错误：跳转到案例标签 [英] Error: Jump to case label
问题描述
我写了一个程序，涉及使用switch语句...但是在编译它显示：
I wrote a program which involves use of switch statements... However on compilation it shows:
错误：跳转到case标签。
Error: Jump to case label.
为什么要这样做？
#include <iostream>
#include <cstdlib>
#include <fstream>
#include <string>
using namespace std;
class contact
{
public:
string name;
int phonenumber;
string address;
contact() {
name= "Noname";
phonenumber= 0;
address= "Noaddress";
}
};
int main() {
contact *d;
d = new contact[200];
string name,add;
int choice,modchoice,t;//Variable for switch statement
int phno,phno1;
int i=0;
int initsize=0, i1=0;//i is declared as a static int variable
bool flag=false,flag_no_blank=false;
//TAKE DATA FROM FILES.....
//We create 3 files names, phone numbers, Address and then abstract the data from these files first!
fstream f1;
fstream f2;
fstream f3;
string file_input_name;
string file_input_address;
int file_input_number;
f1.open("./names");
while(f1>>file_input_name){
d[i].name=file_input_name;
i++;
}
initsize=i;
f2.open("./numbers");
while(f2>>file_input_number){
d[i1].phonenumber=file_input_number;
i1++;
}
i1=0;
f3.open("./address");
while(f3>>file_input_address){
d[i1].address=file_input_address;
i1++;
}
cout<<"\tWelcome to the phone Directory\n";//Welcome Message
do{
//do-While Loop Starts
cout<<"Select :\n1.Add New Contact\n2.Update Existing Contact\n3.Display All Contacts\n4.Search for a Contact\n5.Delete a Contact\n6.Exit PhoneBook\n\n\n";//Display all options
cin>>choice;//Input Choice from user
switch(choice){//Switch Loop Starts
case 1:
i++;//increment i so that values are now taken from the program and stored as different variables
i1++;
do{
cout<<"\nEnter The Name\n";
cin>>name;
if(name==" "){cout<<"Blank Entries are not allowed";
flag_no_blank=true;
}
}while(flag_no_blank==true);
flag_no_blank=false;
d[i].name=name;
cout<<"\nEnter the Phone Number\n";
cin>>phno;
d[i1].phonenumber=phno;
cout<<"\nEnter the address\n";
cin>>add;
d[i1].address=add;
i1++;
i++;
break;//Exit Case 1 to the main menu
case 2:
cout<<"\nEnter the name\n";//Here it is assumed that no two contacts can have same contact number or address but may have the same name.
cin>>name;
int k=0,val;
cout<<"\n\nSearching.........\n\n";
for(int j=0;j<=i;j++){
if(d[j].name==name){
k++;
cout<<k<<".\t"<<d[j].name<<"\t"<<d[j].phonenumber<<"\t"<<d[j].address<<"\n\n";
val=j;
}
}
char ch;
cout<<"\nTotal of "<<k<<" Entries were found....Do you wish to edit?\n";
string staticname;
staticname=d[val].name;
cin>>ch;
if(ch=='y'|| ch=='Y'){
cout<<"Which entry do you wish to modify ?(enter the old telephone number)\n";
cin>>phno;
for(int j=0;j<=i;j++){
if(d[j].phonenumber==phno && staticname==d[j].name){
cout<<"Do you wish to change the name?\n";
cin>>ch;
if(ch=='y'||ch=='Y'){
cout<<"Enter new name\n";
cin>>name;
d[j].name=name;
}
cout<<"Do you wish to change the number?\n";
cin>>ch;
if(ch=='y'||ch=='Y'){
cout<<"Enter the new number\n";
cin>>phno1;
d[j].phonenumber=phno1;
}
cout<<"Do you wish to change the address?\n";
cin>>ch;
if(ch=='y'||ch=='Y'){
cout<<"Enter the new address\n";
cin>>add;
d[j].address=add;
}
}
}
}
break;
case 3 : {
cout<<"\n\tContents of PhoneBook:\n\n\tNames\tPhone-Numbers\tAddresses";
for(int t=0;t<=i;t++){
cout<<t+1<<".\t"<<d[t].name<<"\t"<<d[t].phonenumber<<"\t"<<d[t].address;
}
break;
}
}
}
while(flag==false);
return 0;
}
推荐答案
问题是，在 case 中仍然可见，除非明确的 {} 使用，因为它们不会被初始化，因为初始化代码属于另一个 case 。
The problem is that variables declared in one case are still visible in the subsequent cases unless an explicit { } block is used, but they will not be initialized because the initialization code belongs to another case.
在下面的代码中，如果 foo 等于1，一切正常，但如果它等于2，
In the following code, if foo equals 1, everything is ok, but if it equals 2, we'll accidentally use the i variable which does exist but probably contains garbage.
switch(foo) {
case 1:
int i = 42; // i exists all the way to the end of the switch
dostuff(i);
break;
case 2:
dostuff(i*2); // i is *also* in scope here, but is not initialized!
}
将案例包含在显式块中解决了问题：
Wrapping the case in an explicit block solves the problem:
switch(foo) {
case 1:
{
int i = 42; // i only exists within the { }
dostuff(i);
break;
}
case 2:
dostuff(123); // Now you cannot use i accidentally
}
编辑
为了进一步阐述， switch 语句只是一种特殊的 goto 。这里是一个类似的代码展示同样的问题，但使用 goto 而不是开关：
Edit
To further elaborate, switch statements are just a particularly fancy kind of a goto. Here's an analoguous piece of code exhibiting the same issue but using a goto instead of a switch:
int main() {
if(rand() % 2) // Toss a coin
goto end;
int i = 42;
end:
// We either skipped the declaration of i or not,
// but either way the variable i exists here, because
// variable scopes are resolved at compile time.
// Whether the *initialization* code was run, though,
// depends on whether rand returned 0 or 1.
std::cout << i;
}
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