This question already has answers here:
How to wrap printf() into a function or macro?
(8 answers)
call printf using va_list
(4 answers)
Closed 4 years ago.
I am working on making a version of printf in an embedded system that will print over a variety of interfaces. I am using snprintf to create the string and then sending it out over the interface. My curiosity is what the "..." being passed into the function is. The printf that I am using looks like, this is taken from an embedded printf library
void u_printf(const char* format, ...)
In this case the ... is some number of assorted variables passed into the function. But if I want to write my own version of this how do I get those variables passed in so I can then pass them on to snprintf.
void u_printf(const char* format, ...){
int cx;
char buffer[100];
cx = x_snprintf(buffer, 100, format, ??);
// function to send buffer to interface goes here
}
So how do I go about getting the ... variables passed in to where the ?? mark is? Thanks!
Some quick help revealed that va_list is the answer here. When the ... is present we can grab the arguments passed in by creating va_list va variable. This can then be used or passed to vsnprintf()
Related
This question already has answers here:
An example of use of varargs in C
(4 answers)
Closed 3 months ago.
I'm trying to understand libv4l2, & came across with following function that contains three dots in the argument part. What does this ... mean?
int v4l2_ioctl (int fd, unsigned long int request, ...)
{
void *arg;
va_list ap;
int result, index, saved_err;
int is_capture_request = 0, stream_needs_locking = 0;
Also, my C knowledge is poor. So please, Explain Like I’m 5
I'm expecting to understand that does ... meant in a function,
I've done basic search about these three-dots, but only found something related with C's Macro (not good results with something related with C functions).
My long-term goal is to understand about this v4l2-ioctl function, (not like understand in and out) but understand briefly so I can use this v4l2-ioctl comfortably.
its a variadic function- a function that can take a variable number of arguments
you have to access them with va_start, va_arg and va_end
if u want to read more about it u can here:
http://publications.gbdirect.co.uk/c_book/chapter9/stdarg.html
This question already has answers here:
How do Variable length arguments work in C?
(6 answers)
Closed 1 year ago.
If I have a C function where one argument is "...", how do I determine if more than two arguments are being passed to that function? If the third argument being sent to my_func() would be int arg2, how do I access it?
#include <stdio.h>
int my_func(int arg0, int arg1, ...)
{
/* How can I determine if 2 or 3 arguemnts are being passed to
* this function? How do I access the third argument so I could
* print it out? */
return 0;
}
int main() {
my_func(1,2);
my_func(1,2,3);
return 0;
}
In C, you can't deduce how many arguments are sent to the function from the arguments alone. You need some supporting information, usually passed in one of the arguments themselves. E.g., for the well-known printf, the first argument is a format string, and parsing it will tell the function how many other arguments it should take and their types.
This question already has answers here:
How to achieve function overloading in C?
(14 answers)
Closed 3 years ago.
Is it ok to leave the parameters unused?
I'm trying to run the function like this:
void encoder();
void encoder(int argc, FILE* inputFP, FILE* outputFP);
Is there a need for a second function, one for dealing with stdio and one for dealing with files? When I try to run
void encoder(int argc, FILE* inputFP, FILE* outputFP); without any arguments I get errors:
error: too few arguments to function ‘void encoder(int, FILE*, FILE*)’
encoder();```
Is it ok to leave the parameters unused?
No, all arguments must be present and correct.
You should also be careful that you do not provide different prototypes for a function (or any symbol) within a given application.
If you really want to call it with no arguments, then the best approach is to have two functions and call the one you're after:
void encoder(void) {
encoder2(0, stdin, stdout);
}
void encoder2(int argc, FILE* inputFP, FILE* outputFP) {
/* ... */
}
If you want to have a variable number of arguments, then you could look at using the macros in stdarg.h (like printf() and friends), though you still won't be able to get away with zero arguments, and you'll need to be very careful with your usage.
As you've discovered, no it's not valid.
C lacks the support for default values, so the arguments would be quite useless. The solution to this particular problem is to use magic FILE * "files" that map to the standard input/output channels, i.e. stdin and stdout from <stdio.h>.
This question already has answers here:
Macro returning the number of arguments it is given in C? [duplicate]
(3 answers)
Closed 4 years ago.
It might be impossible to do but can we actually get the number of arguments a function's wants in C?
The goal is to make a function named call that requires a function's pointer and some arguments. But because we can't count the number of arguments in a variadic arguments function, I want to count how many arguments does the function's pointer requires.
Here's how I want to create it:
void func(int a, int b, char* c) {
// Do things
}
int call(void *f, ...) {
// Find how much arguments *f wants and loop over the variadic args
}
int main() {
call(func, 1, 2, "hello");
}
By the way, it needs to be at run-time.
Nothing in the C standard provides a facility for ascertaining the number of arguments a function requires, other than the fact that C implementations are free to provide extensions.
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
source code of c/c++ functions
I was wondering where I can find the C code that's used so that when I write printf("Hello World!"); in my C programm to know that it has to print that string to STDOUT. I looked in <stdio.h>, but there I could only find its prototype int printf(const char *format, ...), but not how it looks like internally.
Here's the GNU version of printf... you can see it passing in stdout to vfprintf:
__printf (const char *format, ...)
{
va_list arg;
int done;
va_start (arg, format);
done = vfprintf (stdout, format, arg);
va_end (arg);
return done;
}
See here.
Here's a link to vfprintf... all the formatting 'magic' happens here.
The only thing that's truly 'different' about these functions is that they use varargs to get at arguments in a variable length argument list. Other than that, they're just traditional C. (This is in contrast to Pascal's printf equivalent, which is implemented with specific support in the compiler... at least it was back in the day.)