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從手頭上的pdf里面的原文復制的……
6.5.3.2 Address and indirection operators
Semantics
3 The unary & operator returns the address of its operand. If the operand has type ‘‘type’’,
the result has type ‘‘pointer to type’’. If the operand is the result of a unary * operator,
neither that operator nor the & operator is evaluated and the result is as if both were
omitted, except that the constraints on the operators still apply and the result is not an
lvalue. Similarly, if the operand is the result of a [] operator, neither the & operator nor the unary * that is implied by the [] is evaluated and the result is as if the & operator
were removed and the [] operator were changed to a + operator. Otherwise, the result is
a pointer to the object or function designated by its operand.
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樓主: lemoncookie
發(fā)表于 2010-10-20 20:20
