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    <p>It appears the program is accessing the same memory location for every array <code>int tab[n]</code> you declared but, as mentioned, it should not work. </p> <p>But I think what's happening here is something like: you are allocating tab[] inside main(), let's say, under the address 0x00000001 (take it as an example only). The array have n integers but no values at all. </p> <p>Then you go into myInput(), declare the array again (same size), in other address, like 0x001F0000, and then set the values, one by one. </p> <p>So, when the function terminates, it frees the allocated memory of its variables, so your array does not exist anymore. </p> <p>But wait, this is C, so when you free the memory, you only tells the heap (or the memory allocator, in general) that the addresses can be used again. You do NOT exactly remove the values from memory. </p> <p>Then you call myDisp() and declare your array again. It appears the memory you just requested has higher priority and then it is somewhat given again to your program. So your array is again instanciated and on the same address. </p> <p>So, even if you did not fill it with values, the memory is read (as it is always valid in C) and the values are still there. </p> <p>Oh, and the array declared inside main()? Nothing happens to that. Try to print its values and I bet you'll not have the correct ones.</p> <p>That is my guess. </p> <p>EDIT: Just to see things happening: try to declare another array after tab (do not rename tab), say tab2, same length, and use it to put your values instead of tab, then let the program run again :)</p>
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    1. POStrange behaviour in a C code under gcc 4.4.3
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    1. COYes, well the array in the main() scope is completely useless. I just rewrote the code as I was given. Should this array declaration be commented of, the code still runs and gives the good values!
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    2. COSure. I'm blamming the memory heap for this situation. In my opinion, as I stated, You declare that array and the heap gives you one address X because it was on top of the heap. Then you free X and it will top the heap again (sounds odd), and then the heap will give you the same address when you declare tab again.
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