是否可以使用break函数退出几个嵌套的for循环?
如果是,你会怎么做呢?你还能控制刹车出多少圈吗?
是否可以使用break函数退出几个嵌套的for循环?
如果是,你会怎么做呢?你还能控制刹车出多少圈吗?
当前回答
虽然这个答案已经提出了,但我认为一个很好的方法是这样做:
for(unsigned int z = 0; z < z_max; z++)
{
bool gotoMainLoop = false;
for(unsigned int y = 0; y < y_max && !gotoMainLoop; y++)
{
for(unsigned int x = 0; x < x_max && !gotoMainLoop; x++)
{
//do your stuff
if(condition)
gotoMainLoop = true;
}
}
}
其他回答
c++不支持命名循环,就像Java和其他语言一样。您可以使用goto,或创建您使用的标志值。在每个循环结束时检查标志值。如果它被设置为true,那么您可以跳出该迭代。
打破嵌套循环的另一种方法是将两个循环分解成一个单独的函数,并在希望退出时从该函数返回。
当然,这也引发了另一个争论,即是否应该显式地从函数的任何地方返回,而不是在函数的末尾。
用一个bool变量打破任意数量的循环,如下所示:
bool check = true;
for (unsigned int i = 0; i < 50; i++)
{
for (unsigned int j = 0; j < 50; j++)
{
for (unsigned int k = 0; k < 50; k++)
{
//Some statement
if (condition)
{
check = false;
break;
}
}
if (!check)
{
break;
}
}
if (!check)
{
break;
}
}
在这段代码中,我们中断了;所有的循环。
你可以使用“goto”来留下嵌套循环 下面是我的原始代码,包括“goto”
int main()
{
string str;
while (cin >> str)
{
if (str == "0")
break;
int sum = 0;
for (auto ch : str)
{
if (ch <= 'z' && ch >= 'a')
sum += (ch - 'a' + 1);
else if (ch >= 'A' && ch <= 'Z')
sum += (ch - 'A' + 1);
else
{
cout << "Fail" << endl;
goto fail;
}
}
cout << sum << endl;
fail:
}
return 0;
}
然而,我可以通过添加函数“计算”来避免“goto”
void calculate(const string &str)
{
int sum = 0;
for (auto ch : str)
{
if (ch <= 'z' && ch >= 'a')
sum += (ch - 'a' + 1);
else if (ch >= 'A' && ch <= 'Z')
sum += (ch - 'A' + 1);
else
{
cout << "Fail" << endl;
return;
}
}
cout << sum << endl;
}
int main()
{
string str;
while (cin >> str)
{
if (str == "0")
break;
calculate(str);
}
return 0;
}
Breaking out of a for-loop is a little strange to me, since the semantics of a for-loop typically indicate that it will execute a specified number of times. However, it's not bad in all cases; if you're searching for something in a collection and want to break after you find it, it's useful. Breaking out of nested loops, however, isn't possible in C++; it is in other languages through the use of a labeled break. You can use a label and a goto, but that might give you heartburn at night..? Seems like the best option though.