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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
Related
This question already has answers here:
Is it better to use C void arguments "void foo(void)" or not "void foo()"? [duplicate]
(6 answers)
Closed 6 months ago.
So I get that a void function won't return a value, while a int one for example will return an integer. So void main(){} doesnt return, but int main(){return 0;} will return.
My question is, what is the difference between these 3 functions. I know the first one wont return a value, the second one will return an integer. But how about the third one? I know it returns an integer because I've tested it, so what does that (void) does? Why is is there?
void main(){}
int main(){return 0;};
int main(void){return 0;};
I'm a begginer so sorry if it sounded confusing... Thanks in advance!!
(The question is about the programming language C)
Between the parenthesis are the parameters to the function.
In a function definition, the proper way to designate a function that takes no parameters is to simply specify void for the parameter list.
An empty parameter list (in a definition) is also a way of saying the function takes no parameters, however this syntax is considered deprecated and should not be used in new programs.
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difference between int* i and int *i
(9 answers)
Why is the asterisk before the variable name, rather than after the type?
(12 answers)
Closed 3 years ago.
I have been trying to learn C as a programming language, and have been trying to solve sample problems on site like LeetCode in C programs. When I was reading over some of the skeleton code that was provided as a function header for a problem on LeetCode that I want to solve in C, the function header had asterisks post fixing some of the types, specifically like this:
int* twoSum(int* nums, int numSize, int target, int* returnSize) {
/* Code goes here */
}
After doing a fair bit of reading, I learned that prefixing a variable with an asterisk when declaring a variable reserves the variable as a pointer, but I have not been able to find anything about what it means when the type specifier itself is post fixed with an asterisk.
The spaces there don't matter.
int* nums is identical to int *nums. So are int * nums and int*nums.
All four of these declare nums as pointer to int.
It's a matter of style preference (though I wouldn't use that last one), with no effect on the generated code.
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:
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()
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What is this strange function definition syntax in C? [duplicate]
(6 answers)
Closed 8 years ago.
When reading some FreeBSD source code (See: radix.h lines 158-173), I found variable declarations that followed the "function heading" in the definition.
Is this valid in ISO C (C99)? when should this be done in production code instead of just declaring the variables within the "function heading?" Why is it being done here?
I refer to the function heading the string that looks like this: int someFunction(int i, int b) {
That looks like K&R (pre-ANSI) style. I don't think it's valid C99, but are they using C99? Joel
I think you are referring to the "old-fashioned" pre-ANSI way of declaring parameters in C. It looked something like this:
int foo(a, b)
int a,
int b
{
/* ... */
}
That may still be valid in C99, and will be accepted by compilers for backward-compatibility reasons, but it should be considered deprecated/archaic.
Er. Maybe I'm misunderstanding your question, but i and b in that snippet are parameters to the function. It's not some compact way of declaring variables within the function, like:
int someFunction() {
int i, b;
When you call someFunction, you pass it those arguments:
someFunction(1, 2); // `i` will be `1` and `b` `2` within `someFunction`