Convert double value to uint_8 type array

Question

I'm new to C development. How can convert I double values to a uint_8 type array.

double test=1234.642462

I want to save as it as

uint8_t data[] = {'1','2','3','4','.','6','4','2','4','6','2'};

Any help is appreciated. Thank you.

Convert it to a string, then convert the string to uint8_t. — Lundin
– Lundin, Commented Sep 13, 2016 at 13:42
Do you want the result as int8_t or uint8_t? Because your question mentions both. — user1350184
– user1350184, Commented Sep 13, 2016 at 13:51

user1350184 · Accepted Answer · 2016-09-13 13:42:31Z

You can use the snprintf function like this:

#include <stdlib.h> int8_t data[20]; double test=1234.642462; snprintf(data, 20, "%f", test);

The 20 character limit should be adjusted to the desired precision, since floating point numbers can be very long.

Community · Accepted Answer · 2017-05-23 11:51:44Z

DBL_MAX is typically about 179769...(303 more digits).0. Usually only the first 17 are consider significant.

DBL_MIN is typically about 0.000(303 more zeros)222507... And again usually only the first 17 are consider significant.

To convert the entire range of double into a non-exponential decimal string showing significant digits can take hundreds of bytes - not a practical approach.

To convert a double into a decimal text representation, stored in char, uint8_t, int8_t, etc., is best done using exponential notation with enough, but not excessive amount of significant digits. Use *printf() family with e.

#include <float.h> // - d . dddd e - expo \0 #define DBL_TEXT_SIZE (1 + 1 + 1 + (DBL_DECIMAL_DIG - 1) + 1 + 1 + 6 + 1) ... uint8_t data[DBL_TEXT_SIZE]; sprintf(data, "%.e", DBL_DECIMAL_DIG - 1, test);

For more details: See Printf width specifier to maintain precision of floating-point value

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