Why type conversion between int and float causes data loss?

Question

Why type conversion between int and float causes data loss?

Walkbitterwyrm Lightwarg 20

In the following program type conversion of pointers from int to float and from floar to int is done but the output is not correct.

#include <stdio.h>

int main()
{
    int a, *ptr_i, *ptr_xi;
    float f, *ptr_f, *ptr_xf;

    a = 65;
    f = 65.5;

    ptr_i = &a;
    ptr_f = &f;
    ptr_xi = ptr_i;
    ptr_xf = ptr_f;

    ptr_f = (float*) ptr_xi;
    ptr_i = (int*) ptr_xf;
            
    printf("The value of \"ptr_i\" is %d\n", *ptr_i);
    printf("The value of \"ptr_f\" is %f\n", *ptr_f);

    return 0;
}

Output:

The value of "ptr_i" is 1115881472

The value of "ptr_f" is 0.000000

Why the output is different?

David Lowndes 2,640 Reputation points MVP

2023-08-25T17:06:49.0766667+00:00

Different from what?
What did you expect the output to be, and why did you expect it to be that?
Walkbitterwyrm Lightwarg 20 Reputation points

2023-08-26T11:43:04.3366667+00:00
David Lowndes---it's different because the int value 65 should be converted to float value 65.00 and the float value 65.6 should get converted to the int value 65.

This type of type cast conversion is possible with variables and shown in the following program.

#include <stdio.h> int main() { int a, i; float b, f; a = 65; b = 65.5; f = (float) a; i = (int) b; printf("The value of 'i' is %d\n", i); printf("The value of 'f' is %.2f\n", f); return 0; }

Output:

The value of 'i' is 65

The value of 'f' is 65.00

But why this output is not seen when type casting is used with pointers?
David Lowndes 2,640 Reputation points MVP

2023-08-26T12:31:28.0366667+00:00

Your understanding is wrong, in your original code there is no conversion, you're just using casts to lie to the compiler. The binary representation of a float and int are different; inappropriate pointer casting won't convert things.
Minxin Yu 13,501 Reputation points Microsoft External Staff

2023-08-29T06:36:31.12+00:00

The answer below gives an explanation. Have you got any chance to check?

1 answer

Your answer

David Lowndes 2,640 Reputation points MVP

2023-08-25T17:06:49.0766667+00:00

Different from what?
What did you expect the output to be, and why did you expect it to be that?
Walkbitterwyrm Lightwarg 20 Reputation points

2023-08-26T11:43:04.3366667+00:00

David Lowndes---it's different because the int value 65 should be converted to float value 65.00 and the float value 65.6 should get converted to the int value 65.

This type of type cast conversion is possible with variables and shown in the following program.

#include <stdio.h> int main() { int a, i; float b, f; a = 65; b = 65.5; f = (float) a; i = (int) b; printf("The value of 'i' is %d\n", i); printf("The value of 'f' is %.2f\n", f); return 0; }

Output:

The value of 'i' is 65

The value of 'f' is 65.00

But why this output is not seen when type casting is used with pointers?
David Lowndes 2,640 Reputation points MVP

2023-08-26T12:31:28.0366667+00:00

Your understanding is wrong, in your original code there is no conversion, you're just using casts to lie to the compiler. The binary representation of a float and int are different; inappropriate pointer casting won't convert things.
Minxin Yu 13,501 Reputation points Microsoft External Staff

2023-08-29T06:36:31.12+00:00

The answer below gives an explanation. Have you got any chance to check?

Answer 1

If the alignment requirements of types int and float are not the same, then this program potentially causes undefined behavior.

If the requirements are the same, then

The value of ptr_i is NOT 1115881472. The value of ptr_i is the address of object f. 1115881472 is the decimal representation of whatever data is in object f when those bits are interpreted as an int.
The value of ptr_f is NOT 0.000000. The value of ptr_f is the address of object a. 0.000000 is the decimal representation of whatever data is in object a when those bits are interpreted as a float.
You printed the values in different formats (one int and one promoted to double). Why would you expect the output to be the same?
You printed different values. Even if you used the same format, the output would be different.
If you want to know how the system arrived at that particular set of output, you need to research how your system stores int and float data.

You initialized your two primary variables to bit patterns appropriate for their defined types. Why do you think telling printf to interpret those patterns as a different type produces anything meaningful? Did you think that changing the type of the pointer used to access the memory would somehow magically convert the data in memory to match the new pointer?

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Why type conversion between int and float causes data loss?

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