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    <p>Your problem (apart from compilation issues noted in my comment to the question) is that pointers are not as small as <code>3</code>. You need:</p> <pre><code>for (p = alph; p &lt; alph + sizeof(alph); p++) </code></pre> <p>for the loop. Note that <code>sizeof()</code> generates a compile-time constant. (In C99 or C2011, that is not always the case; in C++, it is always the case, AFAIK). In this context, <code>sizeof()</code> is a fancy way of writing <code>3</code>, but if you add new characters to the array, it adjusts automatically, whereas if you write <code>3</code> and change things, you have to remember to change the <code>3</code> to the new value too.</p> <hr> <h3>Ruminations on the use of <code>sizeof()</code></h3> <p>As <a href="https://stackoverflow.com/users/596781/kerrek-sb">Kerrek SB</a> points out, <code>sizeof()</code> returns the size of an array in bytes. By definition, <code>sizeof(char) == 1</code>, so in this context, it was safe to use <code>sizeof()</code> on the array. There are also times when it is not safe - or you have to do some extra work. If you had an array of some other type, you can use:</p> <pre><code>SomeType array[] = { 2, 3, 5, 7, 11, 13 }; </code></pre> <p>then the number of elements in the array is <code>(sizeof(array)/sizeof(array[0]))</code>. That is, the number of elements is the total size of the array, in bytes, divided by the size of one element (also in bytes).</p> <p>The other big gotcha is that if you 'pass an array' as a function argument, you can't use <code>sizeof()</code> on it to get the correct size - you get the size of a pointer instead. This is a good reason for not using C-style strings or C-style arrays: use <code>std::string</code> and <code>std::vector&lt;SomeType&gt;</code> instead, not least because you can find their actual size reliably with a member function call.</p>
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    1. POHow to use pointers in for loops to count characters?
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    1. USJonathan Leffler
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    1. COIt's worth noting (especially for a beginner) that `sizeof` does **not** return the array size, but rather the total memory size. It only so happens that the size of one unit is 1.
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      1. USKerrek SB
    2. COcool, now i know about sizeof() :D, so this is an example of implementing dynamic memory
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      1. USPatrick Kennedy
    3. CO@user1177261 no, dynamic memory would be allocated with `new`. `sizeof(alph)` is just a way for the compiler to calculate the size of an array for you. It happens to be `3` in this case (because it holds 3 `char`s and each one is `1` byte), so `alph + sizeof(alph)` is just like `alph + 3`. The advantage of using `sizeof` is that when you change the size of your array, you don't have to go back and redo all the numbers.
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      1. USSeth Carnegie
 

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