Question

I want to pass an argument to a function, which takes a const char **

#include<iostream>

using namespace std;

void testFunc(const char **test){}

string testString = "This is a test string";

int main()
{
    const char *tempC = testString.c_str();

    testFunc(&tempC);

    return 0;
}

This code works fine, But I dont want to go through the temporary variable tempC. I want to pass testString.c_str() directly. Like the following,

int main()
{    
    testFunc(&testString.c_str());

    return 0;
}

But, it shows error,

 error C2102: '&' requires l-value

Is it possible to do it without using the temp variable.

Answer 1

I want to pass testString.c_str() directly.

You can’t. std::string::c_str() returns a const char *. In order to make a pointer to a string, you have to put it in a variable and take its address.

That being said, I’m far more concerned about what function you’re trying to pass this to. In general, a function that takes a const char ** does so for two reasons:

1: It takes an array of strings. You are passing a single string. Usually, C-style functions that take an array need a second parameter that says how many elements are in the array. I hope you’re putting a 1 in there.

2: It is returning a string. In which case what you’re doing is not helpful at all. You should create a const char * as a variable, initialize it to NULL, and then pass a pointer to it as the parameter. It’s value will be filled in by the function.