how can I get a char*[] from python

Question

I use swig wrapped some c++ api function. There is one function, the interface is f(char*[] strs).

How can I pass a valid parameter to this function. This is what I did.

str = ["str","str2"]
f(str)

it will throw error

TypeError: in method 'f', argument 1 of type 'char *[]

check your f function, it takes the char [] list and you given list of string. So call f function with char list. — Bhavesh Odedra
– Bhavesh Odedra, Commented Jul 2, 2014 at 6:12
@Odedra char* is a string in C. you can init a char* [] like this. char *list[] = {"str","str1"} — Crazymooner
– Crazymooner, Commented Jul 2, 2014 at 17:34

Oliver · Accepted Answer · 2014-07-04 02:58:51Z

1

SWIG does not convert arrays to Python lists automatically. Since you are using C++, use std::string and std::vector for your f, then SWIG will make all the necessary conversions automatically (don't forget to include "std_vector.i" and such, see the SWIG docs):

void f(std::vector<std::string> > strs)

If you cannot modify the declaration of f, you can create an %inline wrapper in the .i:

%inline {
    void f(const std::vector<std::string> >& strs) 
    {
        // create tmpCharArray of type char*[] from strs
        const char* tmpCharArray[] = new const char* [strs.size()];
        for (i=0; i<strs.size(); i++) 
              tmpCharArray[i] = strs[i].c_str();
        f(tmpCharArray);
        delete[] tmpCharArray;
     }
}

answered Jul 4, 2014 at 2:58

Oliver

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3 Comments

Oliver Over a year ago

Cool. Let me know if there is any improvement needed for you to upvote, and welcome to SO!

Crazymooner Over a year ago

It is a good enough answer. I find something cooler. this way, I do not even have to write a wrapper, and char** can be used for all the functions which required.

Oliver Over a year ago

Thanks for posting the link.

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