JustBlossom - 1 year ago 351

C Question

**Background:**

I am playing around with bit-level coding (this is not homework - just curious). I found a lot of good material online and in a book called Hacker's Delight, but I am having trouble with one of the online problems.

It asks to convert an integer to a float. I used the following links as reference to work through the problem:

How to manually (bitwise) perform (float)x?

How to convert an unsigned int to a float?

http://locklessinc.com/articles/i2f/

**Problem and Question:**

I thought I understood the process well enough (I tried to document the process in the comments), but when I test it, I don't understand the output.

Test Cases:

float_i2f(2) returns 1073741824

float_i2f(3) returns 1077936128

I expected to see something like 2.0000 and 3.0000.

Did I mess up the conversion somewhere? I thought maybe this was a memory address, so I was thinking maybe I missed something in the conversion step needed to access the actual number? Or maybe I am printing it incorrectly? I am printing my output like this:

`printf("Float_i2f ( %d ): ", 3);`

printf("%u", float_i2f(3));

printf("\n");

But I thought that printing method was fine for unsigned values in C (I'm used to programming in Java).

Thanks for any advice.

`/*`

* float_i2f - Return bit-level equivalent of expression (float) x

* Result is returned as unsigned int, but

* it is to be interpreted as the bit-level representation of a

* single-precision floating point values.

* Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while

* Max ops: 30

* Rating: 4

*/

unsigned float_i2f(int x) {

if (x == 0){

return 0;

}

//save the sign bit for later and get the asolute value of x

//the absolute value is needed to shift bits to put them

//into the appropriate position for the float

unsigned int signBit = 0;

unsigned int absVal = (unsigned int)x;

if (x < 0){

signBit = 0x80000000;

absVal = (unsigned int)-x;

}

//Calculate the exponent

// Shift the input left until the high order bit is set to form the mantissa.

// Form the floating exponent by subtracting the number of shifts from 158.

unsigned int exponent = 158; //158 possibly because of place in byte range

while ((absVal & 0x80000000) == 0){//this checks for 0 or 1. when it reaches 1, the loop breaks

exponent--;

absVal <<= 1;

}

//find the mantissa (bit shift to the right)

unsigned int mantissa = absVal >> 8;

//place the exponent bits in the right place

exponent = exponent << 23;

//get the mantissa

mantissa = mantissa & 0x7fffff;

//return the reconstructed float

return signBit | exponent | mantissa;

}

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Answer Source

Continuing from the comment. Your code is correct, and you are simply looking at the *equivalent* `unsigned integer`

made up by the bits in your IEEE-754 single-precision floating point number. The IEEE-754 single-precision number format (made up of the sign, extended exponent, and mantissa), can be interpreted as a `float`

, or those same bits can be interpreted as an `unsigned integer`

(just the number that is made up by the 32-bits). You are outputting the *unsigned equivalent* for the floating point number.

You can confirm with a simple union. For example:

```
#include <stdio.h>
#include <stdint.h>
typedef union {
uint32_t u;
float f;
} u2f;
int main (void) {
u2f tmp = { .f = 2.0 };
printf ("\n u : %u\n f : %f\n", tmp.u, tmp.f);
return 0;
}
```

**Example Usage/Output**

```
$ ./bin/unionuf
u : 1073741824
f : 2.000000
```

Let me know if you have any further questions. It's good to see that your study resulted in the correct floating point conversion. (also note the second comment regarding truncation/rounding)

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