huridocs/uwazi

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

title: no-return-assign

rule_type: suggestion

One of the interesting, and sometimes confusing, aspects of JavaScript is that assignment can happen at almost any point. Because of this, an errant equals sign can end up causing assignment when the true intent was to do a comparison. This is especially true when using a return statement. For example:

function doSomething() {
    return foo = bar + 2;
}

It is difficult to tell the intent of the return statement here. It's possible that the function is meant to return the result of bar + 2, but then why is it assigning to foo? It's also possible that the intent was to use a comparison operator such as == and that this code is an error.

Because of this ambiguity, it's considered a best practice to not use assignment in return statements.

Rule Details

This rule aims to eliminate assignments from return statements. As such, it will warn whenever an assignment is found as part of return.

Options

The rule takes one option, a string, which must contain one of the following values:

• except-parens (default): Disallow assignments unless they are enclosed in parentheses.
• always: Disallow all assignments.

except-parens

This is the default option. It disallows assignments unless they are enclosed in parentheses.

Examples of incorrect code for the default "except-parens" option:

::: incorrect

/*eslint no-return-assign: "error"*/

function doSomething() {
    return foo = bar + 2;
}

function doSomething() {
    return foo += 2;
}

const foo = (a, b) => a = b

const bar = (a, b, c) => (a = b, c == b)

function doSomething() {
    return foo = bar && foo > 0;
}

:::

Examples of correct code for the default "except-parens" option:

::: correct

/*eslint no-return-assign: "error"*/

function doSomething() {
    return foo == bar + 2;
}

function doSomething() {
    return foo === bar + 2;
}

function doSomething() {
    return (foo = bar + 2);
}

const foo = (a, b) => (a = b)

const bar = (a, b, c) => ((a = b), c == b)

function doSomething() {
    return (foo = bar) && foo > 0;
}

:::

always

This option disallows all assignments in return statements. All assignments are treated as problems.

Examples of incorrect code for the "always" option:

::: incorrect

/*eslint no-return-assign: ["error", "always"]*/

function doSomething() {
    return foo = bar + 2;
}

function doSomething() {
    return foo += 2;
}

function doSomething() {
    return (foo = bar + 2);
}

:::

Examples of correct code for the "always" option:

::: correct

/*eslint no-return-assign: ["error", "always"]*/

function doSomething() {
    return foo == bar + 2;
}

function doSomething() {
    return foo === bar + 2;
}

:::

When Not To Use It

If you want to allow the use of assignment operators in a return statement, then you can safely disable this rule. Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

title: consistent-return

rule_type: suggestion

Unlike statically-typed languages which enforce that a function returns a specified type of value, JavaScript allows different code paths in a function to return different types of values.

A confusing aspect of JavaScript is that a function returns undefined if any of the following are true:

• it does not execute a return statement before it exits
• it executes return which does not specify a value explicitly
• it executes return undefined
• it executes return void followed by an expression (for example, a function call)
• it executes return followed by any other expression which evaluates to undefined

If any code paths in a function return a value explicitly but some code path do not return a value explicitly, it might be a typing mistake, especially in a large function. In the following example:

• a code path through the function returns a Boolean value true
• another code path does not return a value explicitly, therefore returns undefined implicitly

function doSomething(condition) {
    if (condition) {
        return true;
    } else {
        return;
    }
}

Rule Details

This rule requires return statements to either always or never specify values. This rule ignores function definitions where the name begins with an uppercase letter, because constructors (when invoked with the new operator) return the instantiated object implicitly if they do not return another object explicitly.

Examples of incorrect code for this rule:

::: incorrect

/*eslint consistent-return: "error"*/

function doSomething(condition) {
    if (condition) {
        return true;
    } else {
        return;
    }
}

function doSomething(condition) {
    if (condition) {
        return true;
    }
}

:::

Examples of correct code for this rule:

::: correct

/*eslint consistent-return: "error"*/

function doSomething(condition) {
    if (condition) {
        return true;
    } else {
        return false;
    }
}

function Foo() {
    if (!(this instanceof Foo)) {
        return new Foo();
    }

    this.a = 0;
}

:::

Options

This rule has an object option:

• "treatUndefinedAsUnspecified": false (default) always either specify values or return undefined implicitly only.
• "treatUndefinedAsUnspecified": true always either specify values or return undefined explicitly or implicitly.

treatUndefinedAsUnspecified

Examples of incorrect code for this rule with the default { "treatUndefinedAsUnspecified": false } option:

::: incorrect

/*eslint consistent-return: ["error", { "treatUndefinedAsUnspecified": false }]*/

function foo(callback) {
    if (callback) {
        return void callback();
    }
    // no return statement
}

function bar(condition) {
    if (condition) {
        return undefined;
    }
    // no return statement
}

:::

Examples of incorrect code for this rule with the { "treatUndefinedAsUnspecified": true } option:

::: incorrect

/*eslint consistent-return: ["error", { "treatUndefinedAsUnspecified": true }]*/

function foo(callback) {
    if (callback) {
        return void callback();
    }
    return true;
}

function bar(condition) {
    if (condition) {
        return undefined;
    }
    return true;
}

:::

Examples of correct code for this rule with the { "treatUndefinedAsUnspecified": true } option:

::: correct

/*eslint consistent-return: ["error", { "treatUndefinedAsUnspecified": true }]*/

function foo(callback) {
    if (callback) {
        return void callback();
    }
    // no return statement
}

function bar(condition) {
    if (condition) {
        return undefined;
    }
    // no return statement
}

:::

When Not To Use It

If you want to allow functions to have different return behavior depending on code branching, then it is safe to disable this rule. Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

title: consistent-return

rule_type: suggestion

Unlike statically-typed languages which enforce that a function returns a specified type of value, JavaScript allows different code paths in a function to return different types of values.

A confusing aspect of JavaScript is that a function returns undefined if any of the following are true:

• it does not execute a return statement before it exits
• it executes return which does not specify a value explicitly
• it executes return undefined
• it executes return void followed by an expression (for example, a function call)
• it executes return followed by any other expression which evaluates to undefined

If any code paths in a function return a value explicitly but some code path do not return a value explicitly, it might be a typing mistake, especially in a large function. In the following example:

• a code path through the function returns a Boolean value true
• another code path does not return a value explicitly, therefore returns undefined implicitly

function doSomething(condition) {
    if (condition) {
        return true;
    } else {
        return;
    }
}

Rule Details

This rule requires return statements to either always or never specify values. This rule ignores function definitions where the name begins with an uppercase letter, because constructors (when invoked with the new operator) return the instantiated object implicitly if they do not return another object explicitly.

Examples of incorrect code for this rule:

::: incorrect

/*eslint consistent-return: "error"*/

function doSomething(condition) {
    if (condition) {
        return true;
    } else {
        return;
    }
}

function doSomething(condition) {
    if (condition) {
        return true;
    }
}

:::

Examples of correct code for this rule:

::: correct

/*eslint consistent-return: "error"*/

function doSomething(condition) {
    if (condition) {
        return true;
    } else {
        return false;
    }
}

function Foo() {
    if (!(this instanceof Foo)) {
        return new Foo();
    }

    this.a = 0;
}

:::

Options

This rule has an object option:

• "treatUndefinedAsUnspecified": false (default) always either specify values or return undefined implicitly only.
• "treatUndefinedAsUnspecified": true always either specify values or return undefined explicitly or implicitly.

treatUndefinedAsUnspecified

Examples of incorrect code for this rule with the default { "treatUndefinedAsUnspecified": false } option:

::: incorrect

/*eslint consistent-return: ["error", { "treatUndefinedAsUnspecified": false }]*/

function foo(callback) {
    if (callback) {
        return void callback();
    }
    // no return statement
}

function bar(condition) {
    if (condition) {
        return undefined;
    }
    // no return statement
}

:::

Examples of incorrect code for this rule with the { "treatUndefinedAsUnspecified": true } option:

::: incorrect

/*eslint consistent-return: ["error", { "treatUndefinedAsUnspecified": true }]*/

function foo(callback) {
    if (callback) {
        return void callback();
    }
    return true;
}

function bar(condition) {
    if (condition) {
        return undefined;
    }
    return true;
}

:::

Examples of correct code for this rule with the { "treatUndefinedAsUnspecified": true } option:

::: correct

/*eslint consistent-return: ["error", { "treatUndefinedAsUnspecified": true }]*/

function foo(callback) {
    if (callback) {
        return void callback();
    }
    // no return statement
}

function bar(condition) {
    if (condition) {
        return undefined;
    }
    // no return statement
}

:::

When Not To Use It

If you want to allow functions to have different return behavior depending on code branching, then it is safe to disable this rule. Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/

For more information visit Source: http://eslint.org/docs/rules/