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    <p>I think the main thing here is that conditionals do short-circuiting, which is a form of control flow. What helps is to re-write the code to make that explicit. This sort of normalization is common when doing program analysis. Some ad-hoc normalization (not formal and a machine wouldn't generate this, but it gets the point across) would make your code look like the following:</p> <pre><code>public static boolean validatePassword(String password) { int len = password.length(); //evaluate 'len &lt; 8 || len &gt; 20' bool cond1 = len &lt; 8; if (!cond1) cond1 = len &gt; 20; //do the if test if (cond1) return false; boolean hasLetters = false; boolean hasDigits = false; //for loops are equivalent to while loops int i = 0; while(i &lt; len) { if (!Character.isLetterOrDigit(password.charAt(i))) return false; //evaluate 'hasDigits || Character.isDigit(password.charAt(i))' bool hasDigitsVal = hasDigits; if (!hasDigitsVal) hasDigitsVal = Character.isDigit(password.charAt(i)); //hasDigits = ... hasDigits = hasDigitsVal //evaluate 'hasLetters || Character.isLetter(password.charAt(i))' bool hasLettersVal = hasLetters; if (!hasLettersVal) hasLettersVal = Character.isLetter(password.charAt(i)); //hasLetters = ... hasLetters = hasLettersVal; i++; } //evaluate 'hasDigits &amp;&amp; hasLetters' bool cond2 = hasDigits; if (cond2) cond2 = hasLetters; //return ... return cond2; } </code></pre> <p>Notice how the <code>||</code> and <code>&amp;&amp;</code> operators essentially just add <code>if</code> statements to the code. Also notice that you now have 6 <code>if</code> statements and one <code>while</code> loop! Maybe that is the 7 you were looking for?</p> <hr> <p>About multiple exit points, that's a red herring. Consider each function as having one exit node, the end of the function. If you have multiple <code>return</code> statements, each <code>return</code> statement would draw an edge to that exit node.</p> <pre><code>void foo() { if (cond1) return a; if (cond2) return b; return c; } </code></pre> <p>The graph would look like this, where <code>-----val----&gt; EXIT</code> means exiting the function with a value of <code>val</code>:</p> <pre><code>START -&gt; cond1 ------------------------a------------&gt; EXIT | | cond2 ------------------------b----------------+ | | return -----------------------c----------------| </code></pre> <p>If you re-write the code, then you just basically add another "pre-return" node that then goes to the exit node:</p> <pre><code>void foo() { int val; if (cond1) { val= a; } else { if (cond2) { val= b; } else { val= c; } } return val; } </code></pre> <p>Now it looks like this:</p> <pre><code>START -&gt; cond1 ---&gt; val=a --------------------------&gt; return ----val----&gt; EXIT | | cond2 ---&gt; val=b ------------------------------+ | | + -----&gt; val=c ------------------------------+ </code></pre> <p>It's still as complex, and the code is just uglier.</p>
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    1. POCyclomatic Complexity in piece of code with multiple exit points
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    1. CO"I wouldn't say it has multiple exit points" what do you mean by that? I used to thing that everytime you return you add an exit point (e.g. `if (somecondition) return;`)
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    2. CO@tmh: i updated my answer to show what I mean. thinking about multiple `return` statements is a red herring IMO
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