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    <p>If by "shallow copy", you mean that after assignment of a <code>struct</code> containing an array, the array would point to the original <code>struct</code>'s data, then: it can't. Each element of the array has to be copied over to the new <code>struct</code>. "Shallow copy" comes into the picture if your struct has pointers. If it doesn't, you <em>can't</em> do a shallow copy.</p> <p>When you assign a <code>struct</code> containing an array to some value, it cannot do a shallow copy, since that would mean assigning to an array, which is illegal. So the only copy you get is a deep copy.</p> <p>Consider:</p> <pre><code>#include &lt;stdio.h&gt; struct data { char message[6]; }; int main(void) { struct data d1 = { "Hello" }; struct data d2 = d1; /* struct assignment, (almost) equivalent to memcpy(&amp;d2, &amp;d1, sizeof d2) */ /* Note that it's illegal to say d2.message = d1.message */ d2.message[0] = 'h'; printf("%s\n", d1.message); printf("%s\n", d2.message); return 0; } </code></pre> <p>The above will print:</p> <pre><code>Hello hello </code></pre> <p>If, on the other hand, your <code>struct</code> had a pointer, <code>struct</code> assignment will only copy pointers, which is "shallow copy":</p> <pre><code>#include &lt;stdio.h&gt; #include &lt;stdlib.h&gt; #include &lt;string.h&gt; struct data { char *message; }; int main(void) { struct data d1, d2; char *str = malloc(6); if (str == NULL) { return 1; } strcpy(str, "Hello"); d1.message = str; d2 = d1; d2.message[0] = 'h'; printf("%s\n", d1.message); printf("%s\n", d2.message); free(str); return 0; } </code></pre> <p>The above will print:</p> <pre><code>hello hello </code></pre> <p>In general, given <code>struct T d1, d2;</code>, <code>d2 = d1;</code> is equivalent to <code>memcpy(&amp;d2, &amp;d1, sizeof d2);</code>, but if the struct has padding, that may or may not be copied.</p> <p><em>Edit</em>: In C, you <a href="https://stackoverflow.com/questions/2035066/type-of-an-array/2035255#2035255">can't assign to arrays</a>. Given:</p> <pre><code>int data[10] = { 0 }; int data_copy[10]; data_copy = data; </code></pre> <p>is illegal. So, as I said above, if you have an array in a <code>struct</code>, assigning to the struct <em>has to</em> copy the data element-wise in the array. You don't get shallow copy in this case: it doesn't make any sense to apply the term "shallow copy" to a case like this.</p>
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    1. POIs shallow copy sufficient for structures with char[]?
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    1. USAlok Singhal
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    1. POIs shallow copy sufficient for structures with char[]?
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    1. COYes, I understand the case for pointers but I'm just not sure for arrays.
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    2. CO@jasonline: see my edit. You can't assign to arrays in C, so you can't get a shallow copy if your struct has array(s) (and no pointers).
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    3. CO@Alok: Was typing up my answer. You are, of course, right that you can't assign to arrays, but that's beside the point. A shallow copy is done and not a deep copy, even on arrays.
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