My understanding is that all numbers in JavaScript are double-precision floating-point. How can I convert a double to a float in JavaScript? I understand the precision loss, but that is tolerable.
I need to depict C style floating point numbers
3 Answers
This will work in modern engines:
function doubleToFloat ( d ) {
if ( Float32Array )
return new Float32Array([d])[0];
}
Note that the data type of the result is still number (double precision), but it only uses 32 bits of information. I added a fallback for the case that Float32Array is not available:
function doubleToFloat ( d ) {
if ( Float32Array )
return new Float32Array( [ d ] )[ 0 ];
if ( d === 0 )
return d;
var sign = 2*(d >= 0) - 1;
var b = Math.abs( d ).toString( 2 );
var decimalIndex = b.indexOf( '.' );
var oneIndex = b.indexOf( '1' );
var exponent, mantissa, round, result;
if( decimalIndex === -1 ) {
exponent = b.length - 1;
mantissa = b.substr( 1, 23 );
round = +( mantissa.length === 23 && b[24] === '1' );
result = sign*( parseInt( '1' + mantissa, 2 ) + round )*Math.pow( 2, exponent - mantissa.length );
} else if ( decimalIndex === 1 ) {
exponent = 1 - oneIndex;
if ( oneIndex === 0 ) {
mantissa = '1' + b.substr( 2, 23 );
round = +( mantissa.length === 24 && b[25] === '1' );
result = sign*( parseInt( mantissa, 2 ) + round )*Math.pow( 2, 1 - mantissa.length );
} else {
mantissa = b.substr( oneIndex, 24 );
round = +( mantissa.length === 24 && b[oneIndex + 24] === '1' );
result = sign*( parseInt( mantissa, 2 ) + round )*Math.pow( 2, 1 + exponent - mantissa.length );
}
} else {
exponent = decimalIndex - 1;
mantissa = b.replace( '.', '' ).substr( 1, 23 );
round = +( mantissa.length === 23 && b.replace( '.', '' )[24] === '1' );
result = sign*( parseInt( '1' + mantissa, 2 ) + round )*Math.pow( 2, decimalIndex - mantissa.length - 1 );
}
if ( exponent < -126 )
return 0;
if ( exponent > 127 )
return sign*Infinity;
return result;
}
3 Comments
marc
this gives pretty bad precision
Sebastian Simon
@asehgal Yeah, isn’t that exactly what you wanted?
marc
I didnt want BAD precision. I said a 32 bit representation is desired even if it comes with precision loss.
Another way is:
var doubleValue = Math.random()
var floatValue = ((doubleValue * 1e7) | 0) / 1e7
Comments
Just use Math.fround() - it returns the nearest 32-bit single precision float representation of a number.
1 Comment
Community
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new Float32Array([yourNumber])[0], but it’s not supported in all browsers. Why do you need the lost precision?floats. Are you interfacing with C code?