一道枚举相关的华为面试题

[zw1270644192]

enum ENUM_A
{
x1,
y1,
z1 = 5,
a1,
b1
};

enum ENUM_A enumA = y1;
enum ENUM_A enumB = b1;
请问enumA和enumB的值是多少？

不太懂枚举，给的答案是0 3
不知对不对 为什么是这个答案啊 知道的给讲讲吧 谢谢

[turing-complete]

毫无疑问 1，7

[Greg_han]

应该是1,7

[jernymy]

[Quote=引用 12 楼 zmlovelx 的回复:]
1,7 是对的.
楼主不是被忽悠就是我们被楼主忽悠.
[/Quote]
支持1，7

[liutengfeigo]

[Quote=引用 13 楼 mymtom 的回复:]

3 The identifiers in an enumerator list are declared as constants that have type int and
may appear wherever such are permitted.107) An enumerator with = defines its
enumeration constant as the val……
[/Quote]
这个++

[一叶之舟]

x1 == 0,
y1 == 1,
z1 == 5,
a1 == 6,
b1 == 7

[机智的呆呆]

我怎么感觉应该是1 7哪？

[小坏猪猪]

不对啊，应该是1,7啊

[war10811]

第一个数默认为0，后面的都是它前面那个值+1

0 7

[yxriyin]

华为真心招的都是民工啊
擦

[war10811]

[Quote=引用 3 楼 war10811 的回复:]
第一个数默认为0，后面的都是它前面那个值+1

0 7
[/Quote]
打错是1 7

[gaomingok]

1和7
默认从0开始，下一个值是上一个值+1

[偶是江湖中人]

enum ENUM_A
{
  x1,
  y1,
  z1 = 5,
  a1,
  b1
};
enum ENUM_A
{
  x1 = 0,
  y1 = 1,
  z1 = 5,
  a1 = 6,
  b1 = 7
};

[defy_]

应该是1和7吧。

枚举无初始化，但从第一个标示符开始从0，依次加1赋值，遇到Z1=5，a1同理还是继续加1，就是6，所以b1是7

[zw1270644192]

[Quote=引用 7 楼 weicaijiang 的回复:]
C/C++ code

enum ENUM_A
{
x1,
y1,
z1 = 5,
a1,
b1
};
enum ENUM_A
{
x1 = 0,
y1 = 1,
z1 = 5,
a1 = 6,
b1 = 7
};
[/Quote]

刚刚在VS里运行了一下 答案就是1 7
是下载的文档里的错了
尽信书不如无书呀
谢谢各位了 学习了 已经明白了

[qsfool1987]

1和7

[wori011]

1，7

[帅得不敢出门]

1,7 是对的.
楼主不是被忽悠就是我们被楼主忽悠.

[mymtom]

3 The identifiers in an enumerator list are declared as constants that have type int and
may appear wherever such are permitted.107) An enumerator with = defines its
enumeration constant as the value of the constant expression. If the first enumerator has
no =, the value of its enumeration constant is 0. Each subsequent enumerator with no =
defines its enumeration constant as the value of the constant expression obtained by
adding 1 to the value of the previous enumeration constant. (The use of enumerators with
= may produce enumeration constants with values that duplicate other values in the same
enumeration.) The enumerators of an enumeration are also known as its members.