Android Guides | Samples

Java.Lang.ExceptionInInitializerError.Exception Property

Returns the exception that is the cause of this error.

Syntax

[get: Android.Runtime.Register("getException", "()Ljava/lang/Throwable;", "GetGetExceptionHandler")]
public virtual Throwable Exception { get; }

Remarks

Returns the exception that is the cause of this error.

[Android Documentation]

Requirements

Namespace: Java.Lang
Assembly: Mono.Android (in Mono.Android.dll)
Assembly Versions: 0.0.0.0
Since: Added in API level 1