warning: uninitialized variable //But I have initialized ! C++ Compiler bug?

Question

Iam trying to compile this program but i get warning and when i run vc++ 2010 debugger pops up : ( Here is my code :

#include 
using namespa         


        

Answer 1

Suppose the entire array you pass in is 0. Both loops short-circuit and never execute, both minn and index are uninitialized.

Now if this happens, what should be happening? Set the variables to the values that accomplish just that.

Answer 2

As you say in your comment, yes, you have declared your variables, but you haven't initialized them. Initializing a variable means giving it a value. So in this case, you have told the compiler that you want to create three integers, but you haven't told it what values you want to store in those integers. That would be ok if, for every possible path through your function, index and minn were guaranteed to be given a value, but the problem here is that there is a path through your function where minn and index will never be initialized. First of all, here:

for(i=0;i<num;i++)
{            
    if(mas[i]!=0)minn=mas[i];
    break;        
}

If you have an array of zeros, then minn is never initialized to a value.

Then further down:

for(i=0;i<num;i++) 
  if(mas[i]!=0 && minn>mas[i])        
  {
      minn=mas[i];
      index=i;
  }

first of all, if you had an array of zeros, well what is the value in minn? There is no value. You are asking the compiler to compare mas[i] to a number which doesn't exist. Furthermore, what if mas[i] is always equal to zero? Well now you don't initialize minn or index. Yet at the end of the function, you are attempting to use the value of index to get an integer from the array amd then you return minn (which still equals nothing).

That's the problem you're getting from the compiler. It can see this potential outcome and is warning you that your function can be broken due to these integers never getting a value. To fix it, do what the other lads have suggested and let index and minn equal zero at the start.

Answer 3

But you haven't initialized them : )))) EX: int i,minn=0,index=0; Imagine that you pass num that equals 0, at the end you would be returning uninitialized value of minn and just before that you would set mas[unknown_number]=0; which will probably cause your app to crash since you will be referencing memory that is most likely beyond your scope. You should do a check in the beggining like if(num<1)return -1;

Answer 4

You have declared them, but not initialized them. Simply write int minn = 0, index = 0; to avoid the warning. If you don't initialize a variable, its default value is whatever was at that location in memory already; usually garbage.

The thing is, if num is negative, then neither of the for loops in your min() function will execute, and so minn and index will not have been assigned values. The if(i == num) test also won't break out of the function and prevent this from happening. So the last two lines of the function will have completely undefined results.

Sometimes there really isn't a path for the variables to be used uninitialized, though; sometimes the compiler just isn't quite smart enough to figure out all the subtleties. Just give them an initial value to avoid the warning.

Answer 5

Declaration != initialization. When you declare them the variables have random values. Just initialize them to sensible values like -1 for index and minn to a INT_MAX.