Which functions get called when you send 0 as argument [closed] - c++

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void func_print(int value) {
std::cout << “int” << std::endl;
}
void func_print(void* value) {
std::cout << “void” << std::endl;
}
int main() {
func_print(0);
func_print(NULL);
}
I cant seem to find. My compiler may be broken because it gives me stray errors cant find it what is wrong.
g++: error: -E or -x required when input is from standard input
c.cpp:3:5: error: stray ‘\342’ in program
std::cout << “int” << std::endl;
^
c.cpp:3:5: error: stray ‘\200’ in program
c.cpp:3:5: error: stray ‘\234’ in program
c.cpp:3:5: error: stray ‘\342’ in program
c.cpp:3:5: error: stray ‘\200’ in program
c.cpp:3:5: error: stray ‘\235’ in program
c.cpp:6:5: error: stray ‘\342’ in program
std::cout << “void” << std::endl;
^
c.cpp:6:5: error: stray ‘\200’ in program
c.cpp:6:5: error: stray ‘\234’ in program
c.cpp:6:5: error: stray ‘\342’ in program
c.cpp:6:5: error: stray ‘\200’ in program
c.cpp:6:5: error: stray ‘\235’ in program
c.cpp: In function ‘void func_print(int)’:
c.cpp:3:21: error: expected primary-expression before ‘int’
std::cout << “int” << std::endl;
^
c.cpp: In function ‘void func_print(void*)’:
c.cpp:6:21: error: expected primary-expression before ‘void’
std::cout << “void” << std::endl;
^
c.cpp: In function ‘int main()’:
c.cpp:11:20: error: call of overloaded ‘func_print(NULL)’ is ambiguous
func_print(NULL);
^
c.cpp:2:6: note: candidate: void func_print(int)
void func_print(int value) {
^
c.cpp:5:6: note: candidate: void func_print(void*)
void func_print(void* value) {
All errors here explained with messages though i don't get it exactly what is wrong.

Answer 1

Which functions get called when you send 0 as argument
The call is ambiguous. Neither function is preferred by the overload resolution, since 0 is both an int literal, as well as a pointer literal. An ambiguous call makes the program ill-formed, so a compiler is not required to accept it. This is what the compiler told you:
error: call of overloaded ‘func_print(NULL)’ is ambiguous
std::cout << “void” << std::endl;
This is wrong, because “ (Left Double Quotation Mark) is not a valid character there. You've most likely attempted to write a string literal. A string literal uses the " (Quotation Mark) character, which is similar. This is what the compiler told you:
error: expected primary-expression before ‘void’
It doesnt have any include or headers it should compile by its own (supposedly)
Your supposition is wrong. Besides the problems mentioned earlier, std::cout (or anything else from the std namespace) cannot be used without including standard headers.