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",t};e.buildCustomizationMenuUi=t;function n(e){let t='
",t}function s(e){let n=e.filter.currentValue||e.filter.defaultValue,t='${e.filter.label}
`,e.filter.options.forEach(s=>{let o=s.id===n;t+=``}),t+="${e.filter.label}
`,t+=`ID: swift-code-style/forced-unwrapped
Language: Unknown
Severity: Notice
Category: Code Style
Declaring a variable as optional clearly signals that it may not hold a valid value and could be nil
. Force-unwrapping bypasses that safety and will cause a runtime crash if the value is actually nil
. Even if you check for nil
first, relying on force-unwrapping is still discouraged because it undermines the intent of optionals. A safer and more maintainable approach is to use optional binding or optional chaining to handle values gracefully.
var greeting: String?
// ...
println( \(greeting!)) // Noncompliant; could cause a runtime error
if greeting != nil {
println( \(greeting!)) // Noncompliant; better but still not great
}
var greeting: String?
// ...
if let howdy = greeting {
println(howdy)
}