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POIs there a way to free only a part of the dynamically allocated array in C\C++ (shrink existing array)?
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  1. POIs there a way to free only a part of the dynamically allocated array in C\C++ (shrink existing array)?
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    copied!<p>In a program I allocate a huge multidimensional array, do some number-crunching, then only the first part of that array is of further interest and I'd like to free just a part of the array and continue to work with the data in the first part. I tried using realloc, but I am not sure whether this is the correct way, given I must preserve the data in the array and preferably avoid copying of that chunk in memory.</p> <pre><code>#include &lt;cstring&gt; #include &lt;cassert&gt; #include &lt;iostream&gt; using namespace std; void FillArrayThenTruncate(char* my_array, const int old_size, int* new_size); int main() { const int initial_size = 1024*1024*1024; char* my_array = static_cast&lt;char*&gt;(malloc(initial_size)); assert(my_array); int new_size; FillArrayThenTruncate(my_array, initial_size, &amp;new_size); for(int i = 0; i &lt; new_size; ++i) cout &lt;&lt; static_cast&lt;int&gt;(my_array[i]) &lt;&lt; endl; } void FillArrayThenTruncate(char* my_array, const int old_size, int* new_size) { //do something with my_array: memset(my_array, 0, old_size); for(int i = 0; i &lt; 10; ++i) my_array[i] = i % 3; //cut the initial array *new_size = 10; void* new_array = realloc(my_array, *new_size); cout &lt;&lt; "Old array pointer: " &lt;&lt; static_cast&lt;void*&gt;(my_array) &lt;&lt; endl; cout &lt;&lt; "New array pointer: " &lt;&lt; new_array &lt;&lt; endl; my_array = static_cast&lt;char*&gt;(new_array); assert(my_array != NULL); } </code></pre> <p>UPDATE:<br> * Please do not bother to suggest to use STL. The question is about C array.<br> * Thanks to "R Samuel Klatchko" for pointing out the bug in the code above.</p>
 

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