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The Dictionary is a reference type. This means its bits cannot just be checked. Further, two different Dictionaries may have different orders of keys, even if they are equal.
Example. As we begin, please note that it would be possible to create a generic method abstraction of this logic. This article does not use that syntax. Two Dictionary instances are allocated upon the managed heap at program runtime.
And: Each dictionary has three key-value pairs. The data is equal in every way.
Then, the equality test begins. The two Count properties are checked for equality, then every key from the first Dictionary is looked up in the second Dictionary, and its value is also checked.
C# program that tests Dictionary instances for equality using System; using System.Collections.Generic; class Program { static void Main() { // Create a dictionary and add several elements to it. var dict = new Dictionary<string, int>(); dict.Add("cat", 2); dict.Add("dog", 3); dict.Add("x", 4); // Create another dictionary. var dict2 = new Dictionary<string, int>(); dict2.Add("cat", 2); dict2.Add("dog", 3); dict2.Add("x", 4); // Test for equality. bool equal = false; if (dict.Count == dict2.Count) // Require equal count. { equal = true; foreach (var pair in dict) { int value; if (dict2.TryGetValue(pair.Key, out value)) { // Require value be equal. if (value != pair.Value) { equal = false; break; } } else { // Require key be present. equal = false; break; } } } Console.WriteLine(equal); } } Output True
In this example, the algorithm used has several details. If you do not correctly add every constraint, it will not return correct results. The counts of the two dictionaries must be equal.
This prevents a larger second dictionary from matching a smaller first. Every key must be found, and its value must be equal in both dictionaries. If the key is not found, the match fails.
Summary. It is possible to compare two Dictionary instances for equality using a custom method with key-value pair checking. There are other possible implementations, but this one is fairly resource-efficient and also simple.
Tip: When testing this method, make sure to change keys and values in both dictionaries.
And: As always, incorrect code is worse than no code at all, particularly if you are doing something important.