## How to represent FLOAT number in memory in C

how floating point numbers are stored in memory in c

how to store float value in c

how is a float represented in memory

a floating-point number is actually stored as

memory allocated for float value

how floating point numbers are stored in memory in java

float32 memory

While reading a tutorial I came across how to represent Float number in memory. The tutorial had an example with a floating point number.

float a=5.2 with below Diagram

Can anyone please tell how this 5.2 is converted in to binary and how it is represented in memory in above the above diagram?

As was said, 5.2 is represented as a sign bit, an exponent and a mantissa. How do you encode 5.2?

5 is easy:

101.

The rest, 0.2 is 1/5, so divide `1.00000...`

(hex) by 5 and you get `0.3333333...`

(hex).

*(This can be followed more easily if you consider one bit less: 0.FFFF... → F / 5 = 3, so it is easy to see that 0.FFFF... / 5 = 0.33333.... That one missing bit doesn't matter when dividing by 5, so 1.0000... / 5 = 0.3333... too).*

That should give you

0.0011001100110011001100110011...

Add 5, and you get

101.00110011001100110011... exp 0 (== 5.2 * 2^0)

Now shift it right (normalize it, i.e. make sure the top bit is just before the decimal point) and adjust the exponent accordingly:

1.010011001100110011001100110011... exp +2 (== 1.3 * 2^2 == 5.2)

Now you only have to add the bias of 127 (i.e. `129 = 0b10000001`

) to the exponent and store it:

0 10000001 1010 0110 0110 0110 0110 0110

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:

01000000 10100110 01100110 01100110

Now you only have to decide little or big endian.

This is not exactly how it works, but that is more or less what happens when a number like 5.2 is converted to binary.

**Can anyone explain representation of float in memory?,** A typical single-precision 32-bit floating-point memory layout has the This C code will print binary representation of float on the console. Memory representation of float data type in c (Both in Turbo c compiler and Linux gcc compiler) Float numbers are stored in exponential form i.e. (Mantissa)*10^ (Exponent)

**How floating-point no is stored in memory? - DEV,** Representation. Now we have,. Sign bit 0 because 10.75 is positive number. Exponent value is 130 which is (10000010)� The float value : 10.327000 The sum of float and int variable : 38.327000 Double. Double is also a datatype which is used to represent the floating point numbers. It is a 64-bit IEEE 754 double precision floating point number for the value. It has 15 decimal digits of precision. Here is the syntax of double in C language, double variable_name; Here is an example of double in C language, Example

The value is represented in memory in reverse order, but the confusing point may be that 5.2f is really represented as 5.1999998 due to the accuracy loss of the floating point values.

**How float values are stored in memory,** Real numbers are represented in C by the floating point types float, double, and long double. Just as the integer types can't represent all integers because they fit in a storing doubles instead of floats will take twice as much space in memory. In computer Memory every data is represented in the form of binary bits. i.e. in the form of 0 and 1. To represent floating point numbers i.e. decimal numbers the memory will follow some special rules to store and recognise these numbers. To understand the memory representation of decimal numbers we need to understand the following things –

Representing 5.2 is very simple in binary logic:

8 4 2 1 5 -> 0 1 0 1

For a decimal number:

Take .2 and multiply by 2 (since it is represented in binary).

.2 X 2 = 0.4 -> take the value after the decimal point, don't take the value before the decimal point .4 X 2 = 0.8 .8 X 2 = 1.6 .6 X 2 = 1.2 .2 X 2 = 0.4

and so on...

After this step, take the value before the decimal point from output of the above steps:

.2 X 2 = 0.4 -> take 0 from this for representing in binary form

So the final o/p of 5.2 is:

0101.00110...

**C/FloatingPoint,** Floating-point numbers are stored on byte boundaries in the following format: Address+0 Address+1 Example � Compact Model Example � Large Model Example � Interfacing C programs to PL/M-51 The mantissa represents the actual binary digits of the floating-point number. In memory, this value appears as follows:� Floating Point Numbers Floating point numbers are represented by non-computers (humans) in scientific notation (** represents raising to a power) 4.01 X 10**8 = 401,000,000.0 4.01 X 10**-3 = 0.00401 - 4.01 X 10**8 = -401,000,000.0 -4.01 X 10**-3 = -0.00401 From these examples, it is apparent that a floating point number is represented using

Raw float 5.2:

01000000101001100110011001100110 ^ sign bit

In memory, reverse byte order (as your diagram):

01100110011001101010011001000000 ^ sign bit

**Cx51 User's Guide: Floating-point Numbers,** C provides various floating-point types to represent non-integer number with a decimal point at any position. Double-precision floating-point format is a computer number format, usually occupying 64 bits in computer memory; it represents a wide dynamic range of numeric values by using a floating radix point. Floating point is used to represent fractional values, or when a wider range is needed than is provided by fixed point (of the same bit width

**C Float Types,** On modern architectures, floating point representation almost always follows IEEE memory means floating point numbers can only store a certain number of � Single-precision floating-point format is a computer number format, usually occupying 32 bits in computer memory; it represents a wide dynamic range of numeric values by using a floating radix point. A floating-point variable can represent a wider range of numbers than a fixed-point variable of the same bit width at the cost of precision.

**4.8 — Floating point numbers,** Following figure illustrate how floating point number is stored in memory. Double precision Number - Double. To store double, computer will allocate 8 byte (64 bit) memory. Where, 1 bit for sign, 11 bit for exponent, 52 bit for significant. only difference between double and float representation is the bias value. Here we use 11 bit for exponent.So bias value will be 2 11 - 1 - 1 i.e 2 10 - 1 which is 1023.

**Memory representation of float data type in c,** How is float a=5.2 stored in memory (C/C++)? Converting 5.2 into single precision 32 bits Duration: 4:22
Posted: 19 Oct 2013 Typically, if you need to use floating point numbers, using the double type instead is a better choice because it has more precision and is probably more common. However: The double type uses more memory than the float type. In a large array this difference becomes relevant. Array

##### Comments

- This tutorial seems to be developing a bit more on the subject : kipirvine.com/asm/workbook/floating_tut.htm
- Is the diagram surely correct? h-schmidt.net/FloatApplet/IEEE754.html says something else.
- I have taken this diagram from question number 60 from this link cquestionbank.blogspot.com/2010/01/…
- @phimuemue: yes, the diagram is correct for little-endian. See my answer.
- The exponent counts powers of 2. The number should not change its value, so if you shift the mantissa right one bit (which is equivalent to a division by 2), the exponent must be incremented by 1 (which is equivalent to a multiplication by 2). For 5.2, you do this twice. The top bit must always be 1, and the "decimal/binary point" is right behind it. So you go from 5.2 * 2^0 to 2.6 * 2^1 to 1.3 * 2^2. The value must always be between 1 and 2 (except for special values like 0 or
**very**small values, or NaNs, etc.). This is called normalization. Since the top bit is always 1, it is not stored. - The 'shifting' of exponents is done so you can store exponents smaller than zero. E.g. 0.125 is 1/8. That's stored as Mantissa (1)00..., sign +, Exp 124 (127-3)
- Can you please explain why did you choose
`0.FFFFFFF...`

to divide and also why add 127 ? - 0.FFFF... was chosen because it is only one least signigicant bit lower than 1 and easier to understand. 127 was chosen because that is the bias the IEEE declares for he exponent of single precision floats. The 127 is only added to store the value in the format.
- @Vultrao: The zeroes, the infinities, NaNs and denormals are different, indeed. The format is well defined, but not how to convert from string representation to float.
- Thanks @Phimuemue for your explanation.
- Floats in IEEE-754 are represented with a 8 bit exponent and a 23 bit mantissa. That excludes the hidden bit, which is always 1 (except for denormals or NaNs or 0), so the mantissa (significand) is usually 24 bits, but only 23 are stored. The diagram shown is almost correct for little-endian floats. M should always have a top bit of 1, and the exponent is adjusted correspondingly.
- FWIW, the diagram is correct. It is little-endian, so the topmost byte is at the topmost address.
- this is not a floating point representation, just a normal binary representation.