Uninitialized local variable 'lc' used but I initialized it - c++

Question

I wanted to check whether what I wrote on the programming exam was working at least. And it turned out that it was not. And I do not understand why EXACTLY it does not work.
The task was to write a program with boolean function which should return true state if 2d matrix has only one row which consist entirely of negative element.
Here is the code:
#include "stdafx.h"
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
bool cns();
const int n=5;
int a[n][n];
bool cns() {
int ts;
//!!!!
int lc; //!! I have initiated lc variable but still it does not work !!
//!!!
//standard 2d static quad matrix
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
cout << "a[" << i << "][" << j << "]=" << endl;
cin >> a[i][j];
}
}
//check whether the sum of elements of the row is negative or not
for (int i = 0; i < n; i++) {
ts = 0; //temp sum
for (int j = 0; j < n; j++) {
ts += a[i][j]; //I thought what if sum of elements is negative then the whole row is negative too
if (ts < 0) { //I have just realized that is wrong
lc++; //counter of negative lines, which consist entirely of negative elements
}
}
}
//only one row should be whole negative
if (lc == 1) {
return true;
}
else {
return false;
}
}
int main()
{
int lc;
cout << cns << endl;
return 0;
}
So could you tell me please where I did mistake with variable 'lc' and why compiler tells me "uninitialized local variable 'lc' used"?

Answer 1

You haven't initialized lc, but declared it.
To initialize a variable means giving it an initial value (which you should always do):
int lc = 0;

Answer 2

Initialising a variable is, essentially, giving it an initial value.
Your definition of lc
int lc;
does not initialise it. Since it is a variable of automatic storage duration (i.e. it is local to a block), it is not initialised.
Accessing its value therefore gives undefined behaviour.
The first thing that the code does with lc (within the first set of loops in your code) is
lc++;
Incrementing a variable of type int requires accessing its value, before producing an effect (doing the act of incrementing). Hence undefined behaviour.
The compiler warning is being issued because of that. To eliminate the warning, either initialise it where it is defined. For example;
int lc = 42;
or ensure the first operation is to set it to a valid value
int lc;
// later on the first thing ever done to lc is ...
lc = 47;
People often assume that all variables (of basic types, like int) which are defined without being explicitly initialised will have an initial value of 0 (zero). That is true in some other languages, but not in C++ - at least not in this context (an int of static storage duration IS zero-initialised).

Answer 3

Initialization is not what you have done here. As stated by amc176 you have only declared it.
When you declare variable lc, memory is reserved on the stack. The amount of memory reserved depends on the data type (a char will take up more memory than an int).
However, if you do not provide an initial value for that variable (i.e. initialize it) the initial value of the data type will be exactly what was present in that specific piece of memory. That is why your compiler is complaining.