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    <p><strong>A const member function, is a member function that does not mutate its member variables.</strong> </p> <p><strong>const on a member function does not imply const char *. Which would mean that you can't change the data in the address the pointer holds.</strong> </p> <p>Your example does not mutate the member variables themselves.</p> <p>A const on a member function, will ensure that you treat all of your member variables as const. </p> <p>That means if you have:</p> <pre><code>int x; char c; char *p; </code></pre> <p>Then you will have:</p> <pre><code>const int x; const char c; char * const p; //&lt;-- means you cannot change what p points to, but you can change the data p points to </code></pre> <p>There are 2 types of const pointers. A const member function uses the one I've listed above. </p> <hr> <p><strong>A way to get the error you want:</strong></p> <p>Try changing:</p> <pre><code>char * rep; </code></pre> <p>to:</p> <pre><code>char rep[1024]; </code></pre> <p>And remove this line:</p> <pre><code>rep = new char [strlen(s)+1]; </code></pre> <p>It will throw the error you are expecting (can't modify members because of const keyword)</p> <p>Because there is only 1 type of const array. And that means you cannot modify any of its data. </p> <hr> <p>Now the whole system is actually broken with the following example:</p> <pre><code>class String { private: char rep2[1024]; char* rep; ... String :: String (const char* s) { rep = rep2; strcpy (rep, s); } </code></pre> <p><strong>So the lesson to learn here is that the const keyword on member functions does not ensure that your object will not change at all.</strong> </p> <p>It only ensures that each member variable will be treated as const. And for pointers, there is a big diff between const char * and char * const.</p> <p>Most of the time a const member function will mean that the member function will not modify the object itself, but this is not always the case, as the above example shows. </p>
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    1. POWhy does this const member function allow a member variable to be modified?
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    1. USBrian R. Bondy
    1. USBrian R. Bondy
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    1. COThanks a lot for quick response & Very Neat Delineation. Stack Overflow really rocks
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      1. USmahesh
 

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