CODEWITHSUNDEEP
c++stringpointerscharpass-by-reference
user3552519
user3552519

Reputation: 411

How to use call by reference to get value of a char** type variable from a function in C++?

I am trying to print the following using the code as:

int main() {
        char p[] = "hello";
        char*t = p;
        char**b = &t;
        std::string s = b[0];
        cout<<"string="<<s<<" b[0]="<<b[0];
}

The output I get is as expected:

string=hello b[0]=hello

Now the same thing I want to achieve using a function call in which I pass a variable by reference as:

void func(char**& p) {
        char *k = (char*)malloc(8*sizeof(char));
        k = (char*)"abc";
        p = &k;
}

int main() {
        char**j = NULL;
        func(j);
        std::string s1 = j[0];
        cout<<s1;
}

Now I am getting a null string. What could be the reason and how to solve it ?

Upvotes: 1

Views: 456

Answers (2)

john
john

Reputation: 88027

There is NO WAY to write code like this that follows any sort of standards of good programming. If you are serious about learning how to program you should abandon this mess and rewrite your code to do things the correct way. This is sort of mess is never necessary, even when interfacing with legacy code.

However the following 'works'

void func(char**& p) {
    p = new char*;
    *p = new char[8];
    strcpy(*p, "abc");
}

And take note, I didn't need to use a cast.

Upvotes: 1

Some programmer dude
Some programmer dude

Reputation: 409442

You have (at least) two problems:

  1. The first is that you make p point to the local variable k.
  2. The second is more subtle, and is that you reassign k losing the original memory you allocated and make k point to.

Then the usual spiel about never using arrays of char or pointers to char for strings, when you have std::string. And that you should never use malloc in C++, only new or new[]. And if you need new[] consider using std::vector instead (except for strings of course).

Upvotes: 4

Related Questions