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1.. title:: clang-tidy - bugprone-suspicious-semicolon2 3bugprone-suspicious-semicolon4=============================5 6Finds most instances of stray semicolons that unexpectedly alter the meaning of7the code. More specifically, it looks for ``if``, ``while``, ``for`` and8``for-range`` statements whose body is a single semicolon, and then analyzes9the context of the code (e.g. indentation) in an attempt to determine whether10that is intentional.11 12.. code-block:: c++13 14 if (x < y);15 {16 x++;17 }18 19Here the body of the ``if`` statement consists of only the semicolon at the end20of the first line, and `x` will be incremented regardless of the condition.21 22 23.. code-block:: c++24 25 while ((line = readLine(file)) != NULL);26 processLine(line);27 28As a result of this code, `processLine()` will only be called once, when the29``while`` loop with the empty body exits with ``line == NULL``. The indentation30of the code indicates the intention of the programmer.31 32 33.. code-block:: c++34 35 if (x >= y);36 x -= y;37 38While the indentation does not imply any nesting, there is simply no valid39reason to have an `if` statement with an empty body (but it can make sense for40a loop). So this check issues a warning for the code above.41 42To solve the issue remove the stray semicolon or in case the empty body is43intentional, reflect this using code indentation or put the semicolon in a new44line. For example:45 46.. code-block:: c++47 48 while (readWhitespace());49 Token t = readNextToken();50 51Here the second line is indented in a way that suggests that it is meant to be52the body of the `while` loop - whose body is in fact empty, because of the53semicolon at the end of the first line.54 55Either remove the indentation from the second line:56 57.. code-block:: c++58 59 while (readWhitespace());60 Token t = readNextToken();61 62... or move the semicolon from the end of the first line to a new line:63 64.. code-block:: c++65 66 while (readWhitespace())67 ;68 69 Token t = readNextToken();70 71In this case the check will assume that you know what you are doing, and will72not raise a warning.73