What is the reason for TypeScript disabling unsecure/non-strict compiler rules by default?

Recently, I found myself having to enable a slew of compiler options in my application:

"alwaysStrict": true,
"extendedDiagnostics": true,
"noFallthroughCasesInSwitch": true,
"noImplicitAny", true,
"noImplicitThis", true,
"noImplicitReturns": true,
"noUnusedLocals": true,
"noUnusedParameters": true,
"strict": true,
"strictBindCallApply": true,
"strictFunctionTypes": true,
"strictNullChecks": true,
"strictPropertyInitialization": true

These options are meant for stricter compilation, which leads to more secure and higher quality code. I'm curious as to why they are all disabled by default. More information can be found at: https://www.typescriptlang.org/docs/handbook/compiler-options.html

typescriptsecuritycompilationtranspiler

Answer №1

One way to streamline your coding is by simply including strict: true. This option comes with various compiler settings:

--noImplicitAny
--noImplicitThis
--alwaysStrict
--strictBindCallApply
--strictNullChecks
--strictFunctionTypes
--strictPropertyInitialization

The strict mode
is not active by default. This decision likely facilitates a smoother transition from JavaScript projects to TypeScript without causing too much frustration initially.

extendedDiagnostics
provides detailed diagnostic information for debugging purposes, but may be deemed unnecessary as a default setting.

noFallthroughCasesInSwitch
is dependent on individual coding preferences. There are scenarios where combined cases are valid.

noImplicitReturns
has been considered a stylistic issue. It does not impact type safety since the compiler can infer return types even without explicit declarations.

noUnusedLocals/Parameters
With its functionalities and IDE support, TypeScript allows editors like VS Code to highlight unused locals and parameters as suggestions rather than causing too many compilation errors, especially when using noEmitOnError: true.

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