How do I call a program or code which has main in debug mode?

Question

I have a CPP program (let's call it program 1) which calls another CPP program (Program 2). This Program2 is built in debug mode. That is:

   int main(int argc, char** argv) {
    ///Function body#
     eerot[0] = atof(argv[1]); 
eerot[1] = atof(argv[2]); 
eerot[2] = atof(argv[3]); 
eetrans[0] = atof(argv[4]); //Exception thrown here due to memory clash
     ///Continue
    }

The above fragment of the code describes the structure of my main program in program2. Program2 has neither execution error nor does it throw any exception. Now I want to call this main function of program2 from program1 but I am stuck with the double pointer variable. Because the argv variable is double pointer, and even though I was able to create a string, double reference it and pass it as parameter to program2, I get exceptions. There is no compile error. All the appropriate headers are perfectly defined.

I cannot edit program2 since it was developed by some other team and it is quite difficult to follow. I have changed the name of main function of program 2(say to "calculate()") and called that from main of program1. I have also included directories and header files appropriately to make all functions and classes of program2 visible in program1. Right now only thing I am stuck is with the double pointer argv variable. How can I pass the parameters which I have to pass via the command prompt be passed via another program?

Note that the parameters we pass in command prompt for program2 is of type double. I tried to use arrays but I keep getting exceptions thrown at me. Let me know anything that comes to your mind when trying to solve the problem.

what exceptions are thrown at you? did you try to catch them? — 463035818_is_not_an_ai, May 21 '19 at 19:07
Not exactly. It will run even if I choose different lines for each end of line semicolon. — prashanth kumar, May 21 '19 at 19:08
sorry for my bad wording. What I wanted to say is: please do not put everything on the same line, it just makes it harder to read — 463035818_is_not_an_ai, May 21 '19 at 19:09
@formerlyknownas_463035818 I made a mistake there. I will remove it — prashanth kumar, May 21 '19 at 19:10
This: `int main(int argc, char** argv[])` makes *no* sense. Did you mean `int main(int argc, char** argv)` or `int main(int argc, char* argv[])` ? — Jesper Juhl, May 21 '19 at 19:11
@eerot and eetrans and just variable to get data from the pointer variable argv — prashanth kumar, May 21 '19 at 19:13

score 0 · Accepted Answer · answered May 21 '19 at 19:09

0

I believe the problem is the triple pointer in the main() declaration. You probably meant:

int main(int argc, char** argv) {

or

int main(int argc, char* argv[]) {

answered May 21 '19 at 19:09

donjuedo

2,475
18
28

Thank you. Yeah I made a typo there. I changed it. However my problem is still there – prashanth kumar May 21 '19 at 19:15
@prashanthkumar argv[0] is the invoked program name. argv[1] is the first parameter. What happens if you run the program with an extra parameter or two? If no exception, that would strongly hint at a mistake in the indexes of the parameters used. – donjuedo May 21 '19 at 19:28

score 0 · Answer 2 · answered May 21 '19 at 19:11

0

you want either char **argv ... OR char *argv[], NOT char **argv[]

both of the above are pointers to arrays. What you have is a pointer to a 2D array basically so the way you are accessing the elements ends up exceeding the number of array elements.

answered May 21 '19 at 19:11

Ryan

106
1
4

Sorry about that. I made a typo there. But my problem persists even if I just use **argv or *argv[] – prashanth kumar May 21 '19 at 19:18
argc will be the number of strings entered through command line arguments, maybe add some error checking so you never access argv[argc] or higher? – Ryan May 21 '19 at 20:43

How do I call a program or code which has main in debug mode?

2 Answers2