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POWhy am I able to change the contents of const char *ptr?
 

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  1. POWhy am I able to change the contents of const char *ptr?
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    <p>I passed a pointer <code>ptr</code> to a function whose prototype takes it as <code>const</code>.</p> <pre><code>foo( const char *str ); </code></pre> <p>Which according to my understanding means that it will not be able to change the contents of <code>ptr</code> passed. Like in the case of <code>foo( const int i )</code>. If <code>foo()</code> tries to chnage the value of <code>i</code>, compiler gives error. <br/> But here I see that it can change the contents of <code>ptr</code> easily. <br/> Please have a look at the following code</p> <pre><code>foo( const char *str ) { strcpy( str, "ABC" ) ; printf( "%s(): %s\n" , __func__ , str ) ; } main() { char ptr[ ] = "Its just to fill the space" ; printf( "%s(): %s\n" , __func__ , ptr ) ; foo( const ptr ) ; printf( "%s(): %s\n" , __func__ , ptr ) ; return; } </code></pre> <p>On compilation, I only get a warning, no error:</p> <p><code>warning: passing argument 1 of ‘strcpy’ discards qualifiers from pointer target type</code></p> <p>and when I run it, I get an output instead of <code>Segmentation Fault</code></p> <blockquote> <p>main(): Its just to fill the space<br/> foo(): ABC<br/> main(): ABC<br/></p> </blockquote> <p>Now, my questions is<br/> <strong>1-</strong> What does <code>const char *str</code> in prototype actually means? <br/> Does this mean that function cannot change the contents of <code>str</code>? If that is so then how come the above program changes the value?<br/> <strong>2-</strong> How can I make sure that the contents of the pointer I have passed will not be changed?</p> <p>From "contents of the pointer" in the above stated question, I mean "contents of the memory pointed at by the pointer", not "address contained in the pointer".</p> <h2><strong>Edit</strong></h2> <p>Most replies say that this is because of <code>strcpy</code> and C implicit type conversion. But now I tried this</p> <pre><code>foo( const char *str ) { str = "Tim" ; // strcpy( str, "ABC" ) ; printf( "%s(): %s\n" , __func__ , str ) ; } </code></pre> <p>This time the output is, with no warning from compiler</p> <blockquote> <p>main(): Its just to fill the space<br/> foo(): Tim<br/> main(): Its just to fill the space<br/></p> </blockquote> <p>So apparently, memory pointed to by <code>str</code> is changed to the memory location containing <code>"Tim"</code> while its in <code>foo()</code>. Although I didn't use <code>strcpy()</code> this time.<br/> Is not <code>const</code> supposed to stop this? or my understanding is wrong?</p> <p>To me it seems that even with <code>const</code>, I can change the memory reference and the contents of memory reference too. Then what is the use?</p> <p>Can you give me an example where complier will give me error that I am trying to change a const pointer?</p> <p>Thanks to all of you for your time and effort.</p>
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    1. CO[this q+a](http://stackoverflow.com/questions/162480/const-in-c) may help you and is basically all you need also to understand why `str="Tim"` is ok, while `str[0] = something` wouldn't be --- duplicate-busters haven't caught this (luckly?)
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    2. CO+1 for the pointer to other thread. It was helpful but not as much as the selected answer. I don't think it was a duplicate if you consider the `strcpy` part of the original question. Thanks to all of you.
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