C++ cannot convert 'const char*' to 'std::string*'

Question

I have this code below and I'm getting the error upon compilation:

error: cannot convert 'const char*' to 'std::string*' for argument '1' to 'void sillyFunction(std::string*, int)'

#include <iostream>
#include <string>

using namespace std;
int counter = 0;

void sillyFunction(string * str, int cool=0);

int main(){
    sillyFunction("Cool");
    sillyFunction("Cooler", 1);
    return 0;
}

void sillyFunction(string * str, int cool){
    counter++;
    if (cool){
        for (int i=0; i<counter; i++) cout << *str << endl;
    } else {
        cout << *str << endl;
    }
}

Answer 1

Don't take your parameter in as a string * try just using a const string & instead

EDIT:

std::string and const char* are different types. the std::string already has a conversion from string literals (ex: "Cool" ) to the actual string object. So by passing in the string literal "Cool" you are in a sense passing in a std::string object, not a pointer to one.

The reason I chose to use a const string & is mostly from personal coding practices. This minimizes stack memory usage, and since you are passing in a constant string literal, there is no need for the parameter to be modify-able.

Also don't forget if you change from a string * that you no longer need to dereference it in your cout :

if (cool){
    for (int i=0; i<counter; i++) cout << str << endl;
} else {
    cout << str << endl;
}

Answer 2

change

void sillyFunction(string * str, int cool){
   counter++;
    if (cool){
        for (int i=0; i<counter; i++) cout << *str << endl;
    } else {
        cout << *str << endl;
    }
}

to

void sillyFunction(const char* str, int cool){
    counter++;
    if (cool){
        for (int i=0; i<counter; i++) cout << str << endl;
    } else {
        cout << str << endl;
    }
}

Answer 3

To explain what the problem actually is ...

While the compiler will happily arrange for a char * /C-string to be "converted" to a std::string via the appropriate std::string constructor, that's not what you've asked for.

You have asked for a pointer to an existing std::string object. By definition, what is passed to your function must be the address of an already existing std::string (or descendant) object.

You must understand pointers as a distinct type -- your function takes a pointer-to-std::string "object". While a std::string can be accessed via a pointer to std::string , the pointer itself is not a std::string , nor can it be "converted" to a std::string , nor can it be treated as a pointer-to-char (or vice-versa).

The easiest alternative is indeed a const reference to a std::string ( const std::string & ). const, in this case, because you're not doing anything to modify the string. If you were, that would be a different matter, and you'd have to consider carefully whether you intend for the caller to see your changes.

By doing this, you're saying you want a std::string object (remember, a reference to an object is that object , see C++ FAQ 8.5 in particular), which allows the compiler to invoke the appropriate constructor to create a std::string for you when the function is called with a char * (const or not).

At the same time, if someone passes you an actual std::string , the constructor is avoided and you get the same efficiency as if you had taken a pointer-to-std::string . Win-win.

Alternatively, of course, you can just take a plain std::string , but in that case you always get a copy of the string being passed in, whether it's a C-string or a std::string . Sometimes that's desirable, sometimes not. In your case, you don't do anything but print the string out, making the overhead unnecessary.

Answer 4

You can acheive that by changing the prototype to:

void sillyFunction(string const &str, int cool=0);

char const * could be implicitly converted to a temporary std::string , which in turn could be passed by reference ( std::string const & ). There's no implicit conversion to a pointer ( std::string * ), that's why you get the error.

Answer 5

You can convert from a const char * to a string , but not to a string * .

Perhaps you want your sillyFunction to take a const reference?

void sillyFunction(const string &str, int cool){
    counter++;
    if (cool){
        for (int i=0; i<counter; i++) cout << str << endl;
    } else {
        cout << str << endl;
    }
}

Answer 6

应该

void sillyFunction(const string& str, int cool=0);

Answer 7

If you're going to use pointers for your functions you have to declare the strings somewhere. memory needs to be allocated. so either declare a variable or call new to create memory for the string.

int main(){
    string str = "Cool";
    string str2 = "Cooler";
    sillyFunction(&str);
    sillyFunction(&str2, 1);
    return 0;
}

Answer 8

I have a very simple solution for that, using string copy

char s[20] strcpy(s,const char *p);

Then you have the string pointed by *p in s... it works..