我如何处理迭代的最后一个元素

Question

for(int t(0); t < 10;++t) { cout<<t<<endl;}

I'm just biginer in C++, and want to know how can I take the last elemnt of my "cout...."; in this case my laste element is 9

thx for help ;)

Answer 1

You can extract int t from the for loop :

int t;  
for (t = 0; t < 10; ++t)  
{
    cout << t << endl;
}

Answer 2

int c = 0;
for(int t = 0; t<10; t++)
{
  c = t;
}
cout<<c;

This might be what you are looking for I am not sure I understand your question properly though.The variable c should hold the last element of t when the loop ends.

Answer 3

int t = 9;
cout << t << endl;

Now you have the last element, #9.

Answer 4

ghagha，C ++中的范围从0到n-1运行时，在例如，你必须<10因此0到9的范围为0，因此你的最后一个元素是9.但正如我说可以做任何范围正最后一个元素为-1 ，前提是它遵循常规约定（如果以这种方式编码，则范围可能在1到n之间 ）

Answer 5

It is not clear what you want but in any case your loop contains a bug. Instead of

for(int t(0); t < 10;  t) { cout<<t<<endl;}

should be

for(int t(0); t < 10;  t++) { cout<<t<<endl;} 

that is variable t has to be incremented.

Answer 6

One simple way -

int t = 0;
for (; t < 10; ++t)
   cout << t << ;

Tough the correct way to do it will be (one variable should not have two meanings, ie 1. last element, 2. iterator context) -

int last_element;
for (int t = 0; t < 10; ++t;
{
    cout << t << ;
    last_element = t;
}