My code look like this
fn main() {
// some other codes goes here
let int = 1;
if int == 1 {
let x = "yes";
} else {
let x = "no";
}
if x == "yes" {
// some other codes goes here
println!("yes");
} else if x == "no" {
// some other codes goes here
println!("no")
}
}
When I run this I get this
error[E0425]: cannot find value `x` in this scope
--> src/main.rs:9:8
|
9 | if x == "yes" {
| ^ not found in this scope
error[E0425]: cannot find value `x` in this scope
--> src/main.rs:12:15
|
12 | } else if x == "no" {
| ^ not found in this scope
While searching for a solution I came across this post How do I access a variable outside of an `if let` expression? but could not able to understand the reason or its solution?
Easiest is by far to code it such that it is in scope in the first place. You can assign variable with the result of a statement in a single assignment. If you can make it as a one-liner, it also makes it arguably more readable. If the actual processing is too long, nothing prevents you from just... making it a proper function.
let x = if int == 1 { "yes" } else { "no" };
// rest of the code accessing x.
Alternatively, the compiler will let you declare unassigned variables if you properly assign them later, with all the compile-time safety checks in place. Read the docs on RAII (Resource Acquisition Is Initialization) RAII Docs to see more how it works. In practice, it's just as simple as this:
let x;
if i == 1 {
x = "yes";
}
else {
x = "no";
}
// keep doing what you love
The compiler will error if there's a logic path where x doesn't get initialized, or if it gets initialized as a different type.
Notice you also do not need to declare it as mut, as the first value it gets will stay immutable. Unless you do want to change it afterwards, obviously.
You can't access a variable which is out of scope. But you use a workaround and set the variable in the same scope.
fn main(){
let int = 1;
let x = if int == 1 {
"yes"
} else {
"no"
};
if x == "yes" {
println!("yes");
} else if x == "no" {
println!("no");
}
}
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