logical-assignment-operators
Require or disallow logical assignment operator shorthand
Some problems reported by this rule are automatically fixable by the --fix
command line option
Some problems reported by this rule are manually fixable by editor suggestions
ES2021 为逻辑运算符 ||
、&&
和 ??
引入赋值运算符简写。
在此之前仅允许算术运算符如 +
或 *
(参见 operator-assignment 规则)进行简写。
如果赋值目标和逻辑表达式的左表达式相同,则可以使用该简写。
比如 a = a || b
就可以简写为 a ||= b
。
规则细节
此规则要求或禁止使用逻辑赋值运算符简写。
选项
此规则有一个字符串和一个对象选项。 字符串选项:
"always"
(默认值)"never"
对象选项(仅当字符串选项被设置为 "always"
时可用):
"enforceForIfStatements": false
(默认值)不会检查等价的if
语句"enforceForIfStatements": true
检查等价的if
语句
always
使用此规则与默认的 "always"
选项的错误示例:
/*eslint logical-assignment-operators: ["error", "always"]*/
a = a || b
a = a && b
a = a ?? b
a || (a = b)
a && (a = b)
a ?? (a = b)
使用此规则与默认的 "always"
选项的正确示例:
/*eslint logical-assignment-operators: ["error", "always"]*/
a = b
a += b
a ||= b
a = b || c
a || (b = c)
if (a) a = b
never
使用此规则与 "never"
选项的错误示例:
/*eslint logical-assignment-operators: ["error", "never"]*/
a ||= b
a &&= b
a ??= b
使用此规则与 "never"
选项的正确示例:
/*eslint logical-assignment-operators: ["error", "never"]*/
a = a || b
a = a && b
a = a ?? b
enforceForIfStatements
此选项检查能够表达为逻辑赋值运算符的 if 语句的其他模式。
使用此规则与 ["always", { enforceForIfStatements: true }]
选项的错误示例:
/*eslint logical-assignment-operators: ["error", "always", { enforceForIfStatements: true }]*/
if (a) a = b // <=> a &&= b
if (!a) a = b // <=> a ||= b
if (a == null) a = b // <=> a ??= b
if (a === null || a === undefined) a = b // <=> a ??= b
使用此规则与 ["always", { enforceForIfStatements: true }]
选项的正确示例:
/*eslint logical-assignment-operators: ["error", "always", { enforceForIfStatements: true }]*/
if (a) b = c
if (a === 0) a = b
何时不用
使用逻辑操作符赋值简写是风格上的选择。将此规则关闭可以允许开发者根据具体案例选择更加易读的风格。
Version
This rule was introduced in ESLint v8.24.0.