I have to write a function that accepts a variable number of arguments, similar to the open() function
int open (const char *filename, int flags[, mode_t mode])
I had a look at the va_list functionality, but how can I find out how many arguments the user has supplied? I dont want to have a parameter in the function that indicates this.
Being a sort of newbie, where can I look how the open() function is implemented? How does open() know whether the user has supplied the mode argument or not?
Thanks in advance.
You cannot. The caller must either supply a trailing sentinel argument (e.g. a 0), or explicitly tell the function how many arguments to expect (e.g. the format fields in printf).
open knows to expect the third argument because you will have set the O_CREAT flag in the flags argument.
Copied straight from the GCC page.
Here is a complete sample function that accepts a variable number of arguments. The first argument to the function is the count of remaining arguments, which are added up and the result returned. While trivial, this function is sufficient to illustrate how to use the variable arguments facility.
#include <stdarg.h>
#include <stdio.h>
int
add_em_up (int count,...)
{
va_list ap;
int i, sum;
va_start (ap, count); /* Initialize the argument list. */
sum = 0;
for (i = 0; i < count; i++)
sum += va_arg (ap, int); /* Get the next argument value. */
va_end (ap); /* Clean up. */
return sum;
}
int
main (void)
{
/* This call prints 16. */
printf ("%d\n", add_em_up (3, 5, 5, 6));
/* This call prints 55. */
printf ("%d\n", add_em_up (10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10));
return 0;
}
What you're looking for is variadic function implementations. Here is a doc that will walk you through the entire chain.
The short answer to the question, "How does open know if it was passed two or three arguments?" is: it's magic. The long answer is that the implementation is allowed to do certain things that you are not (within standard C, that is).
Another example is main. It has two valid prototypes:
int main(void);
and
int main(int argc, char *argv[]); /* or equivalent */
The implementation must allow you to use any of the two, and "deduce", which is the one you used.
The answer to the question, "How can I do something similar in my code?", is: you can't, at least not within the constraints imposed by the C standard.
Related
I really wonder how printf executed. Is there a parameter array structure in C? Can i define my custom function like printf?
A special type va_list is used for using variable list arguements. read this.
You could use the va_arg macro. Here's an example
#include <stdio.h> /* printf */
#include <stdarg.h> /* va_list, va_start, va_arg, va_end */
int FindMax (int n, ...)
{
int i,val,largest;
va_list vl;
va_start(vl,n);
largest=va_arg(vl,int);
for (i=1;i<n;i++)
{
val=va_arg(vl,int);
largest=(largest>val)?largest:val;
}
va_end(vl);
return largest;
}
int main ()
{
int m;
m= FindMax (7,702,422,631,834,892,104,772);
printf ("The largest value is: %d\n",m);
return 0;
}
A program conforms to some specific ABI and the calling convention is defined by the abi.
A calling convention defines how parameters passed to a function, usually stored either in registers or/and on the stack.The function then retrieves the parameters accordingly and this is also for variadic functions.
Sure you can define variadic function yourself.
I have this function:
void func(int a, int b, char s, FILE *fp, int flag)
and i want to use the function's args based on the flag. For example:
func(num, NOTHING, NOTHING, file, 1)func(num, othernum, NOTHING, NOTHING, 2)and on the function this_is_function i want to have:
void func(int a, int b, char s, FILE *fp, int flag){
if(flag == 1){
/* use a and file */
}
if(flag == 2){
/* use a,b */
}
/* etc etc */
}
i would like to know if it is possible and how to do this!Thanks in advance :)
If by NOTHING you mean that you really want to omit that argument, I don't think it's possible the way you've outlined it.
The way this typically is done in C is through variable argument lists. That would mean you'd have to reformulate so that the flag goes first, since it decides the rest of the arguments:
void func(int flag, ...)
{
va_list args;
int num, othernum;
FILE *file;
va_start(args, flag);
if(flag == 1)
{
num = va_arg(args, int);
file = va_arg(args, FILE *);
}
else if(flag == 2)
{
num = va_arg(args, int);
othernum = va_args(args, int);
}
va_end(args);
/* Inspect `flag` again, and do things with the values we got. */
}
Then you can use the function like so:
func(1, 42, a_file);
or
func(2, 17, 4711);
This of course requires lots of care, since you're not getting a lot of help from the compiler anymore to match the values provided in the call to what the function expects.
I would recommend restructuring it into different top-level functions instead, that call a common "worker" function with the proper arguments:
func_mode1(42, a_file);
func_mode2(17, 4711);
these could then call func() with the proper flag value, filling in suitable defaults for arguments that don't apply (such as NULL for a non-used file pointer).
You can pass 0 to int and char and NULL for pointers when you call the function and the arguments are not used.
func(num, 0, 0, file, 1)
func(num, othernum, 0, NULL, 2)
You can also use variadic functions. Variadic functions are functions with a variable number of arguments.
Create seperate functions for each of the cases you are trying to handle. This would yield to more readable code. State your intention through your code rather than hack around it.
Your code should work fine and moreover try it out once and check for any errors. I am sure as no error should occur. You can also try using switch case in the func with flag as the choice for switch. I am sure you are not going to use Nothing and othersum in your code.
I have a function
void func(int x, char *str, ...)
{
...
}
I am invoking it as follows:
func(1, "1", "2", "3");
How can I print the values of all the extra arguments (2, 3) in function?
From the man page of STDARG about the use of va_arg to get the next argument:
If there is no next argument, or
if type is not compatible with the
type of the actual next argument (as
promoted according to the default
argument promotions), random errors
will occur.
Hence, unless you want random errors to creep in, you should know the number of arguments beforehand.
Even so, if you want to throw caution to the winds, you could try:
void func(int x,char *str, ...)
{
va_list al;
va_start(al,str);
while(x>0)
{
str=va_arg(al,char *);
--x;
}
while(str != NULL)
{
printf("%s ",str);
str=va_arg(al,char *);
}
va_end(al);
}
With,
func(1,"1","2","3");
I got the output,
2 3 U��WVS�O
If it satisfies your purpose, you could pick out the required number of arguments from this output.
It is customary with variable arguments to pass a string which describes the variable arguments, e.g. printf( char *format_string, ... );
This is a solution - and the customary solution - to your problem.
Pass an additional argument which describes the variable arguments, and then use that information to process the variable arguments.
So, if you receive a printf-like format string and it is "%d%u", you know you have an int, followed by an unsigned int.
lookup va_list in this site. Example: What is ellipsis operator in c
PHP has a func_get_args() for getting all function arguments, and JavaScript has the functions object.
I've written a very simple max() in C
int max(int a, int b) {
if (a > b) {
return a;
} else {
return b;
}
}
I'm pretty sure in most languages you can supply any number of arguments to their max() (or equivalent) built in. Can you do this in C?
I thought this question may have been what I wanted, but I don't think it is.
Please keep in mind I'm still learning too. :)
Many thanks.
You could write a variable-arguments function that takes the number of arguments, for example
#include <stdio.h>
#include <stdarg.h>
int sum(int numArgs, ...)
{
va_list args;
va_start(args, numArgs);
int ret = 0;
for(unsigned int i = 0; i < numArgs; ++i)
{
ret += va_arg(args, int);
}
va_end(args);
return ret;
}
int main()
{
printf("%d\n", sum(4, 1,3,3,7)); /* prints 14 */
}
The function assumes that each variable argument is an integer (see va_arg call).
Yes, C has the concept of variadic functions, which is similar to the way printf() allows a variable number of arguments.
A maximum function would look something like this:
#include <stdio.h>
#include <stdarg.h>
#include <limits.h>
static int myMax (int quant, ...) {
va_list vlst;
int i;
int num;
int max = INT_MIN;
va_start (vlst, quant);
for (i = 0; i < quant; i++) {
if (i == 0) {
max = va_arg (vlst, int);
} else {
num = va_arg (vlst, int);
if (num > max) {
max = num;
}
}
}
va_end (vlst);
return max;
}
int main (void) {
printf ("Maximum is %d\n", myMax (5, 97, 5, 22, 5, 6));
printf ("Maximum is %d\n", myMax (0));
return 0;
}
This outputs:
Maximum is 97
Maximum is -2147483648
Note the use of the quant variable. There are generally two ways to indicate the end of your arguments, either a count up front (the 5) or a sentinel value at the back.
An example of the latter would be a list of pointers, passing NULL as the last. Since this max function needs to be able to handle the entire range of integers, a sentinel solution is not viable.
The printf function uses the former approach but slightly differently. It doesn't have a specific count, rather it uses the % fields in the format string to figure out the other arguments.
In fact, this are two questions. First of all C99 only requires that a C implementation may handle at least:
127 parameters in one function
definition
127 arguments in one function call
Now, to your real question, yes there are so-called variadic functions and macros in C99. The syntax for the declaration is with ... in the argument list. The implementation of variadic functions goes with macros from the stdarg.h header file.
here is a link to site that shows an example of using varargs in c Writing a ``varargs'' Function
You can use the va_args function to retrieve the optional arguments you pass to a function. And using this you can pass 0-n optional parameters. So you can support more then 2 arguments if you choose
Another alternative is to pass in an array, like main(). for example:
int myfunc(type* argarray, int argcount);
Yes, you can declare a variadic function in C. The most commonly used one is probably printf, which has a declaration that looks like the following
int printf(const char *format, ...);
The ... is how it declares that it accepts a variable number of arguments.
To access those argument it can uses va_start, va_arg and the like which are typically macros defined in stdarg.h. See here
It is probably also worth noting that you can often "confuse" such a function. For example the following call to printf will print whatever happens to be on the top of the stack when it is called. In reality this is probably the saved stack base pointer.
printf("%d");
C can have functions receive an arbitrary number of parameters.
You already know one: printf()
printf("Hello World\n");
printf("%s\n", "Hello World");
printf("%d + %d is %d\n", 2, 2, 2+2);
There is no max function which accepts an arbitrary number of parameters, but it's a good exercise for you to write your own.
Use <stdarg.h> and the va_list, va_start, va_arg, and va_end identifiers defined in that header.
http://www.kernel.org/doc/man-pages/online/pages/man3/stdarg.3.html
Converting a C++ lib to ANSI C and it seems like though ANSI C doesn't support default values for function variables or am I mistaken?
What I want is something like
int funcName(int foo, bar* = NULL);
Also, is function overloading possible in ANSI C?
Would need
const char* foo_property(foo_t* /* this */, int /* property_number*/);
const char* foo_property(foo_t* /* this */, const char* /* key */, int /* iter */);
Could of course just name them differently but being used to C++ I kinda used to function overloading.
No, Standard C does not support either. Why do you feel you need to convert your C++ code to C? That could get quite tricky - I'd have thought writing wrappers would be the way to go, if your C++ must be callable from C.
Nevertheless I found a "trick" to do so if you use GCC (edit December 2020) or any compiler compatible with C++2a -yes it works with 'plain C' since it is a pre-compiler trick-.
GCC has a handy ## extension on variadic macro that allows you to simulate a default argument.
The trick has limitations: it works only for 1 default value, and the argument must be the last of you function parameters.
Here is a working example.
#include <stdio.h>
#define SUM(a,...) sum( a, (5, ##__VA_ARGS__) )
int sum (a, b)
int a;
int b;
{
return a + b;
}
main()
{
printf("%d\n", SUM( 3, 7 ) );
printf("%d\n", SUM( 3 ) );
}
In this case, I define SUM as a call to sum with the default second argument being 5.
If you call with 2 arguments (first call in main), it would be prepocessed as:
sum( 3, (5, 7) );
This means:
1st argument is 3
second argument is the result of the sequence (5, 7)... which is
obviously 7!
As gcc is clever, this has no effect on runtime as the first member of the sequence is a constant and it is not needed, it will simply be discarded at compile time.
If you call with only one argument, the gcc extension will remove the VA_ARGS AND the leading coma. So it is preprocessed as:
sum( 3, (5 ) );
Thus the program gives the expected output:
10
8
So, this does perfectly simulate (with the usual macro limitations) a function with 2 arguments, the last one being optional with a default value applied if not provided.
Edit
-a) It does also work with CLANG (and possibly other compilers)
-b) A version that does NOT complain about unused arguments:
#define DEF_OR_ARG(z,a,arg,...) arg
#define SUM(a,...) sum( a, DEF_OR_ARG(,##__VA_ARGS__,__VA_ARGS__,5))
[Edit - October 2020] :
You could also try the new __VA_OPT__ that was standardised with c++2a (and should work in plain C too) instead of ## which is a gcc extension. Typical usage is __VA_OPT__(,) that would add the coma when the argument list is non-empty and otherwise outputs nothing.
[Edit - December 2020] :
So the above trick, with __VA_OPT__, becomes:
#define DEF_OR_ARG(value,...) value
#define SUM(a,...) sum( a, DEF_OR_ARG(__VA_ARGS__ __VA_OPT__(,) 5))
Unlike the 'sequence trick' that might complain for unused variables, this only involves the pre-compiler and is more readable.
When SUM is called with only one argument, ... is empty and __VA_OPT__ does not output anything, thus DEF_OR_ARG(__VA_ARGS__ __VA_OPT(,) 5) becomes DEF_OR_ARG( 5)
When SUM is called with a second argument, ... is this second argument and __VA_OPT__ expands to the value given which is a coma. In that case
DEF_OR_ARG(__VA_ARGS__ __VA_OPT(,) 5) becomes DEF_OR_ARG(second_argument_of_SUM , 5)
Now the expansion of DEF_OR_ARG happens. This one is easy since it considers only the first argument and just discards the rest.
So, when SUM was called with no second argument (first case above), the first argument to DEF_OR_ARG is our default value. In the case there was a second argument to SUM, it becomes the first argument to DEF_OR_ARG that will expand to that and discard the default which is now second argument.
Try this.
#include <stdio.h>
#include <stdarg.h>
/* print all non-negative args one at a time;
all args are assumed to be of int type */
void printargs(int arg1, ...)
{
va_list ap;
int i;
va_start(ap, arg1);
for (i = arg1; i >= 0; i = va_arg(ap, int))
printf("%d ", i);
va_end(ap);
putchar('\n');
}
int main(void)
{
printargs(5, 2, 14, 84, 97, 15, 24, 48, -1);
printargs(84, 51, -1);
printargs(-1);
printargs(1, -1);
return
0;
}
There is a way to support as many default parameters you need, just use a structure.
// Populate structure with var list and set any default values
struct FooVars {
int int_Var1 = 1; // One is the default value
char char_Command[2] = {"+"};
float float_Var2 = 10.5;
};
struct FooVars MainStruct;
//...
// Switch out any values needed, leave the rest alone
MainStruct.float_Var2 = 22.8;
Myfunc(MainStruct); // Call the function which at this point will add 1 to 22.8.
//...
void Myfunc( struct FooVars *MyFoo ) {
switch(MyFoo.char_Command) {
case '+':
printf("Result is %i %c %f.1 = %f\n" MyFoo.int_Var1, MyFoo.char_Command, MyFoo.float_Var2, (MyFoo.float_Var2 + MyFoo.int_Var1);
break;
case '*':
// Insert multiply here, you get the point...
break;
case '//':
// Insert divide here...
break;
}
}
As far as I know ANSI C doesn't directly support function overloading or default arguments. The standard substitute for overloading is adding suffixes to the function name indicating the argument types. For example, in OpenGL, a "3fv" suffix to a function name means the function takes a vector of three floats.
Default arguments can be viewed as a special case of function overloading.
Neither of default values or function overloading exists in ANSI C, so you'll have to solve it in a different way.
You can't so easily since C does not support them. The simpler way to get "fake overloading" is using suffixes as already said... default values could be simulated using variable arguments function, specifying the number of args passed in, and programmatically giving default to missing one, e.g.:
aType aFunction(int nargs, ...)
{
// "initialization" code and vars
switch(nargs)
{
case 0:
// all to default values... e.g.
aVar1 = 5; // ...
break;
case 1:
aVar1 = va_arg(arglist, int); //...
// initialize aVar2, 3, ... to defaults...
break;
// ...
}
}
Also overloading can be simulated using var args with extra informations to be added and passed and extracode... basically reproducing a minimalist object oriented runtime ...
Another solution (or indeed the same but with different approach) could be using tags: each argument is a pair argument type + argument (an union on the whole set of possible argument type), there's a special terminator tag (no need to specify how many args you're passing), and of course you always need "collaboration" from the function you're calling, i.e. it must contain extra code to parse the tags and choose the actual function to be done (it behaves like a sort of dispatcher)
You'll have to declare each C++ overloaded function differently in C because C doesn't do name mangling. In your case "foo_property1" "foo_property2".
i think u can use a function with variable arguments here is my example
#include <stdarg.h>
#include <stdio.h>
void baz( int flag, ... )
{
va_list ap;
char *bar = "baz"; /* default value */
va_start( ap, flag );
if ( flag == 1 )
bar = va_arg( ap, char * );
va_end( ap );
printf( "%s\n", bar );
}
int main( void )
{
baz( 0 );
baz( 1, "foo");
baz( 2 );
baz( 1, "bar");
return 0;
}
the output is
baz
foo
baz
bar
if u look for example man 2 open they say
SYNOPSIS
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
int open(const char *pathname, int flags);
int open(const char *pathname, int flags, mode_t mode);
int creat(const char *pathname, mode_t mode);
int openat(int dirfd, const char *pathname, int flags);
int openat(int dirfd, const char *pathname, int flags, mode_t mode);
but mode is actually a ... argument