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If only enter a number in a for loop he [I + 1] no error yet but will not don't understand why
The code below

# include & lt; Iostream>
# include & lt; Vector>
using namespace std;

Int main ()
{
Vector He;
Int ival.
Cout & lt; <"Please enter a number to other non-numeric characters ends:" & lt; While (cin & gt;> Ival)
He push_back (ival);
If (he) size ()==0)
{
Cout & lt; <"No one element" & lt; return -1;
}
Cout & lt; <"Two adjacent and, in turn, is:" & lt; //use the decltype inference type I
For (the decltype (he. The size ()) I=0; I & lt; He. The size () - 1; I +=2)
{
//for two adjacent and
Cout & lt; //every five lines of output five number
If ((I + 2) % 10==0)
Cout & lt; }
//if the element is an odd number, alone to deal with the last element
If (he. The size () % 2!=0)
Cout & lt; system("pause");
}

CodePudding user response:

 # include & lt; Iostream> 
# include & lt; Vector> 
using namespace std; 

Int main () {
Vector He; 
Int ival. 
Cout & lt; <"Please enter a number to other non-numeric characters ends:" & lt; While (cin & gt;> Ival) 
He push_back (ival); 
If (he) size ()==0) {
Cout & lt; <"No one element" & lt; return -1; 
} 
Cout & lt; <"Two adjacent and, in turn, is:" & lt; //use the decltype inference type I 
//when only enter a number, he. The size () is 1, the following "I & lt; He. The size () - 1 "do not meet, for not executed 
For (the decltype (he. The size ()) I=0; I & lt; He. The size () - 1; I +=2) {
//for two adjacent and 
Cout & lt; //every five lines of output five number 
If ((I + 2) % 10==0) 
Cout & lt; } 
//if the element is an odd number, alone to deal with the last element 
If (he. The size () % 2!=0) 
Cout & lt; system("pause"); 
return 0; 
}

CodePudding user response:

 for (the decltype (he. The size (), I=0; I & lt; He. The size () - 1; I +=2) 
.//when the input a number, he size ()=1, this is true for (I=0; i<0; +=2), I will not enter the circulation, the implementation of {... } at the back of the statement, 
{
Cout & lt; If ((I + 2) % 10==0)//every five lines of output five number 
Cout & lt; }