macOS C++ average small coding problem

[ekoptur]

I am having problem by putting more integers on this. I can type ./a.out 3 15 and it find 9 which is correct but I can only put 2 numbers. How can i change this to put more than 2 number and find the average? for ex. ./a.out 3 15 17 19 24.

///////    newAverage.C    ///////
#include  <iostream>

int main (int argc,  char* argv[])
{    if ( argc != 3 )
     {   std::cerr << "Usage: " << argv[0] << " integer "
                   << std::endl;
         exit(1);
     }

     int a = atoi(argv[1]), b = atoi(argv[2]);

     float z = (a + b)/2.0;
     std::cout << "a is " << a
           << " and b is " << b << std::endl;
     std::cout << "average is " << z << std::endl;
     return 0;
}

Thank you,

 

[subsonix]

Create a loop around argc, loop and get your values from argv, where i is the index.

 

[ekoptur]

Create a loop around argc, loop and get your values from argv, where i is the index.

I am very sorry to bother you, but I am new on C++. Can you show me how to loop it? I understand the part about the values.

Thank you,

 

[lee1210]

Right now you have an explicit check for 2 arguments. So that has to go. The only problem would be 0 arguments, so you can compare argc to 1 and print an error in that case and quit.

From there, you need to loop from 1 to argc - 1:

long int sum = 0;
std::cout << "The average of: ";
for(int pos = 1; pos < argc; pos ++) {
  //Some error checking ought to go on here.
  if(pos != 1) std::cout << ", ";
  std::cout << argv[pos];
  sum += strtol(argv[pos],NULL,10);
}
double average = sum / (argc - 1.);
std::cout << " is: " << average << std::endl;

-Lee

 

[subsonix]

Here's the most simple example building on what you had, if you want to present all input then you need a loop when printing as well.

///////    newAverage.C    ///////
#include  <iostream>

int main (int argc,  char* argv[])
{
        if ( argc < 3 )
        {
                std::cerr << "Usage: " << argv[0] << " integer " << std::endl;
                exit(1);
        }

        int i, sum = 0;
        float avg;

        for( i = 1; i < argc; i++ ) {
                sum += (int)strtol(argv[i], NULL, 0);
        }

        avg = ((float)sum/(argc -1));

        std::cout << "avg: " << avg << std::endl;

        return 0;
}

I used strtol here, but you can just replace that with atoi.

 

[ekoptur]

Right now you have an explicit check for 2 arguments. So that has to go. The only problem would be 0 arguments, so you can compare argc to 1 and print an error in that case and quit.

From there, you need to loop from 1 to argc - 1:

long int sum = 0;
std::cout << "The average of: ";
for(int pos = 1; pos < argc; pos ++) {
  //Some error checking ought to go on here.
  if(pos != 1) std::cout << ", ";
  std::cout << argv[pos];
  sum += strtol(argv[pos],NULL,10);
}
double average = sum / (argc - 1.);
std::cout << " is: " << average << std::endl;

-Lee

hmm si I put it like this ,

/////    newAverage.C    ///////   
#include  <iostream>
      
int main (int argc,  char* argv[])  
{    if ( argc != 3 )
     {   std::cerr << "Usage: " << argv[0] << " integer "
                   << std::endl;
         exit(1);
     }

     long int sum = 0;
     std::cout << "The average of: ";
     for(int pos = 1; pos < argc; pos ++) {
    
     if(pos != 1) std::cout << ", ";
     std::cout << argv[pos];
     sum += strtol(argv[pos],NULL,10);
     }
     double average = sum / (argc - 1.);
     std::cout << " is: " << average << std::endl;
     }

But I am still getting the same problem?

 

[balamw]

if ( argc != 3 )

But I am still getting the same problem?

What does this line do? Do you see it in Lee or subsonix's code?

B

 

[ekoptur]

Here's the most simple example building on what you had, if you want to present all input then you need a loop when printing as well.

///////    newAverage.C    ///////
#include  <iostream>

int main (int argc,  char* argv[])
{
        if ( argc < 3 )
        {
                std::cerr << "Usage: " << argv[0] << " integer " << std::endl;
                exit(1);
        }

        int i, sum = 0;
        float avg;

        for( i = 1; i < argc; i++ ) {
                sum += (int)strtol(argv[i], NULL, 0);
        }

        avg = ((float)sum/(argc -1));

        std::cout << "avg: " << avg << std::endl;

        return 0;
}

I used strtol here, but you can just replace that with atoi.

Oo I see now.
Ok I think I got it.

Thank you to all of you showing me the looping and for your help.

 

[lee1210]

Here's the most simple example building on what you had, if you want to present all input then you need a loop when printing as well.

Oh, hogwash.

#include  <iostream>
#include  <errno.h>

int main (int argc,  char* argv[])
{
  if(argc <= 1) {
    std::cout << "Please enter arguments to average." << std::endl;
    return -1;
  }

  long int sum = 0;
  std::cout << "The average of: ";
  for(int pos = 1; pos < argc; pos ++) {
    long int tmp = strtol(argv[pos],NULL,10);
    if(tmp == 0 && EINVAL == errno) {
      std::cerr << std::endl << "Invalid value " << argv[pos] << " entered." << std::endl;
      continue;
    }
    if(pos != 1) std::cout << ", ";
    std::cout << argv[pos];
    sum += tmp;
  }
  double average = sum / (argc - 1.);
  std::cout << " is: " << average << std::endl;
  return 0;
}

-Lee

 

[ekoptur]

Oh, hogwash.

#include  <iostream>
#include  <errno.h>

int main (int argc,  char* argv[])
{
  if(argc <= 1) {
    std::cout << "Please enter arguments to average." << std::endl;
    return -1;
  }

  long int sum = 0;
  std::cout << "The average of: ";
  for(int pos = 1; pos < argc; pos ++) {
    long int tmp = strtol(argv[pos],NULL,10);
    if(tmp == 0 && EINVAL == errno) {
      std::cerr << std::endl << "Invalid value " << argv[pos] << " entered." << std::endl;
      continue;
    }
    if(pos != 1) std::cout << ", ";
    std::cout << argv[pos];
    sum += tmp;
  }
  double average = sum / (argc - 1.);
  std::cout << " is: " << average << std::endl;
  return 0;
}

-Lee

This one is abit complicated code for me I will need to try learn more about it. Thank you.

 

[subsonix]

Oh, hogwash.

Excuse me? What is hogwash exactly, that you need a loop to present the args or that it's the most simple example (perhaps not in an absolute sense, but that's clearly not what I meant).

I saw the reply, went ahead and made an example. It has nothing to do with your post.

 

[ekoptur]

Here's the most simple example building on what you had, if you want to present all input then you need a loop when printing as well.

///////    newAverage.C    ///////
#include  <iostream>

int main (int argc,  char* argv[])
{
        if ( argc < 3 )
        {
                std::cerr << "Usage: " << argv[0] << " integer " << std::endl;
                exit(1);
        }

        int i, sum = 0;
        float avg;

        for( i = 1; i < argc; i++ ) {
                sum += (int)strtol(argv[i], NULL, 0);
        }

        avg = ((float)sum/(argc -1));

        std::cout << "avg: " << avg << std::endl;

        return 0;
}

I used strtol here, but you can just replace that with atoi.

I am sorry to bother you again.

I have one more question. I changed the strtol with atoi and it didnt work. Do you know the reason why? with strtol, it works perfect but when I type atoi instead of strtol, it doesnt work. Could you tell me what strtol does?

Thank you,

 

[subsonix]

I am sorry to bother you again.

I have one more question. I changed the strtol with atoi and it didnt work. Do you know the reason why? with strtol, it works perfect but when I type atoi instead of strtol, it doesnt work. Could you tell me what strtol does?

Thank you,

strtol does what atoi does. The reason I did not use it is this section of the atoi man page:

IMPLEMENTATION NOTES
The atoi() function is not thread-safe and also not async-cancel safe.

The atoi() function has been deprecated by strtol() and should not be used in new code.

 

[ekoptur]

strtol does what atoi does. The reason I did not use it is this section of the atoi man page:

ok thank you for helping me. I still have alot to learn. good luck to me

 

[lee1210]

Excuse me? What is hogwash exactly, that you need a loop to present the args or that it's the most simple example (perhaps not in an absolute sense, but that's clearly not what I meant).

I saw the reply, went ahead and made an example. It has nothing to do with your post.

Maybe a smiley would have been appropriate? I just meant you don't need a separate loop to display the args since you're already looping to sum. No harm intended, it was meant in good humor.

-Lee

 

[subsonix]

Maybe a smiley would have been appropriate? I just meant you don't need a separate loop to display the args since you're already looping to sum. No harm intended, it was meant in good humor.

-Lee

Ok, it's all good. I just made the most simple example in a few minutes showing how to loop around argc, as requested. I excluded any detailed output, and explained that a loop is required to present argv in it's entirety. No, it doesn't need to be a separate loop.