问题
i have something like
while(playAgain==true)
{
cout<<"new game"<<endl; //i know 'using namespace std;' is looked down upon
while(playerCard!=21)
{
*statements*
if(decision=='n')
{
break
}
...
}
}
but that break only breaks out of the first while loop when I want to break out of both of the loops
回答1:
Don't cook spaghetti and extract your loops into the function:
void foo(...) {
while (...) {
/* some code... */
while (...) {
if ( /* this loop should stop */ )
break;
if ( /* both loops should stop */ )
return;
}
/* more code... */
}
}
this decomposition will also yield cleaner code since instead of hundreds of lines of ugly procedural code, you will have neat functions at different levels of abstraction.
回答2:
Use goto:
while(playAgain==true)
{
cout<<"new game"<<endl; //i know 'using namespace std;' is looked down upon
while(playerCard!=21)
{
*statements*
if(decision=='n')
{
goto label;
}
...
}
}
label:
...
回答3:
There are basically two options to go.
Add the condition check in outer loop.
while ((playAgain==true) && (decision != '\n'))
Simply use
goto
. People are often told never to usegoto
as if it's monster. But I'm not opposed to use it to exit multiple loops. It's clean and clear in this situation.
回答4:
while(playAgain==true && decision !='n' ){
^^ add a condition
cout<<"new game"<<endl;
while(playerCard!=21){
*statements*
if(decision=='n'){
break
}
...
}
}
回答5:
This solution is specific to your case. When the user's decision is 'n', he doesn't want to play again. So just set playAgain
to false
and then break. Outer loop will be broken automatically.
while(playAgain==true){
cout<<"new game"<<endl; //i know 'using namespace std;' is looked down upon
while(playerCard!=21){
*statements*
if(decision=='n'){
playAgain = false; // Set this to false, your outer loop will break automatically
break;
}
}
}
回答6:
If you don't have to avoid goto
statement, you can write
while (a) {
while (b) {
if (c) {
goto LABEL;
}
}
}
LABEL:
来源:https://stackoverflow.com/questions/18670038/how-to-exit-nested-loops