Possible Duplicate:
Why can't variables be declared in a switch statement?
我在下面的代码中有一个奇怪的错误:
char choice=Getchar();
switch(choice)
{
case 's':
cout<<" display tree ";
thetree->displaytree();
break;
case 'i':
cout<<" enter value to insert "<<endl;
cin>>value;
thetree->insert(value);
break;
case 'f' :
cout<< "enter value to find ";
cin>>value;
int found=thetree->find(value);
if(found!=-1)
cout<<" found = "<<value<<endl;
else
cout<< " not found " <<value <<endl;
break;
default:
cout <<" invalid entry "<<endl;;
}
Visual Studio 2010 编译器说:
1>c:\users\daviti\documents\visual studio 2010\projects\2-3-4\2-3-4\2-3-4.cpp(317): error C2361: initialization of 'found' is skipped by 'default' label
1> c:\users\daviti\documents\visual studio 2010\projects\2-3-4\2-3-4\2-3-4.cpp(308) : see declaration of 'found'
我认为我已经正确编写了break和default语句,那么错误在哪里?
最佳答案
您需要将 case 'f':
用范围大括号括起来:
case 'f' :
{
cout<< "enter value to find ";
cin>>value;
int found=thetree->find(value);
if(found!=-1)
cout<<" found = "<<value<<endl;
else
cout<< " not found " <<value <<endl;
break;
}
或将 found
的声明放在 switch
关于c++ - 错误 C2361 : initialization of 'found' is skipped by 'default' label,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10381144/