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V6033. An item with the same key has al…
Analyzer Diagnostics
General Analysis (C++)
General Analysis (C#)
General Analysis (Java)
Diagnosis of micro-optimizations (C++)
Diagnosis of 64-bit errors (Viva64, C++)
MISRA errors
AUTOSAR errors
Additional information
Contents

V6033. An item with the same key has already been added.

May 07 2018

The analyzer detected the following strange situation: items are being added to a dictionary (containers of type 'map', etc.) or set (containers of type 'set', etc.) while having the same keys that are already present in these containers, which will result in ignoring the newly added items. This issue may be a sign of a typo and result in incorrect filling of the container.

Consider the following example with incorrect dictionary initialization:

Map<String, Integer> map = new HashMap<String, Integer>() {{
  put("a", 10);
  put("b", 20);
  put("a", 30); // <=
}};

The programmer made a typo in the last line of the code performing dictionary initialization, as the 'a' key is already in the dictionary. As a result, this dictionary will contain 2 values.

To fix the error, we need to use a correct key value:

Map<String, Integer> map = new HashMap<String, Integer>() {{
  put("a", 10);
  put("b", 20);
  put("c", 30); 
}};

A similar error may occur when initializing a set:

HashSet<String> someSet = new HashSet<String>(){{
  add("First");
  add("Second");
  add("Third");
  add("First"); // <=
  add("Fifth");
}};

A typo results in an attempt to write string 'First' instead of the 'Fourth' key to the 'someSet' set, but since this key is already in the set, it will be ignored.

To fix this error, we need to fix the initialization list:

HashSet<String> someSet = new HashSet<String>(){{
  add("First");
  add("Second");
  add("Third");
  add("Fourth");
  add("Fifth");
}};

This diagnostic is classified as:

You can look at examples of errors detected by the V6033 diagnostic.

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