CODEWITHSUNDEEP
c++
Huỳnh Hoàng Huy
Huỳnh Hoàng Huy

Reputation: 27

Why does this floating-point division produce zero, when the value rounded down isn't 0?

I am practice with C++ and I see some problem:

#include <iostream>
#include <string>
using namespace std;

int main() {
    int a = 2;
    {
        cout << a;
        cout << "\n";
        float a = a / 2;
        cout << "a= a/2 = ";
        cout << a;

    }
    cout << "\n";
    a = 2;
    {
        cout << a;
        cout << "\n";
        float b = a / 2;
        cout << "b= a/2 = ";
        cout << b;

    }
}

This return:

2 
a= a/2 = 0 
2 
b= a/2 = 1

I want to know why a = a/2 = 0 ?

Thank you

Upvotes: 0

Views: 103

Answers (2)

Lazy Ren
Lazy Ren

Reputation: 904

Because you are actually using declared, but never initialized variable when you stated float a = a/2. My computer prints 4.49985e-039 but that could be any number.
You are confusing yourself because you have two int & float variable with same name. Better to choose your name of the variable carefully or you have to track your code to see which is indicating which.
I'll comment on the each line which variable a has been used.

#include <iostream>
#include <string>
using namespace std;

int main() {
  int a = 2;  // int a declared in main scope used;
  {
    cout << a;  // int a declared in main scope; since there is no a declared in local scope.
    cout << "\n";
    float a = a/2; // block scope variable a used without initialization to initialize itself. UB. 
    cout << "a= a/2 = ";  // block scope variable used
    cout << a;  // block scope variable used
  }
  cout << "\n";   a = 2;  // int a declared in main scope; Since it's block scope is within main scope only.
  {
    cout << a;  // int a declared in main scope; since no a has been declared in local scope
    cout << "\n";
    float b = a/2; // int a declared in main scope; since no a has been declared in local scope
    cout << "b= a/2 = "; // int a declared in main scope; since no a has been declared in local scope
    cout << b;
  }
}

Upvotes: 2

templatetypedef
templatetypedef

Reputation: 372664

This is a subtle error. Look at this code:

int a = 2;
{
    float a = a / 2;
}

Outside of the curly braces, the name a refers to int a, the integer declared up top. But inside the curly braces, once you reach the line in which float a is declared, the name a refers to float a inside the braces rather than int a outside the braces.

This is a problem because the line

float a = a / 2;

means "create a new variable named a of type float. Oh, and it needs an initial value. That's okay! Give it the value of float a, divided by two." See the problem here? The variable a is being initialized in terms of itself, so when a / 2 is computed a has not been initialized and the results are undefined.

To fix this, simply give float a a new name.

Upvotes: 7

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