成都传奇手游制作-成都传奇手游制作团队

成都作为中国“手游第一城”，拥有深厚的游戏研发基础和成熟的传奇手游制作生态。以下是成都地区专注于传奇类手游制作的代表性团队和企业概览，结合产业环境与典型案例分析：

一、成都传奇手游制作团队代表

1. 天象互动（赤月科技）

核心产品：《三国志威力加强版》《花千骨》手游（月流水超千万）。

团队背景：初始团队来自盛大、Gameloft等大厂，擅长IP改编与泛娱乐联动。2015年与赤月科技合并后，整合研发与发行能力，成为西南地区标杆企业。

技术特点：重度MMORPG研发经验丰富，注重长线运营与用户留存。

2. 萤火虫科技

代表作品：《旧约之战》（MMORPG）。

融资能力：2013年获CA创投、松禾资本千万级投资，专注“80后”情怀向题材。

定位：中小型团队，以创新型传奇玩法与美术风格见长。

3. 星柒玩科技

核心产品：《新射雕群侠传》（金庸正版授权）、《新仙剑奇侠传之挥剑问情》。

团队规模：500余人团队，研发占比80%，腾讯战略投资持股14.56%。

优势：IP改编能力突出，覆盖武侠、仙侠等泛传奇题材，兼顾放置卡牌与社交玩法。

4. 掌沃无限（网龙持股）

代表作品：《三国战神》《风暴魔域》。

研发特色：深耕ARPG与魔幻题材，擅长高打击感战斗系统设计。

5. 墨龙科技（三七互娱关联）

经典案例：页游时代《秦美人》（月流水5000万），转型手游后延续传奇类研发基因。

团队关联：三七互娱投资的成都首因科技、龙游天下等子公司，均涉及三国/传奇题材研发。

二、产业生态支持体系

成都手游产业以“研发驱动”为核心，形成以下优势生态：

1. 政策扶持

# 8.1

Compound Expressions

Now that we've covered the fundamental types, operators, and expressions, it's time to talk about compound expressions.

A compound expression is an expression that involves more than one operator. You've already seen these. For example:

c++

int x { 2 + 3 }; // 2 + 3 is a compound expression

But what happens when we combine multiple operators in a single expression? For example:

c++

int x { 2 + 3 * 4 };

In mathematics, operators follow a precedence order that tells us how to resolve expressions that contain multiple operators. The precedence rules for mathematics state that multiplication and division have higher precedence than addition and subtraction, so we do those operations first. Thus the above expression is resolved as follows:

`2 + (3 * 4) = 2 + 12 = 14`

In C++, operator precedence rules are built into the language, and they are the same as the precedence rules in mathematics. The operators in the above expression are resolved in the same way as in mathematics.

Operator Precedence

Operators in C++ are evaluated in order of precedence. A table of precedence can be found here:

Note: The precedence table contains many operators that you have not learned about yet, so don't worry if most of it doesn't make sense. Refer back to this table as you learn about more operators.

Operators with higher precedence are evaluated before operators with lower precedence.

Associativity

If two operators with the same precedence level are adjacent to each other in an expression, the operator's associativity tells the compiler whether to evaluate the operators from left to right or from right to left. The associativity for each operator is also listed in the precedence table.

For example, operator `*` and operator `/` have the same precedence, and they associate left-to-right. So:

c++

int x { 4 / 2 * 3 };

Resolves as:

`(4 / 2) * 3 = 2 * 3 = 6`

If it were evaluated right-to-left, we'd end up with `4 / (2 * 3) = 4 / 6 = 0`, which is probably not what we intended.

Parenthesization

Parenthesis can be used to force an expression to evaluate a subexpression first. Parenthesized subexpressions are evaluated before non-parenthesized expressions with higher precedence than the other operators.

For example, in the expression `2 + 3 * 4`, multiplication has higher precedence than addition, so it evaluates as `2 + (3 * 4)`. But what if we wanted the addition to happen first? We would use parenthesis:

c++

int x { (2 + 3) * 4 }; // evaluates to 20

Expressions with side effects

It is possible for expressions to have side effects. These are things that happen when an expression is evaluated that are not part of the result.

c++

int x { 5 };

int y { --x }; // this expression has the side effect of decrementing x

Quiz time

Question #1

c++

int x { 3 }; // 3

int y { 4 };

std::cout << (x + y + 1); // 8

Question #2

c++

int x { 3 };

int y { 4 };

std::cout << (x = y); // 4

Question #3

c++

int x { 3 };

int y { 4 };

std::cout << (x += y); // 7

Question #4

c++

int x { 3 };

int y { 4 };

std::cout << (x /= y); // 0

Question #5

c++

int x { 3 };

int y { 4 };

std::cout << (x == y); // 0 (false)

Question #6

c++

int x { 3 };

int y { 4 };

std::cout << (x != y); // 1 (true)

Question #7

c++

int x { 3 };

int y { 4 };

std::cout << (x <= y); // 1 (true)

Question #8

c++

int x { 3 };

int y { 4 };

std::cout << (x > y); // 0 (false)

Question #9

c++

int x { 1 };

int y { 2 };

std::cout << (x = (y = 3)); // 3

Question #10

c++

int x { 1 };

int y { 2 };

std::cout << (x = y = 3); // 3

Question #11

c++

int x { 1 };

int y { 2 };

std::cout << (x /= y + 1); // 1 / 3 = 0

Question #12

c++

int x { 3 };

int y { 4 };

std::cout << (x += y -= 2); // 5