What is the smallest integer n (stored as a floating point value) such that n + 1.0 - n will produce 0.0 for: a) Single-precisio
n IEEE 754 floating point?
1 answer:
Answer: The FP-s are stored as sign (1.)111111111111 - number of bits.
Also 24 bits resolution there can be 23 zeros
Explanation:
The –1·2²⁴ might be stored as -1 · 1.00000000000000000000000(1 ←the 2⁰), it depends on how the FFP “engine” manages this, it may also be code specific a n+1–n does return 1 but 1–n+n does not. you should carry out a test for a specific compiler/computer
This is what javascript does with double (k+=1; n-=1) e.g
the (k+=2; n-=2) should be used to pass ±0x20000000000000
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Answer:
Check the explanation
Explanation:
To solve the question above, we will be doing some arithmetic and bit wise operations on enum values. in addition to that, we are asserting the array with INTCOUNT and also printing the size of array i.e 12 and we are adding BIG_COUNT enum value to 5 and the result is 25.
#include<stdio.h>
enum EnumBits
{
ONE = 1,
TWO = 2,
FOUR = 4,
EIGHT = 8
};
enum Randoms
{
BIG_COUNT = 20,
INTCOUNT = 3
};
int main()
{
// Basic Mathimatical operations
printf("%d\n",(ONE + TWO)); //addition
printf("%d\n",(FOUR - TWO)); //subtraction
printf("%d\n",(TWO * EIGHT)); //multiplication
printf("%d\n",(EIGHT / TWO)); //division
// Some bitwise operations
printf("%d\n",(ONE | TWO)); //bitwise OR
printf("%d\n",(TWO & FOUR)); //bitwise AND
printf("%d\n",(TWO ^ EIGHT)); //bitwise XOR
printf("%d\n",(EIGHT << 1)); //left shift operator
printf("%d\n",(EIGHT >> 1)); //right shift operator
// Initialize an array based upon an enum value
int intArray[INTCOUNT]; // declaring array
// Have a value initialized be initialized to a static value plus
int someVal = 5 + BIG_COUNT; //adding constant and BIG_COUNT variable
printf("%d\n",someVal); //value will be 25
printf("%d",sizeof(intArray)); //value will be 12
return 0;
}
Kindly check the attached image below for the Code and Output Screenshots:
