Assigning an integer to float and comparison in C/C++

Consider the below C++ program and predict the output.

#include <iostream>
using namespace std;

int main()
{
    float f = 0xffffffff;
    unsigned int x = 0xffffffff; // Value 4294967295
    if (f == x)
        cout << "true";
    else
        cout << "false";
    return 0;
}

The output of above program is false if the “IEEE754 32-bit single float type” is used by compiler. If we define:

float f = 0xffffffff;

We are basically trying to assign a 32-bit integer (signed or unsigned) to a 32-bit float. The compiler will first convert the integer 0xffffffff to a nearest 32-bit float, and the memory representation of the float f is not the same as the integer 0xffffffff. We can see the above values by printing f and x.

#include <iostream>
using namespace std;

int main()
{
    float f = 0xffffffff;
    unsigned int x = 0xffffffff;
    cout << "f = " << f << endl;
    cout << "x = " << x << endl;
    return 0;
}

Output :

f = 4.29497e+09
x = 4294967295

Even if we copy the memory directly, for example, we have an integer (value equal to 0xffffffff), and we copy over the content (memory values). Since the 0xffffffff in IEEE754 is not a valid float number, so if you compare this invalid representation to itself, it is not equal.

unsigned int x = 0xffffffff;
memoryCopy(&f, &x, sizeof(x));

This article is contributed by Rishav Raj. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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