网友您好， 请在下方输入框内输入要搜索的题目：
题目内容
（请给出正确答案）
2-5 在下面的枚举类型中，Blue的值是多少？ enum COLOR { WHITE， BLACK = 100， RED， BLUE， GREEN = 300 };
参考答案和解析
102
更多 “2-5 在下面的枚举类型中，Blue的值是多少？ enum COLOR { WHITE， BLACK = 100， RED， BLUE， GREEN = 300 };” 相关考题
考题
设有说明var color:(red,green,yellow,blue);a:boolean;下面语句正确的是( )。
Aolor:=‘green‘;Bwriteln(green);Cwriteln(color);Da:=color=red;
考题
在下列程序段中,enumcolor{red,yellow,blue=4,green,white}c1,c2;c1=yellow;c2=white;枚举变量c2的值是()
A、1B、3C、5D、6
考题
（ 18 ）下列枚举类型的定义中，包含枚举值 3 的是A ） enum test {RED, YELLOW, BLUE, BLACK};B ） enum test {RED, YELLOW=4, BLUE, BLACK};C ） enum test {RED=-1, YELLOW,BLUE, BLACK};D ） enum test {RED, YELLOW=6, BLUE, BLACK};
考题
已知枚举类型声明语句为： enumCOLOR{WHITE，YELLOW，GREEN=5，RED，BLACK=10}； 则下列说法中错误的是A．枚举常量YELLOW的值为1B．枚举常量RED的值为6C．枚举常量BLACK的值为10D．枚举常量WHITE的值为1
考题
以下选项中不能正确把c1定义成结构体变量的是A．typedef struct {int red: int green: int blue; } COLOR; COLOR c1;B．struct color c1 {int red int green: int blue; };C．struct color {int red , int green : int blue : )cl;D．struct {int red; int green; int blue } c1 ;
考题
publicclassBall{publicenumColor{RED,GREEN,BLUE};publicvoidfoo(){／／insertcodehere{System.out.println(c);}}}Whichcodeinsertedatline14causesthefoomethodtoprintRED,GREEN,andBLUE?()
A.for（Colorc：Color.values（））B.for（Colorc=RED；c=BLUE；c++）C.for（Colorc；c.hasNext（）；c.next（））D.for（Colorc=Color[0]；c=Color[2]；c++）E.for（Colorc=Color.RED；c=Color.BLUE；c++）
考题
下列枚举类型的定义中，包含枚举值3的是（ ）。A.enumtest{RED，YELLOW，BLUE，BLACK}；B.enumtest{RED，YELLOW=4，BLUE，BLACK)；C.enumtest{RED=-1，YELLOW，BLUE，BLACK}；D.enumtest{RED，YELLOW=6，BLUE，BLACK)；
考题
以下选项中不能正确把cl定义成结构体变量的是A．typedef struct { int red; int green; int blue; } COLOR; COLOR cl;B．struct color cl { int red; int green; int blue; }C．struct color { int red; int green; int blue; } cl;D．struct { int red; int green; int blue; } cl;
考题
下列枚举类型的定义中，包含枚举值3的是A．enum test{RED，YELLOW，BLUE，BLACK}；B．enum test{RED，YELLOW=4，BLUE，BLACK}；C．enum test{RED=-1，YELLOW，BLUE，BLACK}；D．erium test{RED，YELLOW=6，BLUE，BLACK}；
考题
以下选项中不能正确把c1定义成结构体变量的是A．typedef struct { int red； int green；； int blue； }COLOR； COLOR cl；B．struct color cl { int red； int green； int blue； };C．struet color { int red； int green； int blue； }c1；D．struct { int red； int green； int blue； }cl；
考题
在下列枚举类型的定义中，包含枚举值3的是A．entire test(RED，YELLOW，BLUE，BLACK)；B．enum test(RED，YELLOW=4，BLUE，BLACK)；C．enum test(RED=-1，YELLOW，BLUE，BLACK)；D．enum test(RED，YELLOW=6，BLUE，BLACK)；
考题
已知枚举类型声明语句为： enum COLOR{WHITE,YELLOW,GREEN＝5,RED,BLACK＝10}； 则下列说法中错误的是A．枚举常量YELLOW的值为1B．枚举常量RED的值为6C．枚举常量BLACK的值为10D．枚举常量WHITE的值为1
考题
对于这样的一个枚举类型：enum Color:byte{Red,Green,Blue,Orange}string[] ss=Enum.GetNames(typeof(Color));byte[] bb=Enum.GetValues(typeof(Color));试写一段程序显示出枚举类型中定义的所有符号名称以及它们对应的数值。
考题
已知枚举类型声明语句为： enum COl,OR{WHITE，YELLOW，GREEN=5，RED，BLACK=10}； 则下列说法中错误的是( )。A.枚举常量YELLOW的值为1B.枚举常量RED的值为6C.枚举常量BLACK的值为10D.枚举常量WHITE的值为1
考题
Which of the following is the correct pin out for T568B?()
A. Orange/white, orange, green/white, blue, blue/white, green, brown/white, brownB. Blue, blue/white, orange/white, orange, brown/white, brown, green/white, greenC. Green/white, green, orange/white, blue, blue/white, orange, brown/white, brownD. Orange/white, orange, brown/white, brown, green/white, green, blue, blue/white
考题
下列选项中不能正确定义结构体的是_______。A．typedef structB．struct color cl {int red； {int red； int green； int green； int blue； int blue； }COLOR； }； COLOR cl；C．struct colorD．struct {int red； {int red； int green; int green； int blue； int blue； }cl； }cl；
考题
以下选项中能正确把c1定义成结构体变量的是( )。A．typedef struct { int red； int red； int green； int blue； }COLOR； COLOR c1；B．struct color c1 { int red int red； int green int blue； }；C．stmctcolor { int red， int green； int blue； }c1；D．struct { int red； int green； int blue； }c1；
考题
在下列枚举类型的定义中，包含枚举值3的是A．entire test(RED，YELLOW，BLUE，BLACK)；B．entire test(RED，YELLOW，BLUE，BLACK)；C．enum test(RED=-1，YELLOW，BLUE，BLACK)；D．enum test(RED，YELLOW=6，BLUE，BLACK)；
考题
在下列程序段中，枚举变量 c1的值是（）。enum color { red,yellow,blue=4,green,white}c1; c1=yellow; c1=white;A 1B 3C 5D 6
考题
共用题干
第三篇Ways to Create Colors in a PhotographThere are two ways to create colors in a photograph.One method called additive,starts with three basic colors and adds them together to produce some other color. The second method,called subtractive,starts with white light(a mixture of all colors in the spectrum)and,by taking away some or all other colors,leaves the one desired.In the additive method,separate colored lights combine to produce various other colors.The three additive primary colors are green,red and blue(each providing about one-third of the wavelengths in the total spectrum).Mixed in varying proportions,they can produce all colors.Green and red light mix to produce yellow,red and blue light mix to produce magenta(a purplish pink);green and blue mix to produce cyan(a bluish green).When equal parts of all three of these primary-colored beams of light overlap（重叠）,the mixture appears white to the eye.In the subtractive process colors are produced when dye（染料）absorbs some wavelengths and so passes on only part of the spectrum.The subtractive primaries are cyan,magenta and yellow; these primaries or dyes absorb red,green and blue wavelengths respectively,thus subtracting them from white light. These dye colors are the complementary colors to the three additive primaries of red,green and blue.Properly combined,the subtractive primaries can absorb all colors of light, producing black.But,mixed in varying proportions,they too can produce any color in the spectrum.Whether a particular color is obtained by adding colored lights together or by subtracting some light from the total spectrum,the result looks the same to the eye.The additive process was employed for early color photography. But the subtractive method,while requiring complex chemical techniques,has turned out to be more practical and is the basis of all modern color films.Which of the following is NOT a pair of additive and subtractive primary colors?A:Yellow and blue. B: Magenta and green.C:Black and white. D:Cyan and red.
考题
共用题干
第三篇Ways to Create Colors in a PhotographThere are two ways to create colors in a photograph.One method called additive,starts with three basic colors and adds them together to produce some other color. The second method,called subtractive,starts with white light(a mixture of all colors in the spectrum)and,by taking away some or all other colors,leaves the one desired.In the additive method,separate colored lights combine to produce various other colors.The three additive primary colors are green,red and blue(each providing about one-third of the wavelengths in the total spectrum).Mixed in varying proportions,they can produce all colors.Green and red light mix to produce yellow,red and blue light mix to produce magenta(a purplish pink);green and blue mix to produce cyan(a bluish green).When equal parts of all three of these primary-colored beams of light overlap（重叠）,the mixture appears white to the eye.In the subtractive process colors are produced when dye（染料）absorbs some wavelengths and so passes on only part of the spectrum.The subtractive primaries are cyan,magenta and yellow; these primaries or dyes absorb red,green and blue wavelengths respectively,thus subtracting them from white light. These dye colors are the complementary colors to the three additive primaries of red,green and blue.Properly combined,the subtractive primaries can absorb all colors of light, producing black.But,mixed in varying proportions,they too can produce any color in the spectrum.Whether a particular color is obtained by adding colored lights together or by subtracting some light from the total spectrum,the result looks the same to the eye.The additive process was employed for early color photography. But the subtractive method,while requiring complex chemical techniques,has turned out to be more practical and is the basis of all modern color films.It can be inferred from the passage that white can be produced by______.A:mixing different proportions of magenta,green and redB:mixing equal proportions of green,red and blueC:mixing different proportions of cyan, yellow and blueD:mixing equal proportions of black,red and green
考题
Which of the following is the correct pin out for T568B?（）A、Orange/white, orange, green/white, blue, blue/white, green, brown/white, brownB、Blue, blue/white, orange/white, orange, brown/white, brown, green/white, greenC、Green/white, green, orange/white, blue, blue/white, orange, brown/white, brownD、Orange/white, orange, brown/white, brown, green/white, green, blue, blue/white
考题
enum color{red，green，yellow=5，white，black}；定义了一枚举类型。编译程序为值表中各标识符分配的枚举值依次为（）。A、1、2、3、4、5B、0、1、5、2、3C、0、1、5、6、7D、3、4、5、6、7
考题
单选题Which of the following is the correct pin out for T568B?（）A
Orange/white, orange, green/white, blue, blue/white, green, brown/white, brownB
Blue, blue/white, orange/white, orange, brown/white, brown, green/white, greenC
Green/white, green, orange/white, blue, blue/white, orange, brown/white, brownD
Orange/white, orange, brown/white, brown, green/white, green, blue, blue/white
考题
单选题public class Ball { public enum Color { RED， GREEN， BLUE }； public void foo（） { // insert code here { System.out.println（c）； } } } Which code inserted at line 14 causes the foo method to print RED， GREEN， and BLUE？（）A
for（ Color c ： Color.values（））B
for（ Color c = RED； c = BLUE； c++）C
for（ Color c； c.hasNext（） ； c.next（））D
for（ Color c = Color[0]； c = Color[2]； c++）E
for（ Color c = Color.RED； c = Color.BLUE； c++）
考题
单选题enum color{red，green，yellow=5，white，black}；定义了一枚举类型。编译程序为值表中各标识符分配的枚举值依次为（）。A
1、2、3、4、5B
0、1、5、2、3C
0、1、5、6、7D
3、4、5、6、7
热门标签
最新试卷
