Cody

Problem 1165. Convert double scalar to half-precision floating point (IEEE 754r)

Solution 358434

Submitted on 23 Nov 2013 by Alfonso Nieto-Castanon
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Test Suite

Test Status Code Input and Output
1   Pass
%% assert(double2half(2^(-24)) == uint16(1)) % Smallest number

n = -15

2   Pass
%% assert(double2half(2^(-25)) == uint16(1)) % Rounds up to smallest number

n = -15

3   Pass
%% assert(double2half(2^(-26)) == uint16(0)) % Rounds down to zero

n = -15

4   Pass
%% assert(bin2dec('0 01111 0000000000') == double2half(1))

n = 0

5   Pass
%% assert(bin2dec('0 01111 0000000001') == double2half(1 + 2^(-10)))

n = 0

6   Pass
%% assert(bin2dec('1 10000 0000000000') == double2half(-2))

n = 1

7   Pass
%% assert(bin2dec('0 11110 1111111111') == double2half(65504))

n = 15

8   Pass
%% assert(bin2dec('0 00001 0000000000') == double2half(2^(-14)))

n = -14

9   Pass
%% assert(bin2dec('0 00000 1111111111') == double2half(2^(-14) - 2^(-24)))

n = -15

10   Pass
%% assert(bin2dec('0 00000 0000000001') == double2half(2^(-24)))

n = -15

11   Pass
%% assert(bin2dec('0 00000 0000000000') == double2half(0))

n = -15

12   Pass
%% assert(bin2dec('1 00000 0000000000') == double2half(-0))

13   Pass
%% assert(bin2dec('0 11111 0000000000') == double2half(inf))

n = 15

14   Pass
%% assert(bin2dec('1 11111 0000000000') == double2half(-inf))

n = 15

15   Pass
%% assert(bin2dec('0 01101 0101010101') == double2half(0.33325))

n = -2