Passing anonymous parameters to functions in C [closed] - c

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I can pass arguments to a function in C in the following manner without any warnings or errors using gcc on ubuntu-64bit:
void func(char* test){
// do something here
}
void main() {
func((char*) "smaps");
}
I did not find a lot of examples about passing anonymous parameters to functions in C apart from one comment on a similar question about C++. I am thinking of using this technique (a lot) in my code which would be compiled on multiple AMD64/ARM devices as they become available (with mostly debian-based OSes). My question is, is this compiler-specific?
Output of gcc --version:
$ gcc --version
gcc (Ubuntu 6.3.0-12ubuntu2) 6.3.0 20170406
UPDATE: Even though I think 4386427's was good enough for me, I would explain a bit more about the question as it was put on hold. I have a script which makes several calls to a specific function which takes in a char** as argument. I was of the view that I would need to explicitly declare a char** separately and pass it to the function by name every time I made a call to the function. I thought it would be more convenient to declare a string and pass it to the function at the same time; something like:
func( (char**) {"first_string", "second_string"} ).
My initial concern was is this allowed in C generally or is this a compiler-specific feature? 4386427's answer suggests that it is not compiler-specific, however I should be careful that the char** passed to the function would be immutable.

Answer 1

All parameters in C are passed by value. No information about the original object from which the value came is passed to the called function. In this sense, all C arguments are anonymous; no information about the identifier used in the calling function is passed.
In func((char*) "smaps");, the string literal "smaps" is converted to a pointer to char. Only the value of the pointer is passed to func. This is standard C.
(If you do want to pass information about an object to a function, you must do that manually. For example, you can take the address of an object and pass the resulting pointer to the function. Or you can pass the number of elements in an array to a function, along with the address of the first element.)
(In C, string literals such as "smaps" are automatically converted to a pointer to their first character, so you do not need to manually convert them with a char * cast. [This automatic conversion does not occur when the string literal is the operand of sizeof, _Alignof, or & or when it is used to initialize an array.])

Answer 2

It depends on what you are doing in the function. As long as you don't try to modify what test is pointing to, there is no problem. To indicate that it can't be changed, it is good to add const.
Like this the code is fine:
#include <stdio.h>
void func(const char* test){
printf("%s\n", test);
}
int main(void) {
func("smaps1");
func("smaps2");
func("smaps3");
func("smaps4");
return 0;
}
But if you try to change the value like:
void func(char* test){
test[0] = 'A';
printf("%s\n", test);
}
you have undefined behavior because modifying a string literal is undefined behavior.