Super basic code: Why is my loop not breaking? - c++

Question

for(int i=0;i<50;i++,size++)
{
cin >> inputnum[i];
cout << size;
if(inputnum[i] == '.')
{
break;
}
}
The break breaks the input stream but the size keeps outputting.
The output of size is 012345678910111213...474849.
I tried putting size++ inside the loop but it made no difference. And size afterwards will be equal to 50, which means it went through the full loop.
I forgot to explain that I added the cout << size within the loop to debug/check why it outputted to 50 after the loop even if I only inputted 3 numbers.

Answer 1

I suspect that inputnum is an array of int (or some other numeric type). When you try to input '.', nothing actually goes into inputnum[i] - the cin >> inputnum[i] expression actually fails and puts cin into a failed state.
So, inputnum[i] is not changed when inputting a '.' character, and the break never gets executed.
Here's an slightly modified version of your code in a small, complete program that demonstrates using !cin.good() to break out of the input loop:
#include <iostream>
#include <ostream>
using namespace std;
int main()
{
int inputnum[50];
int size = 0;
for(int i=0;i<50;i++,size++)
{
cin >> inputnum[i];
if (!cin.good()) {
break;
}
}
cout << "size is " << size << endl;
cout << "And the results are:" << endl;
for (int i = 0; i < size; ++i) {
cout << "inputnum[" << i << "] == " << inputnum[i] << endl;
}
return 0;
}
This program will collect input into the inputnum[] array until it hits EOF or an invalid input.

Answer 2

What is inputnum ? Make sure t's a char[]!! with clang++ this compiles and works perfectly:
#include <iostream>
int main() {
int size = 0;
char inputnum[60];
for(int i=0;i<50;i++,size++) {
std::cin >> inputnum[i];
std::cout << size;
if(inputnum[i] == '.') {
break;
}
}
return 0;
}
(in my case with the following output:)
a
0a
1s
2d
3f
4g
5.
6Argento:Desktop marinos$
Your code seams OK as long as you're testing char against char in your loop and not something else.. Could it be that inputnum is some integral value ? if so, then your test clause will always evaluate to false unless inputnum matches the numerical value '.' is implicitly casted to..
EDIT
Apparently you are indeed trying to put char in a int[]. Try the following:
#include <iostream>
int main() {
using namespace std;
int size = 0;
int inputnum[50];
char inputchar[50];
for(int i=0;i<50;i++,size++) {
cin >> inputchar[i];
inputnum[i] = static_cast<int>(inputchar[i]); // or inputnum[i] = (int)inputchar[i];
cout << size << endl; // add a new line in the end
if(inputchar[i] == '.') break;
}
return 0;
}
Then again this is probably a lab assignment, in a real program I'd never code like this. Tat would depend on the requirements but I'd rather prefer using STL containers and algorithms or stringstreams. And if forced to work at a lower-level C-style, I'd try to figure out to what number '.' translates to (simply by int a = '.'; cout << a;`) and put that number directly in the test clause. Such code however might be simple but is also BAD in my opinion, it's unsafe, implementation specific and not really C++.