Write a program that takes two int agruments in command line arguments and then print their sum

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1.00/5 (1 vote)

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I tried this code but it is throwing the below error:

int main(int argc, char **argv)
     {
       int total=0;
        for(int i=0;i<argv;++i)>
         {
             total=total+atoi(argc[i]);
         }
         cout<<total<<endl;
       return 0;
     }

Error: Pointer type needed instead of int.

Posted 22-Feb-11 23:47pm

centinnel

Updated 23-Feb-11 2:41am

CPallini

v4

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CPallini 23-Feb-11 8:40am

And you, please, don't bother posting such pointless comments.

[no name] 23-Feb-11 6:23am

On the contrary. Most people don't mind helping a beginner. He did not ask us to do his homework and obviously simply got confused by the types of argc and argv. char** is something I also would not want to encounter right at the beginning

CPallini 23-Feb-11 8:42am

Small sidenote: requirements are for two integers, you should check that too.

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OriginalGriff · Accepted Answer · 2011-02-22T23:56:00

Solution 1

OK: Some slight problems here:
argc contains the number of arguments, not argv. Change to:

C#

for(int i=1;i<argc;++i)

and argv contains the arguments themselves, not argc. Change to:

C#

total=total+atoi(argv[i]);

And all should be well.

[edit]And take out the > symbol on the end of your for loop! (If it is really there...) - OriginalGriff[/edit]

Nuri Ismail correctly suggested: the loop must start from 1 (int i = 1) and 'argc' must be at least 3, because 'argv[0]' is reserved for the process name. Example cmd line: 'my_prog.exe 5 6' -> argv[0] = my_prog.exe, argv[1] = 5 and argv[2] = 6. :thumbsup:

I have updated my response to illustrate that.

It was also (very correctly) pointed out that you should check your inputs for:
Number of arguments - argc should be equal to three or you have a problem!
Valid digts - is the argument actually a number?

If you need help with that, please ask again in another question!

Posted 22-Feb-11 23:56pm

OriginalGriff

Updated 23-Feb-11 1:27am

v3

Comments

[no name] 23-Feb-11 6:14am

And perhaps he should check that argc really equals two, no more, no less. As it is now, you could also pass 200 arguments. Also, it might be worth a thought what happens when I pass %64§ and +xdfg# as arguments.

OriginalGriff 23-Feb-11 6:15am

Or at least a minimum of two.
I did consider that, but as he is a complete beginner I thought "just fix the problem he has now, don't confuse him." :laugh:

Nuri Ismail 23-Feb-11 6:40am

Griff, the loop must start from 1 (int i = 1) and 'argc' must be at least 3, because 'argv[0]' is reserved for the process name. Example cmd line: 'my_prog.exe 5 6' -> argv[0] = my_prog.exe, argv[1] = 5 and argv[2] = 6.

OriginalGriff 23-Feb-11 6:43am

Good point :blush: My bad - it's been a while since I played with C command line arguments...
I'd 5 that if I could...

Nuri Ismail 23-Feb-11 6:54am

No problem, normally I wouldn't notice such a detail, but nearly a month ago I wrote a small cmd line app which works with arguments.

BTW, all of the remarks from your answer are good and valid. Maybe you should update the answer in order to reflect this little detail about the argv[0], because I'm not sure if the OP will take care to read the comments. :)

Best Regards,
Nuri

OriginalGriff 23-Feb-11 7:28am

Good idea - done.

Nuri Ismail 23-Feb-11 8:09am

Thank you, Griff! 5+

Sergey Alexandrovich Kryukov 23-Feb-11 22:31pm

Anyway, 5.
--SA