How can I get the print of python function from cpp

Question

I use cpp to call python function, I have compiled the program without errors, but why can not I see the print result in python function. here is the cpp codes:

#include<python2.7/Python.h>
....
using namespace std;
int main()
{
    Py_Initialize();
    PyRun_SimpleString("import sys");
    PyRun_SimpleString("import os");
    PyRun_SimpleString("import string");
    PyRun_SimpleString("sys.path.append('./')"); 
    PyObject * pModule = NULL;
    PyObject * pFunc = NULL;
    PyObject * pClass = NULL;
    PyObject * pInstance = NULL;
    pModule = PyImport_ImportModule("download");
    if(!pModule)
    {
        std::cout << "there is no this file." << std::endl;
    }
    pFunc= PyObject_GetAttrString(pModule, "geturl");
    if(!pFunc)
    {
        std::cout << "there is no this func." << std::endl;
    }
    std::string url = "www";
    PyObject* args = Py_BuildValue("ss", url.c_str());
    PyEval_CallObject(pFunc, args);
    Py_DECREF(pFunc);
    Py_Finalize();
    return 0;
}

here is the download.py file in the same dir

def geturl(url):
    print(url)
    print("hello")

here is result, without errors nor print:

root@cvm-172_16_20_84:~/klen/test/cpppython # g++ t.cpp -o printurl -lpython2.7
root@cvm-172_16_20_84:~/klen/test/cpppython # ./printurl 
root@cvm-172_16_20_84:~/klen/test/cpppython #

How can I see the print, has the function geturl run successfully? Thanks

Mikel Rychliski · Accepted Answer · 2021-01-31 20:47:42Z

1

The PyEval_CallObject function is encountering a Python exception before it reaches your print statements. Adding error handling to this call (call PyErr_Print on NULL return value) will show the exception being raised:

TypeError: geturl() takes exactly 1 argument (2 given)

The root cause is the format string:

Py_BuildValue("ss", url.c_str());

You are creating a tuple of two values and passing that as the arguments geturl(). You need to pass a tuple with only one value. You are also invoking undefined behavior here because you haven't provided a second string pointer.

Resolve the issue by passing a tuple with only one value:

Py_BuildValue("(s)", url.c_str());

answered Jan 31, 2021 at 20:47

Mikel Rychliski

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