a; {
}
// option 2
// this declaration can be put in a header, but is redundant in this case,
// since the definition exposes a prototype already if both appear in a
// translation unit prior to the call.
void f(float a);
void f(float a) {
}
Option 2 should be preferred if you have the choice because it gets rid of the old style definition up front. If such contradicting function types for a function appears in the same translation unit, the compiler will usually tell you (but is not required). If such contradictions appear over multiple translation units, the error will possibly go unnoticed and can result in hard to predict bugs. It is best to avoid these old style definitions.

This is the so caller K&R style or old-style declaration.
Note, that this declaration is significantly different from the modern declaration. K&R declaration does not introduce a prototype for the function, meaning that it doesn't expose the types of the parameters to the outside code.

There is no difference, it is just that that is the old syntax for function declarations in C -- it was used pre ANSI. Never write such code unless you plan to give it to your friends from the 80's. Also, never depend upon implicit type assumptions (as another answer seems to suggest)

While the old syntax for function definition still works (with warnings, if you ask your compiler), using them does not provide function prototypes.
Without function prototypes the compiler will not check if the functions are called correctly.
#include <stdio.h>
int foo(c)
int c;
{ return printf("%d\n", c); }
int bar(x)
double x;
{ return printf("%f\n", x); }
int main(void)
{
foo(42); /* ok */
bar(42); /* oops ... 42 here is an `int`, but `bar()` "expects" a `double` */
return 0;
}
When the program is run, the output on my machine is
$ gcc proto.c
$ gcc -Wstrict-prototypes proto.c
proto.c:4: warning: function declaration isn’t a prototype
proto.c:10: warning: function declaration isn’t a prototype
$ ./a.out
42
0.000000

Its just the same but old fashion. You probably found it is some old, legacy code.

Old or not, I would argue what is old and what not.. like the pyramids are ancient, but none of today’s so-called scientists have a clue how they were made. Looking back, old programs still work today without memory leaks, but these "new" programs tend to fail more than often. I see a trend here.
Probably they saw functions as structs which have an executable body. Knowledge of ASM is needed here to solve the mystery.
Edit, found a macro which indicates you do not need to supply argument names at all.
#ifndef OF /* function prototypes */
# ifdef STDC
# define OF(args) args
# else
# define OF(args) ()
# endif
#endif
#ifndef Z_ARG /* function prototypes for stdarg */
# if defined(STDC) || defined(Z_HAVE_STDARG_H)
# define Z_ARG(args) args
# else
# define Z_ARG(args) ()
# endif
#endif
Here is an usage example, library is zlib-1.2.11.
ZEXTERN int ZEXPORT deflate OF((z_streamp strm, int flush));
So my second guess would be for function overloading, otherwise these arguments had no use. One concrete function, and now infinite amount of functions with same name.

Related

Why does a function that returns int not need a prototype?

I tried this program on DevC++ that would call two different types of function, one returning int and one returning char. I'm confused why the int function doesn't need a prototype, while the char one and any other type of function does.
#include <stdio.h>
//int function1();
char function2() ;
int main (){
int X = function1() ;
char Y = function2() ;
printf("%d", X) ;
printf ("%c", Y) ;
return 0 ;
}
int function1(){
return 100 ;
}
char function2(){
return 'B' ;
}
The output:
100B
If I remove the prototype for the char function, it results in:
[Error] conflicting types for 'function2'
[Note] previous implicit declaration of 'function2' was here
In the old days of C any function that was not declared explicitely was supposed to return int type when you call it.
If the compiler then finds the function implementation and sees an int return type, everything is fine.
If the function returns anything else than int you get the error message as you saw with the second function.
This implicit int type declaration was removed from the standard with C99. Now you should at least get a warning from your compiler when you use a function without prototype.
If you did not get any diagnostic message for first funcion, you should turn up warning level in your compiler or switch to at least C99 mode instead of ancient versions.
Edit:
Empty parameter lists in funcion definitions is a deprecated feature as well.
You should not use it.
Always provide prototype for every function with return type and parameter list.
If a function is used before it is declared, the usage becomes an implicit declaration of the function. When a function f is implicitly defined, the definition given to it is int f(), i.e. a function which accepts an unspecified number of arguments and returns an int.
This implicit definition of a function matches the actual definition of function1 but not function2. So calling function1 this way gives no error but attempting to call function2 this way results in the implicit definition not matching the actual definition, giving an error.
This behavior goes back to pre-standardized versions of C where all objects (and a function's return type) had a default type of int if not declared. This was still present in the C89 standard but removed in the C99 standard, although some compilers such as gcc still support this obsolescent usage as an extension.
It's just an ancient relic from when C was first designed. It was actually removed as early as C99, but many compilers still support this type of declaration. But it's not recommended to use it.
I'm not sure if there were any real rationale behind it, but C was heavily inspired by the language B. And in B you did not have to specify the return type for functions. That actually made perfect sense, because there was only one type, word.
In the same way you did not have to specify the type of variables either. You only specified if it had automatic or static storage. And that's where the completely useless keyword auto in C comes from. It does not mean the same as in C++. It just means "not static".

What happens to the unspecified arguments in function()? [duplicate]

This question already has answers here:
What does an empty parameter list mean? [duplicate]
(5 answers)
Accessing the parameters passed to a function with an empty parameter list in C
(2 answers)
Closed 2 years ago.
I have been reading the difference between function() and function(void) in C, and I came to know that
an empty parameter list in a function declaration indicates that the
function takes an unspecified number of parameters
So I ran this code:
#include <stdio.h>
void fun();
int main(void)
{
fun(12, 13.22, 1234567890987654321, "wow", 'c');
}
void fun()
{
printf("What happened to those arguments?");
}
I don't understand why C allows this. All the posts I've read so far related to this say things like it's bad practice to use it, it is obsolescent, etc. Also from the above code, I think the type of those arguments is also unspecified. So I just want to know the reason behind "unspecified arguments of unspecified type":
What can be done with those arguments?
Is it possible to access those arguments within the function?
The reason for supporting the notation is historical. Before the first C standard (C89/C90), you couldn't use prototypes in C; prototypes were one of the biggest and most important features of Standard C. All function declarations, therefore, were written the 'empty parentheses' style (when they were written at all; most functions that returned int were not declared at all). The type void was also added in C89/C90, though some compilers supported it before the standard was finalized.
Because it was crucial to the success of C89/C90 that existing code should mostly continue to work, the empty parentheses style had to be allowed by the standard. So, your code might have been written in pre-standard C as:
#include <stdio.h>
int fun(); /* This declaration would probably have been absent */
int main(void)
{
fun(12, 13.22, 1234567, "wow", 'c');
return 0; /* This was required until C99 to give reliable exit status */
}
fun(i, d, l, s, c) /* No return type - implicitly returns int */
long l; /* Defined out of sequence - bad style, but legal */
char c; /* Passed as int; converted to char in function */
char *s; /* Should define all pointer arguments */
double d; /* No definition of i; it was an int by default */
{
printf("This is what happened to those arguments:\n");
printf("i = %d\n", i);
printf("d = %f\n", d);
printf("l = %ld\n", l);
printf("s = [%s]\n", s);
printf("c = %c\n", c);
/* No return statement - don't use the value from the function */
}
For the curious: you could omit the char *s; line in the function definition, and it still compiled and produced the same output. It was a bad idea to try that, though. You could replace the line int fun(); with static fun(); and the code compiles cleanly when no diagnostics are requested.
You get no warnings even now if you compile this file (old31.c) with GCC 9.3.0 using:
$ gcc -std=c90 -o old31 old31.c
$
Your example as written is skirting around the backwards compatibility provisions. Using void means it was new code (it would not have been valid in many pre-standard C compilers because it used void). And new code should not exploit the backwards-compatibility provisions without a good reason. That was true in 1991 as well as in the current millennium (but in 1991, there were a lot more good reasons to exploit the backwards-compatibility provisions). Good pre-standard code usually listed all parameters in the order they were used. Omitted definitions and out of sequence definitions were not entirely satisfactory.
You asked:
What can be done with those arguments?
In the code in the question, nothing can be done with the arguments. The caller pushes the values onto the stack, and pops them off when the function returns. The called function is unaware of their existence and can do nothing with them.
Is it possible to access those arguments within the function?
No — not using any standard mechanism.
There is a difference between a function declaration and a function definition when there is an empty parameter list.
Section 6.7.6.3p14 of the C standard states:
An identifier list declares only the identifiers of the parameters of
the function. An empty list in a function declarator that is
part of a definition of that function specifies that the
function has no parameters. 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.
What this means is that this declaration:
void fun();
Means fun takes an unknown number of parameters. While this definition:
void fun()
{
printf("What happened to those arguments?");
}
Means that fun takes no parameters. So this function call:
fun(12, 13.22, 1234567890987654321, "wow", 'c');
Is invalid and invoked undefined behavior because the number of parameters in the call don't match the actual number of parameters. This is spelled out in section 6.5.2.2p6 regarding the function call operator ():
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.
As for why this is allowed, it is legacy behavior that goes back to pre-standardized versions of C where the type of variables and the return type of functions defaulted to int and the method of declaring functions differed from what they are now.

Why do we need prototype in function?

I'm studying C and my book is explaining how to "prototype a function" so we can use it before we define it.
The point is that I can't imagine a situation in which is needed to use a function before have defined it, why can't we just define it at the beginning?
Can you please provide an example in which is strictly necessary to prototype a function (if it exists)?
Consider the program below:
#include <stdio.h>
void add(int, int);
int main() {
add(5, 3);
return 0;
}
void add(int a, int b) {
printf("%d", a+b);
}
Here, when you call add(5, 3) from main(), the compiler knows what add() is - the parameters it takes and the return type, thanks to the statement void add(int, int) in line 3. If this statement is commented out, then the compiler won't know what add() is and would give you an error. Thus, line 3 tells the compiler that you have defined the function add() later, which makes it perfectly eligible to be used in main(). Alternately, if you write the add() function before main(), then the compiler knows what add() is when you call it; so a function prototype is not required in this case.
Hope this is useful.
One use-case is when you want to take a function pointer to a function defined in another source file. (Or hand-written in asm). Taking a function-pointer always requires the function to be declared ahead of time, either by a definition or a prototype.
C89 (and earlier) allows implicit declaration of functions (use them even if there's no definition or prototype earlier in the compilation unit (this .c + anything you #include).
So you can compile this in a file by itself, in C89 (or with a warning from most compilers in C99 / C11 mode)
int foo() {
return bar( 123 ); /* implicit declaration of int bar() */
}
But you can't use &bar as a function pointer
/* int bar(int); necessary if you haven't defined int bar(int x) { ... } */
void *take_fptr() {
return &bar; // error: 'bar' undeclared from GCC
}
You can also construct cases where a declaration is needed to get args passed correctly, e.g. if you want your function to take a float arg, you need it to be declared, either by a definition or prototype. The "Default Argument Promotions" in the ISO C standard will promote float to double when passing an arg to an unprototyped function, or to the ... part of a variadic function like printf.
(That's why %f prints a double, not a float, and why %lf is also a double if it's supported at all. Also why functions like putchar(int) take their char arg as an int: default promotions of narrow integer types up to int. Some parts of the C standard library date back to very early C, before the language supported prototypes!)
Or for example, if your function is expecting a long long, but the caller writes bar( 123 ), the caller will only infer int, which may be passed differently from a long long. But you can fix that by writing bar( 123LL ), or bar( (long long)123 )
Or if the return type isn't int (or void), then you also need the compiler to know about the function via a declaration. Implicitly-declared functions are assumed to return int. Depending on the calling convention and architecture, you might get away with that for a short return type, or on some machines even for a char*, e.g. on 32-bit machines where intptr_t is int. (Early C relied on that.)
But in general, e.g. on machines with 64-bit pointers, implicit-int return isn't going to work. Nor will it work for a double return value, or a struct.
The definition of the function may not be known at that time. Imagine a program that calls malloc to allocate memory. At compile time, nobody has any idea what allocator the program will be using at run time. So how can you define the function then?

C function weird syntax [duplicate]

I'm relatively new to C. I've come across a form of function syntax I've never seen before, where the parameter types are defined after that parameter list. Can someone explain to me how it is different than the typical C function syntax?
Example:
int main (argc, argv)
int argc;
char *argv[];
{
return(0);
}
That's the old-style syntax for parameter lists, which is still supported. In K&R C you could also leave off the type declarations and they would default to int. i.e.
main(argc, argv)
char *argv[];
{
return 0;
}
would be the same function.
What's also interesting is the calling convention difference of functions with, and functions without a prototype. Consider an old style definition:
void f(a)
float a; {
/* ... */
}
In this case, the calling convention is that all arguments are promoted before being passed to the function (for example, a float argument is first promoted to double, before being passed). So if f receives a double but the parameter has type float (which is perfectly valid) the compiler has to emit code that converts the double to a float prior to executing the function's body.
If you include a prototype, the compiler does not do such automatic promotions anymore and any data passed is converted to the types of the parameters of the prototype as if by assignment. So the following is not legal and results in undefined behavior:
void f(float a);
void f(a)
float a; {
}
In this case, the function's definition would convert the submitted parameter from double (the promoted form) to float because the definition is old-style. But the parameter was submitted as a float, because the function has a prototype. For example, clang gives
main.c:3:9: warning: promoted type 'double' of K&R function parameter is not compatible with the parameter type 'float' declared in a previous prototype [-Wknr-promoted-parameter]
Your options of solving the contradictions are the two following:
// option 1
void f(double a);
void f(a)
float a; {
}
// option 2
// this declaration can be put in a header, but is redundant in this case,
// since the definition exposes a prototype already if both appear in a
// translation unit prior to the call.
void f(float a);
void f(float a) {
}
Option 2 should be preferred if you have the choice because it gets rid of the old style definition up front. If such contradicting function types for a function appears in the same translation unit, the compiler will usually tell you (but is not required). If such contradictions appear over multiple translation units, the error will possibly go unnoticed and can result in hard to predict bugs. It is best to avoid these old style definitions.
This is the so caller K&R style or old-style declaration.
Note, that this declaration is significantly different from the modern declaration. K&R declaration does not introduce a prototype for the function, meaning that it doesn't expose the types of the parameters to the outside code.
There is no difference, it is just that that is the old syntax for function declarations in C -- it was used pre ANSI. Never write such code unless you plan to give it to your friends from the 80's. Also, never depend upon implicit type assumptions (as another answer seems to suggest)
While the old syntax for function definition still works (with warnings, if you ask your compiler), using them does not provide function prototypes.
Without function prototypes the compiler will not check if the functions are called correctly.
#include <stdio.h>
int foo(c)
int c;
{ return printf("%d\n", c); }
int bar(x)
double x;
{ return printf("%f\n", x); }
int main(void)
{
foo(42); /* ok */
bar(42); /* oops ... 42 here is an `int`, but `bar()` "expects" a `double` */
return 0;
}
When the program is run, the output on my machine is
$ gcc proto.c
$ gcc -Wstrict-prototypes proto.c
proto.c:4: warning: function declaration isn’t a prototype
proto.c:10: warning: function declaration isn’t a prototype
$ ./a.out
42
0.000000
Its just the same but old fashion. You probably found it is some old, legacy code.
Old or not, I would argue what is old and what not.. like the pyramids are ancient, but none of today’s so-called scientists have a clue how they were made. Looking back, old programs still work today without memory leaks, but these "new" programs tend to fail more than often. I see a trend here.
Probably they saw functions as structs which have an executable body. Knowledge of ASM is needed here to solve the mystery.
Edit, found a macro which indicates you do not need to supply argument names at all.
#ifndef OF /* function prototypes */
# ifdef STDC
# define OF(args) args
# else
# define OF(args) ()
# endif
#endif
#ifndef Z_ARG /* function prototypes for stdarg */
# if defined(STDC) || defined(Z_HAVE_STDARG_H)
# define Z_ARG(args) args
# else
# define Z_ARG(args) ()
# endif
#endif
Here is an usage example, library is zlib-1.2.11.
ZEXTERN int ZEXPORT deflate OF((z_streamp strm, int flush));
So my second guess would be for function overloading, otherwise these arguments had no use. One concrete function, and now infinite amount of functions with same name.

Warning/error "function declaration isn't a prototype"

I have a library I created,
File mylib.c:
#include <mylib.h>
int
testlib() {
printf("Hello, World!\n");
return (0);
}
File mylib.h:
#include <stdio.h>
extern int testlib();
In my program, I've attempted to call this library function:
File myprogram.c:
#include <mylib.h>
int
main (int argc, char *argv[]) {
testlib();
return (0);
}
When I attempt to compile this program I get the following error:
In file included from myprogram.c:1
mylib.h:2 warning: function declaration isn't a prototype
I'm using: gcc (GCC) 3.4.5 20051201 (Red Hat 3.4.5-2)
What is the proper way to declare a function prototype?
In C int foo() and int foo(void) are different functions. int foo() accepts an arbitrary number of arguments, while int foo(void) accepts 0 arguments. In C++ they mean the same thing. I suggest that you use void consistently when you mean no arguments.
If you have a variable a, extern int a; is a way to tell the compiler that a is a symbol that might be present in a different translation unit (C compiler speak for source file), don't resolve it until link time. On the other hand, symbols which are function names are anyway resolved at link time. The meaning of a storage class specifier on a function (extern, static) only affects its visibility and extern is the default, so extern is actually unnecessary.
I suggest removing the extern, it is extraneous and is usually omitted.
Quick answer: change int testlib() to int testlib(void) to specify that the function takes no arguments.
A prototype is by definition a function declaration that specifies the type(s) of the function's argument(s).
A non-prototype function declaration like
int foo();
is an old-style declaration that does not specify the number or types of arguments. (Prior to the 1989 ANSI C standard, this was the only kind of function declaration available in the language.) You can call such a function with any arbitrary number of arguments, and the compiler isn't required to complain -- but if the call is inconsistent with the definition, your program has undefined behavior.
For a function that takes one or more arguments, you can specify the type of each argument in the declaration:
int bar(int x, double y);
Functions with no arguments are a special case. Logically, empty parentheses would have been a good way to specify that a function takes no arguments, but that syntax was already in use for old-style function declarations, so the ANSI C committee invented a new syntax using the void keyword:
int foo(void); /* foo takes no arguments */
A function definition (which includes code for what the function actually does) also provides a declaration. In your case, you have something similar to:
int testlib()
{
/* code that implements testlib */
}
This provides a non-prototype declaration for testlib. As a definition, this tells the compiler that testlib has no parameters, but as a declaration, it only tells the compiler that testlib takes some unspecified but fixed number and type(s) of arguments.
If you change () to (void) the declaration becomes a prototype.
The advantage of a prototype is that if you accidentally call testlib with one or more arguments, the compiler will diagnose the error.
(C++ has slightly different rules. C++ doesn't have old-style function declarations, and empty parentheses specifically mean that a function takes no arguments. C++ supports the (void) syntax for consistency with C. But unless you specifically need your code to compile both as C and as C++, you should probably use the () in C++ and the (void) syntax in C.)
Try:
extern int testlib(void);

Resources