Question

On the Hammond machines, a single precision floating point (REAL*4) number uses how many bits of memory?

• A. 4
• B. 7
• C. 32
• D. 22

Hint:

The number in types like ‘REAL*4‘ or ‘INTEGER*8‘ always specifies the size in bytes.
To find the size in bits, simply multiply that number by 8:
- ‘REAL*4‘ $\rightarrow 4 \times 8 = 32$ bits.
- ‘REAL*8‘ $\rightarrow 8 \times 8 = 64$ bits (double precision).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In computer architecture and programming languages, data types are allocated specific amounts of memory (measured in bytes or bits) to store values with a defined precision.
Key Formula or Approach:
The memory size of a data type can be converted between bytes and bits using the relation:
\[ \text{Size in bits} = \text{Size in bytes} \times 8 \]

Step 2: Detailed Explanation:
Let us determine the size of a single-precision floating-point number:
- In Fortran, a single-precision floating-point number is declared as ‘REAL‘ or explicitly as ‘REAL*4‘.
- The numeric suffix in ‘REAL*4‘ specifies that the variable is allocated 4 bytes of memory.
- Converting this size to bits:
\[ \text{Size in bits} = 4 \text{ bytes} \times 8 \text{ bits/byte} = 32 \text{ bits} \]
- This 32-bit representation typically follows the IEEE 754 standard for floating-point arithmetic, where:
- 1 bit is used for the sign.
- 8 bits are used for the exponent.
- 23 bits are used for the significand (fraction/mantissa).
Therefore, a ‘REAL*4‘ variable uses exactly 32 bits of memory.

Step 3: Final Answer:
A single precision floating point (REAL*4) number uses 32 bits of memory.