Why the results of these two code fragments are not the same?


 
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# 1  
Old 02-06-2019
Why the results of these two POSIX shell code fragments, are not the same?

Code 1:

Code:
#!/bin/sh

for arg1 in "$@"
do
    counter=0

   for arg2 in "$@"
   do

     if [ "$arg2" = "$arg1" ] && [ $counter -eq 0 ]
     then
       counter=$((counter+1))

       continue 
    fi


   if [ "$arg2" = "$arg1" ]
   then
    
     echo
     echo "Error: Two or more arguments are the same."
     echo
     echo "Exiting..."
     echo

     exit 1
    fi

  done

done

exit 0

It outputs:
Code:
[john@manjaromatepc Downloads]$ ./script.sh ~/Documents/ ~/Downloads/
[john@manjaromatepc Downloads]$

Code 2:

Code:
#!/bin/sh


for arg1 in "$@"
do
  
  counter=0

  for arg2 in "$@"
  do


    if [ $counter -eq 0 ]
    then
  
      counter=$((counter+1))

      continue 
   fi


  if [ "$arg2" = "$arg1" ]
  then
    
    echo
    echo "Error: Two or more arguments are the same."
    echo
    echo "Exiting..."
    echo

    exit 1
  fi

 done

done

exit 0

It outpouts:

Code:
 
[john@manjaromatepc Downloads]$ ./script.sh ~/Documents/ ~/Downloads/

Error: Two or more arguments are the same.

Exiting...

[john@manjaromatepc Downloads]$


Last edited by johnprogrammer; 02-06-2019 at 10:49 PM.. Reason: The second code fragment has a logical mistake
# 2  
Old 02-06-2019
I found the issue, the second code fragment has a logical mistake.
# 3  
Old 02-06-2019
Maybe...

Since both of your scripts are comparing every element of the operand list to every element of the operand list, one might expect that there would be at least two cases where the operands would be identical. Those cases would be when you're comparing the first operand to the first operand (i.e., $HOME/Documents/ to $HOME/Documents/) and when you're comparing the second operand to the second operand (i.e., $HOME/Downloads/ to $HOME/Downloads/).

Maybe there is a logical mistake in both scripts???

What is the logic in both of your scripts behind the variable named counter? Why does it matter what the value of $counter is when trying to determine whether or not two command-line arguments are the same?
This User Gave Thanks to Don Cragun For This Post:
# 4  
Old 02-07-2019
The POSIX shell code detects if there are two or more, same command line arguments to the script.


An example with C++03.

Assuming we have a vector (array) with some strings:

Code:
#include <iostream>
#include <string>
#include <vector>


int main()
{
  using namespace std;

  string stringArray[]={"some", "to", "words", "to", "use", "for", "testing", "purposes"};

  vector<string> myvector(stringArray, stringArray+ sizeof(stringArray)/sizeof(*stringArray) );


  Here is the interesting part:

  for(vector<string>::size_type i= 0; i< myvector.size(); ++i)
  {
    for(vector<string>::size_type j= i+ 1; j< myvector.size(); ++j)
    {
       if(myvector[j]== myvector[i])
       {
         cout<< "Two or more words are the same\n\n";
         
         return 0;
       }
    }
  }

}

This is what I am trying to do with the POSIX shell code I posted.
# 5  
Old 02-07-2019
But, the code you've written in your shell scripts acts more like your C++ code with the inner loop changed from:
Code:
    for(vector<string>::size_type j= i+ 1; j< myvector.size(); ++j)

to:
Code:
    for(vector<string>::size_type j= 0; j< myvector.size(); ++j)

There doesn't appear to be any attempt to keep from checking one element against itself in your command-line argument vector in either of your shell scripts.

I repeat: "What is the logic in both of your scripts behind the variable named counter? Why does it matter what the value of $counter is when trying to determine whether or not two command-line arguments are the same?"
# 6  
Old 02-07-2019
Quote:
Originally Posted by Don Cragun
But, the code you've written in your shell scripts acts more like your C++ code with the inner loop changed from:
Code:
    for(vector<string>::size_type j= i+ 1; j< myvector.size(); ++j)

to:
Code:
    for(vector<string>::size_type j= 0; j< myvector.size(); ++j)


Yes you are right.



Quote:
There doesn't appear to be any attempt to keep from checking one element against itself in your command-line argument vector in either of your shell scripts.

I repeat: "What is the logic in both of your scripts behind the variable named counter? Why does it matter what the value of $counter is when trying to determine whether or not two command-line arguments are the same?"

When the shell script starts, both arg1 and arg2 have the same value.

So I use the counter with value 0, to determine this case, and advance the second for-loop to the next argument, and from there, I check if arg2=arg1.


If the second loop ends, the first loop progresses to the next argument.

The second loop restarts again with the first argument, and proceeds until it meets the first loop, where arg2=arg1 and counter -eq 0.

arg2 then proceeds to the next argument, until it ends, if no duplicate arguments found.

Then the first loop progresses to the next argument. etc.

The code that works, is the first I posted, and is this:

Code:
#!/bin/sh

for arg1 in "$@"
do
    counter=0


   for arg2 in "$@"
   do

     if [ "$arg2" = "$arg1" ] && [ $counter -eq 0 ]
     then
       counter=$((counter+1))

       continue 
    fi


   if [ "$arg2" = "$arg1" ]
   then
    
     echo
     echo "Error: Two or more arguments are the same."
     echo
     echo "Exiting..."
     echo

     exit 1
    fi

  done

done

exit 0

# 7  
Old 02-07-2019
Maybe the entire code could be condensed down to (tested with bash --posix)

Code:
TMPARR=( $@ )
for ((i=0; i<$#; i++))
  do    for ((j=i+1; j<$#; j++))
          do    [ ${TMPARR[i]} = ${TMPARR[j]} ] && { echo same arguments; exit 1; }
          done
  done

This User Gave Thanks to RudiC For This Post:
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