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copied!<p>Concerning the assignment operators, the C++ standard says the following (C++03 §5.17/1):</p> <blockquote> <p>There are several assignment operators... <strong>all require a modifiable lvalue as their left operand</strong></p> </blockquote> <p>An array is not a modifiable lvalue.</p> <p>However, assignment to a class type object is defined specially (§5.17/4):</p> <blockquote> <p>Assignment to objects of a class is defined by the copy assignment operator.</p> </blockquote> <p>So, we look to see what the implicitly-declared copy assignment operator for a class does (§12.8/13):</p> <blockquote> <p>The implicitly-defined copy assignment operator for class X performs memberwise assignment of its subobjects. ... Each subobject is assigned in the manner appropriate to its type:<br> ...<br> -- if the subobject is an array, each element is assigned, in the manner appropriate to the element type<br> ...</p> </blockquote> <p>So, for a class type object, arrays are copied correctly. Note that if you provide a user-declared copy assignment operator, you cannot take advantage of this, and you'll have to copy the array element-by-element.</p> <hr> <p>The reasoning is similar in C (C99 §6.5.16/2):</p> <blockquote> <p>An assignment operator shall have a modiﬁable lvalue as its left operand.</p> </blockquote> <p>And §6.3.2.1/1:</p> <blockquote> <p>A modiﬁable lvalue is an lvalue that does not have array type... [other constraints follow]</p> </blockquote> <p>In C, assignment is much simpler than in C++ (§6.5.16.1/2):</p> <blockquote> <p>In simple assignment (=), the value of the right operand is converted to the type of the assignment expression and replaces the value stored in the object designated by the left operand.</p> </blockquote> <p>For assignment of struct-type objects, the left and right operands must have the same type, so the value of the right operand is simply copied into the left operand.</p>

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