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copied!<p><code>sizeof</code> operator statically evaluates to the size of the thing you are passing to it. A <code>char*</code> is a pointer which, depending on the machine architecture has a specific size (4 bytes on 32 bit systems and 8 bytes on 64 bit machines). To accomplish what you are trying to do, I suggest you use the <code>string</code> type which you can use by adding <code>#include <string></code> along with <code>using namespace std;</code> to your source file.</p> <pre><code>string line; cin >> line; cout << line.length() << endl; </code></pre> <p>It's less error prone and easier to use.</p> <p>By the way, the thing you've tried to do is really dangerous. In fact, when you use <code>cin >> myArray</code>, you should have already allocated some memory for <code>myArray</code> which you haven't done. This will cause memory corruption which might crash your program and possibly put it to buffer overrun attacks.</p> <p>A simple array in C++ has no idea about its size. You can use <code>sizeof</code> <b>only if the array is statically allocated</b> and you are using <code>sizeof</code> on the array itself, not another pointer to it, for example this won't work as you might expect:</p> <pre><code>int x[5]; int *a = &x[0]; // a[i] is now the same as x[i] but: cout << sizeof(x) << endl; // prints 20, assuming int is 32 bits long cout << sizeof(a) << endl; // prints 4, assuming a pointer is 32 bits long </code></pre> <p>Note that the total size of the array is printed on the first line, not the element count. You can use <code>sizeof(x)/sizeof(*x)</code> to find out element count in static arrays. This thing is not possible for dynamically allocated arrays using <code>new</code>. In fact C++ arrays are very error prone and you should take extreme care when working with them and you'd better use <code>vector</code> and <code>string</code> in most cases instead.</p>

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