[Clean Code] 04장. 네이밍 리팩토링 실습

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// 실습 대상: 압축된 변수명의 주문 처리 코드
public class OrderProcessor {
    private int c = 0;
    private double t = 0.0;

    public void p(Map<String, Object> o) {
        String s = (String) o.get("status");
        if (s.equals("pending")) {
            double a = (Double) o.get("amount");
            int q = (Integer) o.get("quantity");
            String pid = (String) o.get("productId");

            if (check(pid, q)) {
                o.put("status", "confirmed");
                t += a;
                c++;
            }
        }
    }

    public boolean check(String id, int q) {
        return getStock(id) >= q;
    }

    private int getStock(String productId) { return 100; }
}

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// 리팩토링 결과: 의도가 드러나는 이름
public class OrderProcessor {
    private int confirmedOrderCount = 0;
    private double totalRevenue = 0.0;

    public void processOrder(Map<String, Object> order) {
        String status = (String) order.get("status");
        if (status.equals("pending")) {
            double orderAmount = (Double) order.get("amount");
            int requestedQuantity = (Integer) order.get("quantity");
            String productId = (String) order.get("productId");

            if (isStockSufficient(productId, requestedQuantity)) {
                order.put("status", "confirmed");
                totalRevenue += orderAmount;
                confirmedOrderCount++;
            }
        }
    }

    public boolean isStockSufficient(String productId, int requestedQuantity) {
        return getAvailableStock(productId) >= requestedQuantity;
    }

    private int getAvailableStock(String productId) { return 100; }
}