Why compiler is calling first overloaded function not second one?

Question

Consider following program.

#include <iostream>
void fun(const char a[]) {
    std::cout<<"function 1\n"; 
    std::cout<<a<<'\n';
}
void fun(char *a) {
    std::cout<<"function 2\n"; 
    std::cout<<a<<'\n';
}
int main() { 
    fun("string");
}

What should I change in the program so that second function gets called instead of first one?

Andrew Malta · Accepted Answer · 2015-05-17 08:07:38Z

1

Well you are passing in a const char * when you pass in "string"

what you could do is declare a char * and set it to something and then pass it in and it will go to the second function.

"string" is a const char* which will go to the first function

answered May 17, 2015 at 8:07

Andrew Malta

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3 Comments

deviantfan Over a year ago

@meet To do what exactly? Again, "string" is const, you can´t change that.

Destructor Over a year ago

@deviantfan: Ok, I understood. you mean to say modifying string literal is UB in C++ because it is stored in read only memory. right?

deviantfan Over a year ago

@meet While that´s true, why is this relevant? First, it was a question about overloading (only), now you actually want to modify it ... Again, tell us what you really want, not how do you think you could do it.

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