## DEV Community is a community of 700,720 amazing developers

We're a place where coders share, stay up-to-date and grow their careers.

# How floating-point no is stored in memory?

Vishal Chovatiya
Software Developer⌨, Fitness Freak🏋, Hipster🕴, Blogger👨‍💻, Productivity Hacker⌚, Technical Writer✍️, Tech talker👨‍🎤, Leader👨‍🔬, Always a Student👨‍🎓, Incomplete🔍 & Learning Junkie📚.
Originally published at vishalchovatiya.com Updated on ・6 min read

This is cross-post from my blog.

This article is just a simplification of the IEEE 754 standard. Here, we will see how floating-point no stored in memory, floating-point exceptions/rounding, etc. But if you will want to find more authoritative sources then go for

Floating-point numbers stored by encoding significand & the exponent (along with a sign bit)

• Above line contains 2-3 abstract terms & I think you will unable to understand the above line until you read further.

## Floating point number memory layout

```+-+--------+-----------------------+
| |        |                       |
+-+--------+-----------------------+
^    ^                ^
|    |                |
|    |                +-- significand(width- 23 bit)
|    |
|    +------------------- exponent(width- 8 bit)
|
+------------------------ sign bit(width- 1 bit)```

A typical single-precision 32-bit floating-point memory layout has the following fields :

1. sign
2. exponent
3. significand(AKA mantissa)

#### Sign

• The high-order bit indicates a sign.
• `0` indicates a positive value, `1` indicates negative.

#### Exponent

• The next 8 bits are used for the exponent which can be positive or negative, but instead of reserving another sign bit, they're encoded such that `1000 0000` represents `0`, so `0000 0000` represents `-128` and `1111 1111` represents `127`.
• How does this encoding work? go to exponent bias or see it in next point practically.

#### Significand

• The remaining 23-bits used for the significand(AKA mantissa). Each bit represents a negative power of 2 countings from the left, so:
```01101 = 0 * 2^-1 + 1 * 2^-2 + 1 * 2^-3 + 0 * 2^-4 + 1 * 2^-5
= 0.25 + 0.125 + 0.03125
= 0.40625```

OK! We are done with basics.

## Let's understand practically

• So, we consider very famous float value `3.14`(PI) example.
• Sign: Zero here, as PI is positive!

#### Exponent calculation

• `3` is easy: `0011` in binary
• The rest, `0.14`
```0.14 x 2 = 0.28, 0

0.28 x 2 = 0.56, 00

0.56 x 2 = 1.12, 001

0.12 x 2 = 0.24, 0010

0.24 x 2 = 0.48, 00100

0.48 x 2 = 0.96, 001000

0.96 x 2 = 1.92, 0010001

0.92 x 2 = 1.84, 00100011

0.84 x 2 = 1.68, 001000111

And so on . . .```
• So,` 0.14 = 001000111...`If you don't know how to convert decimal no in binary then refer this float to binary.
• Add `3`, ` 11.001000111... with exp  0 (3.14 * 2^0)`
• Now shift it (normalize it) and adjust the exponent accordingly ` 1.1001000111... with exp +1 (1.57 * 2^1)`
• Now you only have to add the bias of `127` to the exponent `1` and store it(i.e. `128` = `1000 0000`) ` 0     1000 0000     1100 1000 111...`
• Forget the top `1` of the mantissa (which is always supposed to be `1`, except for some special values, so it is not stored), and you get: ` 0     1000 0000     1001 0001 111...`
• So our value of `3.14` would be represented as something like:
```    0 10000000 10010001111010111000011
^     ^               ^
|     |               |
|     |               +--- significand = 0.7853975
|     |
|     +------------------- exponent = 1
|
+------------------------- sign = 0 (positive)
```
• The number of bits in the exponent determines the range (the minimum and maximum values you can represent).

#### Summing up significand

• If you add up all the bits in the significand, they don't total `0.7853975`(which should be, according to 7 digit precision). They come out to `0.78539747`.
• There aren't quite enough bits to store the value exactly. we can only store an approximation.
• The number of bits in the significand determines the precision.
• 23-bits gives us roughly 6 decimal digits of precision. 64-bit floating-point types give roughly 12 to 15 digits of precision.

Strange! But fact

• Some values cannot represent exactly no matter how many bits you use. Just as values like 1/3 cannot represent in a finite number of decimal digits, values like 1/10 cannot represent in a finite number of bits.
• Since values are approximate, calculations with them are also approximate, and rounding errors accumulate.

## Let's see things working

```#include <stdio.h>
#include <string.h>

/* Print binary stored in plain 32 bit block */
void intToBinary(unsigned int n)
{
int c, k;
for (c = 31; c >= 0; c--)
{
k = n >> c;
if (k & 1)  printf("1");
else        printf("0");
}
printf("\n");
}

int main(void)
{
unsigned int m;
float f = 3.14;

/* See hex representation */
printf("f = %a\n", f);

/* Copy memory representation of float to plain 32 bit block */
memcpy(&m, &f, sizeof (m));
intToBinary(m);

return 0;
}```
• This C code will print binary representation of float on the console.
```f = 0x3.23d70cp+0
01000000010010001111010111000011```

## Where the decimal point is stored?

• The decimal point not explicitly stored anywhere.