trie: reduce allocs in recHash

[MariusVanDerWijden]

Reduces allocations in recHash.
I noticed during profiling, that a lot of allocations (3.8% of total) is spent on hexToCompact. This uses the existing HexToCompactInPlace to reduce allocations.
Since we reset the key anyway after this operation, it should be fine imo

ROUTINE ======================== github.com/ethereum/go-ethereum/trie.keybytesToHex in github.com/ethereum/go-ethereum/trie/encoding.go
 150455888  150455888 (flat, cum)  3.22% of Total
         .          .     97:func keybytesToHex(str []byte) []byte {
         .          .     98:    l := len(str)*2 + 1
 150455888  150455888     99:    var nibbles = make([]byte, l)
         .          .    100:    for i, b := range str {
         .          .    101:        nibbles[i*2] = b / 16
         .          .    102:        nibbles[i*2+1] = b % 16
         .          .    103:    }
         .          .    104:    nibbles[l-1] = 16
(pprof) list hexTo    
Total: 4677552141
ROUTINE ======================== github.com/ethereum/go-ethereum/trie.hexToCompact in github.com/ethereum/go-ethereum/trie/encoding.go
 178393826  178393826 (flat, cum)  3.81% of Total
         .          .     37:func hexToCompact(hex []byte) []byte {
         .          .     38:    terminator := byte(0)
         .          .     39:    if hasTerm(hex) {
         .          .     40:        terminator = 1
         .          .     41:        hex = hex[:len(hex)-1]
         .          .     42:    }
 178393826  178393826     43:    buf := make([]byte, len(hex)/2+1)
         .          .     44:    buf[0] = terminator << 5 // the flag byte
         .          .     45:    if len(hex)&1 == 1 {
         .          .     46:        buf[0] |= 1 << 4 // odd flag
         .          .     47:        buf[0] |= hex[0] // first nibble is contained in the first byte
         .          .     48:        hex = hex[1:]

Benchmarks:

BenchmarkHexToCompact-24    	94361824	        21.67 ns/op	       4 B/op	       1 allocs/op
BenchmarkHexToCompactInPlace-24    	285524278	         4.462 ns/op	       0 B/op	       0 allocs/op

[holiman]

Since we reset the key anyway after this operation, it should be fine imo

Please elaborate a bit, it's not obvious to me. What reset?

[MariusVanDerWijden]

In the stacktrie, we reset the key with st.key = st.key[:0]

[fjl]

Very cool find.