我是Golang的新手,我发现switch case
声明不需要break
声明来停止评估案例.
那么,我想知道如何在go中实现这种穿透行为?
我有一个switch语句,它接受字母等级并返回相应的GPA; 但是,它会为字母(A,B,C,D和F)抛出一个无法找到符号的错误!我已经检查了javaDocs的指导,但找不到问题.导致此错误的原因是什么?
switch (grade) {
case A: nv[i] = 4; //nv = numerical value
break;
case B: nv[i] = 3;
break;
case C: nv[i] = 2;
break;
case D: nv[i] = 1;
break;
case F: nv[i] = 0;
break;
}
Run Code Online (Sandbox Code Playgroud) b)将以下switch语句转换为if ... else语句.
#include <stdio.h>
void main()
{
char option;
int a;
printf("a.Addition\n b.Subtraction\n c.Multiplication\n d.Division\n");
printf("Choose your option : ");
scanf("%c",&option);
switch(option)
{
case 'a':
case 'A': a=20+10;
printf("Addition process result:%d",a);
break;
case 'b':
case 'B': a=20-10;
printf("Subtraction process result:%d",a);
break;
case 'c':
case 'C': a=20*10;
printf("Multiplication process result:%d",a);
break;
case 'd':
case 'D': a=20/10;
printf("Division process result:%d",a);
break;
default: printf("Invalid option");
}
printf("\nEnd of program");
}
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如何将它从SWITCH转换为IF ... ELSE?这是我考试的修订问题谢谢
是否可以在开关盒环路中放置一个功能?因为我试过这个只是为了探索更多这个循环.虽然我尝试了其他方法,但我仍然存在问题.谁能帮我?
#include <stdio.h>
int main(void)
{
int choice;
switch(choice);
{
case 1:
{
int GetData()
{
int num;
printf("Enter the amount of change: ");
scanf("%d%*c", &num);
return (num);
}
int getChange (int change,int fifty,int twenty,int ten,int five)
{
int num = change;
fifty = num/50;
num %= 50;
twenty = num/20;
num %= 20;
ten = num/10;
num %= 10;
five = num/5;
num %= 5;
return (fifty, twenty, ten, five);
}
int main()
{
int change, fifty, twenty, ten, …
Run Code Online (Sandbox Code Playgroud) 我写了一个实现简单计算器的程序.但是,它没有编译.编译器说有22个错误,我不知道为什么.
期望的行为:
具体问题或错误:
在任何发生编译错误cin
,cout
,endl
,case
和break
#include <iostream>
int main()
{
float area, r, l, h, b;
int choice;
cout<<"\n area of?";
cout<<"\n[1]square \n[2]rectangle \n[3]circle \n[4]triangle"<<endl;
cin>>choice;
switch(choice);
{
case 1:
cout<<"enter length"<<endl;
cin>>l;
area=l*l;
cout<<area<<endl;
break;
case 2:
cout<<"enter height"<<endl;
cin>>h;
cout<<"enter length"<<endl;
cin>>l;
area=l*h;
cout<<area<<endl;
break;
case 3:
cout<<"enter radius"<<endl;
cin>>r;
area=r*r*3.14;
cout<<area<<endl;
break;
case 4:
cout<<"enter height"<<endl;
cin>>h;
cout<<"enter breadth"<<endl;
cin>>b;
area=h*b*0.5;
cout<<area<<endl;
break;
}
return …
Run Code Online (Sandbox Code Playgroud) 我试图看看我是否可以制作一个fizzbuzz c ++ switch语句.我收到一个错误,说我在const表达式中不可用.这是否意味着我无法使这件事发挥作用?还是有工作吗?这是我的代码.
#include <iostream>
using namespace std;
int main() {
for(int = 1; 1 <= 100; i++){
switch(true){
case(i % 3 == 0 & i % 5 == 0):
cout << "fizzbuzz" << endl;
break;
case(i % 3 == 0):
cout << "fizz" << endl;
break;
case(i % 5 == 0):
cout << "fizz" << endl;
break;
default:
cout << i << endl;
}
}
}
Run Code Online (Sandbox Code Playgroud) 我正在尝试使用以下方法检查值switch
:
private void btnInput1Rste_Click(object sender, EventArgs e)
{
switch (sender == btnInput1Rste)
{
case "1": currentButtonPressedRste = 1;
break;
}
}
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它给出的错误如下:
can't convert type 'string' to 'bool'
但是,当我尝试将其转换为布尔值时,它说:
a constant value is expected
.
我如何解决它?
当它工作时,它应该检查3个值.(不是这个switch
)
我试图使用函数中的switch语句计算数组中的元素数(仅限数字).老实说,我不确切知道这段代码会是什么样子.但这是我到目前为止所拥有的
<script language="JavaScript">
//an array of numbers
var number = [1,"o",2,3,"a",0];
//a switch statement
switch (number) {
//Not sure what would go here....
break;
}
//display result of count
alert(count)
</SCRIPT>
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我试图在接口中实现常量,但为什么在 switch 情况下访问时会出错?
如果我只使用string
ininterface
而不是常量,APP_STATUS
那么它就可以正常工作。
例子:
// Gives an error
interface InconsistenciesData {
type: typeof APP_STATUS.INCONSISTENCIES;
data: Inconsistency[];
}
// Works fine
interface InconsistenciesData {
type: 'INCONSISTENCIES';
data: Inconsistency[];
}
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下面是我的代码片段。
export const APP_STATUS = {
CONFIRMED: 'CONFIRMED',
INCONSISTENCIES: 'INCONSISTENCIES',
SUCCESS: 'SUCCESS',
ERROR: 'ERROR',
LOADING: 'LOADING',
OK: 'OK'
}
interface InconsistenciesLoading {
type: typeof APP_STATUS.LOADING;
}
interface InconsistenciesError {
type: typeof APP_STATUS.ERROR;
}
interface InconsistenciesSuccess {
type: typeof APP_STATUS.SUCCESS;
}
interface InconsistenciesData …
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