C++ Program Not Outputting as expected (using .at, if/else, getline) - c++

Question

For some reason my prototype program isn't outputting as expected.
My text file: (separated by tabs)
NameOne NameTwo NameThree 56789
My source code:
#include <iostream>
#include <string>
#include <fstream>
using namespace std;
int main()
{
string name1, name2, name3, name4, name5,fullName;
ifstream inFile;
string attendance;
int index;
inFile.open("test2.dat");
getline(inFile, name1, '\t');
getline(inFile, name2, '\t');
getline(inFile, name3, '\t');
if (name3.at(0) == 0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
{
attendance = name3;
fullName = name1 + ' ' + name2;
}
else
{
getline(inFile, name4, '\t');
if (name4.at(0) == 0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
{
attendance = name4;
fullName = name1 + ' ' + name2 + ' ' + name3;
}
else
{
getline(inFile, name5, '\t');
if (name5.at(0) == 0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
{
attendance = name5;
fullName = name1 + ' ' + name2 + ' ' + name3 + ' ' + name4;
}
else
{
fullName = name1 + ' ' + name2 + ' ' + name3 + ' ' + name4 + ' ' + name5;
inFile >> attendance;
}
}
}
cout << endl << fullName << endl << attendance << endl << endl;
system("pause");
return 0;
}
Expected output :
NameOne NameTwo NameThree
56789
Actual output:
NameOne NameTwo
NameThree
For some reason it's storing the string NameThree into attendance and outputting that.
I was expecting to store NameFour into attendance.

Answer 1

Change this:
if (name.at(0) == 0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
To this:
if (isdigit(name.at(0)))
You have two different errors in your code:
The if statement is not properly structured (I'm surprised it even compiles)
In order to check if a character is the 0 digit, for example, you should compare it with '0' (not 0)

Answer 2

You need to fix your if statements:
if (isdigit(name.at(0)))
for the if statement as already stated in another answer.
Now for your stated questiom:
"For some reason it's storing the string NameThree into attendance and outputting that. I was expecting to store NameFour into attendance."
This is because the expression in your first if passes.
name3.at(0) == 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
unfortunately evaluates to true.
The rest of your code (getting name4 etc) is in the else block of that if.
Therefore, everything after the if is executed (and attendance = name3; is executed) and the rest of your code is not run until the end of the else block (and the following cout).
Remember, an if works like this:
if (expr)
{
//run if expr == true
}
else
{
//run if expr == false
}
and don't forget that 0 is false, and everything else is mapped to true when cast to boolean.

Answer 3

In case anyone is wondering how
if (name4.at(0) == 0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
even compiles, it's because the C (and by extension C++) language defines , as a binary operator, like / or ->. It has the lowest precedence, left-associativity, and its result is the value on the right-hand side. So the above expression is evaluated as:
(name4.at(0) == 0, 1) => 1
(1, 2) => 2
...
(8, 9) => 9
if(9) => true
(Adding to this, C++ even allows you to override operator,. It's strongly discouraged.)
Before you get too livid at this, bear in mind that this was how inline functions were implemented in the early, early days of C++, before there were native compilers and you used a tool called cfront to convert your C++ to C to then feed a C compiler. For example, if you wrote:
inline int List::read() const
{
int x = *head++;
if(head == end)
head = begin;
return x;
}
Then, when you actually called the function, cfront would insert something like
(__inl_x = *obj->head, ++obj->head, obj->head = (obj->head == obj->end?
obj->begin: obj->head), __inl_x)
into the calling expression. Note that the last thing in the ,-separated list is the return value from the "inline" function. Needless to say, there were a lot of language constructs that weren't allowed in inline functions (much like there are many that aren't allowed in constexpr functions these days.)