Infendro
ff0e17b5ad
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It turns out signed math is more performant
50 lines
1.7 KiB
Kotlin
50 lines
1.7 KiB
Kotlin
package com.infendro.bytes.intarray
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import com.infendro.bytes.ByteOrder
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import com.infendro.bytes.ByteOrder.BIG_ENDIAN
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import com.infendro.bytes.ByteOrder.LITTLE_ENDIAN
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fun IntArray.copyInto(
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destination: ByteArray,
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destinationOffset: Int = 0,
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startIndex: Int = 0,
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endIndex: Int = size,
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order: ByteOrder = BIG_ENDIAN,
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) {
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val length = endIndex - startIndex
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require(length >= 0)
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require(startIndex in 0..(size - length))
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require(destinationOffset in 0..(destination.size - (length * 4)))
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when (order) {
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BIG_ENDIAN -> {
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repeat(length) { i ->
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val offset = destinationOffset + (i * 4)
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val value = this[startIndex + i]
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destination[offset] = (value shr 24).toByte()
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destination[offset + 1] = (value shr 16).toByte()
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destination[offset + 2] = (value shr 8).toByte()
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destination[offset + 3] = value.toByte()
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}
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}
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LITTLE_ENDIAN -> {
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repeat(length) { i ->
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val offset = destinationOffset + (i * 4)
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val value = this[startIndex + i]
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destination[offset] = value.toByte()
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destination[offset + 1] = (value shr 8).toByte()
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destination[offset + 2] = (value shr 16).toByte()
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destination[offset + 3] = (value shr 24).toByte()
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}
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}
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}
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}
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fun IntArray.copyInto(
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destination: UByteArray,
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destinationOffset: Int = 0,
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startIndex: Int = 0,
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endIndex: Int = size,
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order: ByteOrder = BIG_ENDIAN,
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) = copyInto(destination.asByteArray(), destinationOffset, startIndex, endIndex, order)
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