Shifts move bits left or right; rotations wrap them around. Today you add both, and pin the two subtleties: the shift count is taken modulo 32, and the right shift comes in signed and unsigned forms.
Execute i32.shl, i32.shr_s, i32.shr_u, i32.rotl, and i32.rotr with the count masked to 5 bits.
A shift slides the bits of a value; a rotation slides them and wraps the ones that fall off the end back around. The first thing to pin is the count: WebAssembly masks the shift amount modulo the width, so an i32 shift uses only the low 5 bits of the count. That is why 1 shl 33 equals 1 shl 1 equals 2 - the 33 is reduced to 1. Skip the mask and a large shift count gives undefined or zero results depending on the host language, so it must be explicit.
The second subtlety is that the right shift splits by sign. shr_s is arithmetic: it copies the sign bit down from the top, so -8 shr_s 1 stays negative at -4. shr_u is logical: it fills with zeros, so 0x80000000 shr_u 31 brings the top bit down to 1. Left shift (shl) has only one form. Rotations (rotl, rotr) never lose bits at all - 0x80000000 rotl 1 moves the top bit around to the bottom, giving 1. These five round out i32’s bit manipulation.
// The shift amount is taken modulo 32 (mask with 0x1F). shr_s is an// arithmetic (sign-extending) shift; shr_u is logical (zero-fill).case 0x74: // i32.shlb := popI32(); a := popI32(); k := uint(b) & 31stack.Push(I32(a << k))// 0x75 shr_s (arithmetic), 0x76 shr_u (logical), 0x77 rotl, 0x78 rotr