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Comparison between inconvertible types

ID: js/comparison-between-incompatible-types
Kind: problem
Security severity: 
Severity: warning
Precision: high
Tags:
   - reliability
   - correctness
   - external/cwe/cwe-570
   - external/cwe/cwe-571
Query suites:
   - javascript-security-and-quality.qls

Click to see the query in the CodeQL repository

In JavaScript, equality operators (==, !=, ===, !==) and relational operators (<, <=, >, >=) can be applied to values of arbitrary types. However, if the operands cannot be converted to a common type, the result of the comparison will always be trivially true (for equality) or false (for inequality). Such comparisons are often due to a typo or a misunderstanding of the language semantics.

Recommendation

Inspect the comparison carefully to check whether it is due to a typo. If one of the operands is a constant, replace it with a constant of the right type. Otherwise, introduce appropriate function calls to convert the operands to a common type.

Example

The following code attempts to check whether the global variable window is defined:

if (typeof window !== undefined)
	console.log("Running in a browser.");

However, this test is ineffective: typeof always returns a string, never undefined, so the if condition will always evaluate to true. Instead, the result of typeof should be compared to the string literal "undefined":

if (typeof window !== "undefined")
	console.log("Running in a browser.");

As another example, consider the following code snippet, which is meant to check whether the string "hello" occurs in the string held in variable text.

if (text.indexOf("hello" >= 0))
	console.log("Found it.");

Note, however, that the test has been mistyped: the closing parenthesis of the call to index should come before the operator >=, not after it. As it stands, this code performs a greater-or-equal comparison between the constant string "hello" and the number zero, which evaluates to false. This value is then passed to indexOf, which converts it to the string "false" and returns the first index at which this string occurs in text (or -1 if it does not occur at all).

To fix this issue, the test should be rebracketed like this:

if (text.indexOf("hello") >= 0)
	console.log("Found it.");

References

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