refactor(span/source-type): make SourceType factories const (#5241)

I found myself wanting this while writing `no-unused-vars` and while using oxc
in some downstream personal projects.
This commit is contained in:
DonIsaac 2024-08-27 01:19:53 +00:00
parent f5e05db302
commit 94f60e7748

View file

@ -69,7 +69,7 @@ impl SourceType {
}
#[must_use]
pub fn with_script(mut self, yes: bool) -> Self {
pub const fn with_script(mut self, yes: bool) -> Self {
if yes {
self.module_kind = ModuleKind::Script;
}
@ -77,7 +77,7 @@ impl SourceType {
}
#[must_use]
pub fn with_module(mut self, yes: bool) -> Self {
pub const fn with_module(mut self, yes: bool) -> Self {
if yes {
self.module_kind = ModuleKind::Module;
} else {
@ -87,7 +87,7 @@ impl SourceType {
}
#[must_use]
pub fn with_typescript(mut self, yes: bool) -> Self {
pub const fn with_typescript(mut self, yes: bool) -> Self {
if yes {
self.language = Language::TypeScript;
}
@ -95,7 +95,7 @@ impl SourceType {
}
#[must_use]
pub fn with_typescript_definition(mut self, yes: bool) -> Self {
pub const fn with_typescript_definition(mut self, yes: bool) -> Self {
if yes {
self.language = Language::TypeScriptDefinition;
}
@ -103,7 +103,7 @@ impl SourceType {
}
#[must_use]
pub fn with_jsx(mut self, yes: bool) -> Self {
pub const fn with_jsx(mut self, yes: bool) -> Self {
if yes {
self.variant = LanguageVariant::Jsx;
}
@ -111,7 +111,7 @@ impl SourceType {
}
#[must_use]
pub fn with_always_strict(mut self, yes: bool) -> Self {
pub const fn with_always_strict(mut self, yes: bool) -> Self {
self.always_strict = yes;
self
}