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    <p>According to <a href="http://en.cppreference.com/w/cpp/language/operator_precedence" rel="nofollow">C++ Operator Precedence</a>:</p> <ol> <li><p>"*" has the same precedence as <em>prefix</em> "++" but must be avaluated <strong>rigth to left</strong>.</p> <p>printf("%s ", ++*p);</p></li> </ol> <p>So first <code>*p</code> is evaluated, then <code>++(*p)</code>, leading to the second character in the first string.</p> <ol> <li><p>"*" has <strong>less</strong> precedence than <em>suffix</em> "++".</p> <p>printf("%s ", *p++);</p></li> </ol> <p>So first <code>p</code> is incremented, <strong>but</strong> it is a <em>post-increment</em>. The value returned from the operation <em>is the original one</em>. This way, the <code>*</code> operates over the original pointer, that pointed to the second char on the first string.</p> <p>Note that, this time, <code>++</code> is operating over <code>p</code>, and not over <code>*p</code>.</p> <ol> <li><p>Since "2", <code>p</code> points to the second string. When you do <code>++*p</code> you are now pointing to the second character of the second string ("s"). As you are again using a <em>pre-increment</em>, the value passed to <code>printf</code> is already changed.</p> <p>printf("%s ", ++*p);</p></li> </ol> <p>I may get clearer if you do a little change and print the pointer value aswell (ignore the warnings):</p> <pre><code>printf("%s [%p]\n", ++*p, p ); printf("%s [%p]\n ", *p++, p ); printf("%s [%p]\n ", ++*p, p ); nowledge [0x7fff6f5519e0] nowledge [0x7fff6f5519e8] s [0x7fff6f5519e8] </code></pre>
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    1. CONice answer, but aren't those prints UB since you're not guaranteed in what order the function arguments are evaluated?
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    2. COThanks fljx .It is a really good explanation.
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    3. CO@JoeFish - Yes, you are right. Those expressions have side effects, and that's why I said to ignore the compiler warnings. In a production code we must not rely in such expressions. Thanks for asking!
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