C++ double pointer array to float conversion

Question

What is a correct way to convert double to float in c++. Is the conversion implicit?

Question 1: Consider double d = 5.0; and float f;

Which one is correct?

f = d;
f = (float)d;
f = static_cast<float>(d);

Question 2: Now consider we have

char *buffer = readAllBuffer(); 
double *d = (double*)(buffer + offset);
float f;

Which one is now correct?

f = d[0];
f = (float)d[0];
f = static_cast<float>(d[0]);

Thanks in advance!

None of them are really correct considering a double can store more than a float can so can you loose precision or the value completely — NathanOliver
– NathanOliver, Commented Jun 4, 2018 at 13:46
The two questions are really the same. You start with a double value in either case and want to convert to a float value. In other words, how you obtain the double is independent of converting it to float. — Qaz
– Qaz, Commented Jun 4, 2018 at 13:47
@NathanOliver Loosing precision is not problem here. which one gives me the closest number? Since I have to convert double to float? — mohsen_og
– mohsen_og, Commented Jun 4, 2018 at 13:48
@Mogi - they all will give you the same resulting f value. — Sander De Dycker
– Sander De Dycker, Commented Jun 4, 2018 at 13:49
@sebrockm: Not necessarily; only if the original array is read-only and the program attempts to modify data pointed at by d. — Bathsheba
– Bathsheba, Commented Jun 4, 2018 at 13:54

Bathsheba · Accepted Answer · 2018-06-04 13:58:36Z

6

They all boil down to the same thing, and the use of arrays is a red herring. You can indeed write

float f = d;

Some folk argue that a static_cast makes code more readable as it sticks out so clearly. It can also defeat warnings that some compilers might issue if a less long-winded form is used.

Naturally of course since a double is a superset of float, you might lose precision. Finally, note that for

float f1 = whatever;
double d1 = f1;
float f2 = d1;

, the C++ standard insists that f1 and f2 must be the same value.

edited Jun 4, 2018 at 13:58

answered Jun 4, 2018 at 13:47

Bathsheba

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2 Comments

Jeka Over a year ago

static_cast helps to wipe out warning message about truncating from double to float. I belive this is the main reason for using it, not readability

Bathsheba Over a year ago

@Jeka: Indeed you're correct. I have all such warnings explicitly suppressed in all my builds.

Ben Voigt · Accepted Answer · 2018-06-04 14:17:54Z

0

You do have one major issue. This is not allowed:

double *d = (double*)(buffer + offset);

It violates strict aliasing and quite possibly alignment requirements. Instead you need to use memcpy:

double d;
memcpy(&d, buffer + offset, sizeof d);
float f = d;

Either of the cast alternative can be substituted for the last line, the important change is from dereferencing an pointer with incorrect type and alignment to making a bytewise copy.

answered Jun 4, 2018 at 14:17

Ben Voigt

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1 Comment

Jarod42 Over a year ago

It depends of the real type at buffer + offset BTW. (can be a double* casted to char* and restore back to double*) (Name of the function suggest it is not the case though).

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