UNIX for Dummies Questions & Answers

I have created the following script to add 2 numbers together :
lazyadd2 script :
#!/bin/bash
#I have added the she-bang script above as I felt that this would be a good idea.
#The script takes in 2 arguments that are added together
num3=`expr $num1 + $num2`
echo $num3

I make the following variable assignments :

num1=1
num2=2

I type the following after script creation and relevant variable assignment :

$ sh lazyadd2

The following output is obtained :
+

I type the following to retrieve a value :
$ echo $num3

The result of this is blank (ie: the computer outputs no result).

What I expected to happen was that I should type the above to retrieve the sum of the values $num1 and $num2, so that "echo $num3" returns 3.

I know that this is *really simple* - so I'm sure the mistake is obvious....

Thanks for any help

what you are doing is right
can you show us the script?

In order to make things easier to follow (even thought I'm only dealing with simple scripts), the following are the scripts that are related to each other :

SCRIPT 1 :
#!/bin/bash
num3=`expr $1 + $2`
echo $num3

SCRIPT 2 :
#!/bin/bash
num3=`expr $num1 + $num2`
echo $num3

Here are the inputs and outputs for SCRIPT 1 :
$ sh lazyadd1 1 2
3
Here are the inputs and outputs for SCRIPT 2 :
$ num1=1
$ num2=2
$ sh lazyadd2
+
$ echo $num3

I have a problem with the second script - I think that it should not ouput a '+' at all! Also, when I echo for a num3 call, it should output the sum of the previous num assignments, namely '3'.

Anyone have any ideas as to why this is occurring?
Thanks.

It's happening because when you run the lazyadd2 script, it is unaware of what you want the values of $num1 and $num2 to be. It is only going to know $1, $2, $3, etc. You can do this:

$ num1=1
$ num2=2
$ . lazyadd2

Or do it in the lazyadd2 script itself:

Code:

#!/bin/bash
num1=$1
num2=$2
num3=`expr $num1 + $num2`
echo $num3

...and run it the same way you run lazyadd1 (i.e., lazyadd2 1 2).

try this below
num1=1
num2=2
num3=`expr {$num1} + {$num2}`
echo $num3

In response to Glenn Arndt's response, am I correct in stating that, if I declare a value of num1 and num2 outside of the script, I will find that these values are not automatically accessible from within the script?

I'm guessing that the answer to the last question is yes.
In regards to the script lines you advocate :
#!/bin/bash
num1=$1
num2=$2
num3=`expr $num1 + $num2`
echo $num3
I can see how that works - but am still interested in if the script can access variables written outside of it.

When typing the below into the script, could anyone tell me the significance of the 3rd line? I suppose that this is just a different way of evoking a shell script :

$ num1=1
$ num2=2
$ . lazyadd2

Finally, in regards to gkrishnag, I did try the below :
num1=1
num2=2
num3=`expr {$num1} + {$num2}`
echo $num3

But here's what I get :
$ num3=`expr {$num1} + {num2}`
expr: non-numeric argument
$ echo $num3

That is, what you suggested is not recognised as valid.

An explanation of the dot is here: https://www.unix.com/aix/31018-pb-script-execution-variables.html

I believe gkrishnag meant ${num1} not {$num2}.