Losing a variable defined in an IF statement when the statement is finished

[Anonymous]

Hi,

Here is the problem:

I am losing the value of a variable I assign in an if statement. I don't know why this would happen.

See example:

INT test = INPUT "enter a number"

if test==5 {

string number = "the number is 5"

}

As soon as i get out from that if I lose the value number. What am I doing wrong here?

Thanks,

Marc

[Sean570 · Accepted solution]

I don't think you are losing the number value, just the string value.

try something like:

INT test = INPUT "enter a number"
string number = ""
if test==5 {
    $number = "the number is 5"
}

// send message to make sure value sticks
message info $number 

[Anonymous · Accepted solution]

I found the problem:

I was defining the variable in the if statement:

Resolution:

define the variable outside

update the variable in the if statement

Thank you all

[lokesh.kalia · Accepted solution]

Hi @Anonymous Yes, the curly brackets {} define the scope of any variable declared within them. That is, the variable is destroyed after the closing } bracket. As you've said, for a variable to be used both within the scope defined by the pair of curly brackets and afterwards, it must be defined before that scope begins.

[ChristopherMarion]

@Anonymous & @lokesh.kalia 

Great information to know.  I'll make sure to remember that rule.

Cheers,