convert a float to a short

[Partho]

I have a float variable which I need to add to a short variable. How do
I do this? Do I have to typecast or is there a way around? I tried
typecasting the float to a short, but that gives me a 0 or +/-1, which
is unacceptable.

Any pointers to how to get over this?

 

[Chris Dollin]

I have a float variable which I need to add to a short variable. How do
I do this?

myShort += (short) myFloat;

looks like a plausible first cut. Of course the float will be truncated
to a short value.

Do I have to typecast or is there a way around?

You'll have to cast, since adding a float to an integer will get you
a floaty value, and to unfloat something floaty needs an explicit
conversion - a cast.

I tried
typecasting the float to a short, but that gives me a 0 or +/-1, which
is unacceptable.

I'm /guessing/ that's because the float has values like 1.7, which truncate
to 1 but that you'd rather treat as 2? If so, round it.

[I do wonder what you're doing that involves adding floating-point numbers
to shorts.]

Any pointers to how to get over this?

Heavens, don't use /pointers/ for this. I'd keep anything with sharp edges
away from /my/ shorts.

 

[Ian Collins]

I have a float variable which I need to add to a short variable. How do
I do this? Do I have to typecast or is there a way around? I tried
typecasting the float to a short, but that gives me a 0 or +/-1, which
is unacceptable.

Any pointers to how to get over this?

Just assign it, assuming it's in the range of your short.

 

[Keith Thompson]

I have a float variable which I need to add to a short variable. How do
I do this? Do I have to typecast or is there a way around? I tried
typecasting the float to a short, but that gives me a 0 or +/-1, which
is unacceptable.

Yes, casting (not "typecasting") should do the trick. For example:

float f = 123.0;
short s = 42;
short t;
t = s + (short)f;
/* or */
t = s = (int)f;

(The latter is equivalent, since values of type short are promoted to
int anyway.)

You say the conversion is giving you 0 or +/-1. What float value are
you starting with?

(Also, it's usually better to use double rather than float.)

 

[pete]

I have a float variable which I need to add to a short variable.
How do
I do this? Do I have to typecast or is there a way around? I tried
typecasting the float to a short, but that gives me a 0 or +/-1, which
is unacceptable.

Any pointers to how to get over this?

It depends on what the type of the result of the addition
is supposed to be.

(short)42 + 42.42f is 84
(short)42 + 42.42f is 84.419998

/* BEGIN new.c */

#include <stdio.h>

int main(void)
{
float fv = 42.42f;
short sv = 42;

printf("(short)42 + 42.42f is %hd\n", (short)(sv + fv));
printf("(short)42 + 42.42f is %f\n", sv + fv);
return 0;
}

/* END new.c */

 

[John Smith]

I have a float variable which I need to add to a short variable. How do
I do this? Do I have to typecast or is there a way around? I tried
typecasting the float to a short, but that gives me a 0 or +/-1, which
is unacceptable.

Any pointers to how to get over this?

No cast is required in the simplest case. The short is
implicitily converted to a float when the two types are combined
in an expression. Note use of %f to display the result.

#include <stdio.h>

int main(void)
{
short a = 55;
float b = 12.6;

printf("%f\n", a + b);

return 0;
}

 

[Partho]

Hi all:

I tried this:

since a float is a 32 bit number, and is of the form 1.15, for ex., the
simplest technique is to scale the float by 2^15, and then type cast to
short:

short s = (short) (f * 2^15);

this puts the lower 16 bits of the float into the short var.

correct me if I am wrong.

 

[Keith Thompson]

I tried this:

since a float is a 32 bit number, and is of the form 1.15, for ex., the
simplest technique is to scale the float by 2^15, and then type cast to
short:

short s = (short) (f * 2^15);

this puts the lower 16 bits of the float into the short var.

correct me if I am wrong.

Please don't top-post. See <http://www.caliburn.nl/topposting.html>.

Yes, you're wrong. For starters, "^" is the bitwise xor operator in
C; 2^15 is 13, not 32767. C has no exponentiation operator. (One way
to represent 2**15 is 1<<15; I'm using "**" here as an exponentiation
operator, though it doesn't exist in C.)

Your original question was:

| I have a float variable which I need to add to a short variable. How do
| I do this? Do I have to typecast or is there a way around? I tried
| typecasting the float to a short, but that gives me a 0 or +/-1, which
| is unacceptable.

What exactly do you mean by "of the form 1.15"?

This business about scaling by 2**15 is either new information that
you should have told us about in the first place, or a considerable
misunderstanding on your part.

Based on your original question, we assumed that, for example, you had
a short variable with a value of 5, and a float variable with a value
of 10.0, and you wanted to add them to yield a short value of 15 (or
whatever actual values you happen to have). If so, don't worry about
how the float is represented internally; it's just a number. A cast
(an explicit conversion) on a number merely converts the number to
another type with the same value (with truncation in the case of
floating-to-integer).

If at all possible, show us some actual code that illustrates your
problem, ideally a small, complete, self-contained program that we can
compile and run ourselves. Have it display the actual values of your
short and float variables (use printf with "%d" for short, "%g" for
float).