CodeQL documentation

Expression has no effect

ID: go/useless-expression
Kind: problem
Security severity: 
Severity: warning
Precision: very-high
   - maintainability
   - correctness
   - external/cwe/cwe-480
   - external/cwe/cwe-561
Query suites:
   - go-security-and-quality.qls

Click to see the query in the CodeQL repository

An expression that has no effects (such as changing variable values or producing output) and occurs in a context where its value is ignored possibly indicates missing code or a latent bug.


Carefully inspect the expression to ensure it is not a symptom of a bug.


The following example shows a named type Timestamp that is an alias for int, representing time stamps expressed as the number of seconds elapsed since some epoch. The addDays method returns a time stamp that is a given number of days after another time stamp, without modifying that time stamp.

However, when addDays is used in function test, its result is discarded, perhaps because the programmer mistakenly assumed that addDays updates the time stamp in place.

package main

import "fmt"

type Timestamp int

func (t Timestamp) addDays(d int) Timestamp {
	return Timestamp(int(t) + d*24*3600)

func test(t Timestamp) {
	fmt.Printf("Before: %s\n", t)
	fmt.Printf("After: %s\n", t)

Instead, the result of addDays should be assigned back into t:

package main

import "fmt"

func testGood(t Timestamp) {
	fmt.Printf("Before: %s\n", t)
	t = t.addDays(7)
	fmt.Printf("After: %s\n", t)
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