CodeQL documentation

Serialization check bypass

ID: cs/serialization-check-bypass
Kind: problem
Security severity: 7.8
Severity: warning
Precision: medium
Tags:
   - security
   - external/cwe/cwe-20
Query suites:
   - csharp-security-extended.qls
   - csharp-security-and-quality.qls

Click to see the query in the CodeQL repository

Fields that are deserialized should be validated, otherwise the deserialized object could contain invalid data.

This query finds cases where a field is validated in a constructor, but not in a deserialization method. This is an indication that the deserialization method is missing a validation step.

Recommendation

If a field needs to be validated, then ensure that validation is also performed during deserialization.

Example

The following example has the validation of the Age field in the constructor but not in the deserialization method:

using System;
using System.Runtime.Serialization;

[Serializable]
public class PersonBad : ISerializable
{
    public int Age;

    public PersonBad(int age)
    {
        if (age < 0)
            throw new ArgumentException(nameof(age));
        Age = age;
    }

    [OnDeserializing]
    void ISerializable.GetObjectData(SerializationInfo info, StreamingContext context)
    {
        Age = info.GetInt32("age");  // BAD - write is unsafe
    }
}

The problem is fixed by adding validation to the deserialization method as follows:

using System;
using System.Runtime.Serialization;

[Serializable]
public class PersonGood : ISerializable
{
    public int Age;

    public PersonGood(int age)
    {
        if (age < 0)
            throw new ArgumentException(nameof(age));
        Age = age;
    }

    [OnDeserializing]
    void ISerializable.GetObjectData(SerializationInfo info, StreamingContext context)
    {
        int age = info.GetInt32("age");
        if (age < 0)
            throw new SerializationException(nameof(Age));
        Age = age;  // GOOD - write is safe
    }
}

References

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