Is it safe to return a pointer to a local variable in Go?
Jul 22, 2025 am 01:47 AM
No, returning a pointer to a local variable is not safe in Go. Because local variables are allocated on the stack, the memory of the function is no longer valid after returning, resulting in the returned pointer becoming a dangling pointer. 1. When returning local variable pointers, the variables are out of scope as the function exits, and memory may be reused; 2. The Go compiler sometimes allocates variables to the heap through escape analysis, but this is unreliable; 3. Safe ways include using new() allocation, returning values instead of pointers, or creating heap-allocated values through structures or slice literals. Therefore, you should actively avoid such problems when writing code, rather than relying on the compiler to ensure the security.
No, it's not safe to return a pointer to a local variable in Go. The reason is that local variables are allocated on the stack, and once the function returns, that stack memory is no longer valid. If you return a pointer to it, you end up with a dangling pointer that points to invalid memory.
What happens if you return a pointer to a local variable?
When you do something like this:
func badFunc() *int {
x := 10
return &x
} You're returning a pointer to a variable ( x ) that lives only as long as badFunc is running. Once the function exits, x goes out of scope, and the memory it occurred may be reused for other purposes.
The Go compiler actually catches some of these cases and issues a warning or error during compilation. For example, it might say "escapes to heap" or silently move the variable to the heap to avoid an immediate crash. But relying on that behavior isn't a good practice.
So while your program might seem to work sometimes (especially under light load or simple testing), it can fail unpredictably when that memory is reused.
When is it OK to return a pointer?
It's fine to return a pointer if the value being pointed to is either:
- Allocated on the heap (eg, using
new()or by taking the address of a struct literal) - Passed into the function from elsewhere (like a parameter)
Here's a correct example:
func goodFunc() *int {
x := new(int)
*x = 10
Return x
}Or even shorter:
func goodFunc() *int {
return &[]int{10}[0]
}In both of these cases, the data survives the function call because it's not tied to the function's stack frame.
Why doesn't Go prevent this completely?
Go does try to detect when a local variable escapes to the heap and automatically moves it there. This is part of its escape analysis. So in many cases, the program won't crash — but that doesn't mean it's safe or correct.
You should treat returning points to locals as undefined behavior, even if the compiler tries to help.
How to avoid this mistake
If you're writing a function that needs to return a pointer, consider:
- Using
new()explicitly - Returning a value instead of a pointer if possible
- Creating the value inside a struct or slice literal, which often gets heap-allocated automatically
For example:
func createValue() *int {
return &[]int{42}[0] // Safe
}Or:
func createValue() *int {
val := new(int)
*val = 42
return val
}These are more explicit and safer than referencing a local variable and taking its address.
Basically that's it. Although the compiler sometimes helps you with the bottom of the problem, don't rely on it. It is reliable to be careful when writing code.
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