假设数据是32位整数（a、X、Y和Z是32位长），我用下面的c代码实现了RV32IM汇编程序。我验证了我的工作，结果是正确的。

#define W 20  

int main() {
    int X[W] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19};
    int Y[W] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
    int Z[W];
    int a=2;
    iaXpY(a, X, Y, Z);
    return 0;
};

void iaXpY(int a, int *X, int *Y, int *Z) {
    register int i=0;

    for (int i = 0; i < W; i++) {
        Z[i] = a*X[i] + Y[i];
    }
}

我对32位数据的研究

# iaXpY program, implementation in RV32IM assembly
    # for(i=0; i<W, i=i++) Z[i] = a*X[i] + Y[i];
    
    .data
    X: .word 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19
    Y: .word 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
    Z: .word 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
    W: .word 20                    
    a: .word 2  
    
    .text
    # a0 = a
    # a1 = X base address
    # a2 = Y base address
    # a3 = Z base address
    # s1 = W
    # t0 = i
    # t1 = a*X[i] + Y[i]
    
    main:
        lw  a0, a         # a0 = a
        lw  s1, W         # s1 = W  
        
        la  a1, X         # a1 = &X
        la  a2, Y         # a2 = &Y
        la  a3, Z         # a3 = &Z
        
        jal ra, iaXpY     # jump to for loop
        
        li  a7, 10        # end program
        ecall
    
    iaXpY:
        add  t0, x0, x0   # i = 0   
    loop:
        # scheduled to avoid data hazards
        lw   t1, 0(a1)    # t1 = X[i]
        addi a1, a1, 4    # ++X
        mul  t1, t1, a0   # t1 = a*X[i]
        lw   t2, 0(a2)    # t2 = Y[i]
        addi a2, a2, 4    # ++Y
        add  t1, t1, t2   # t1 = a*X[i] + Y[i]
        addi t0, t0, 1    # i++
        sw   t1, 0(a3)    # Z[i] = a*X[i] + Y[i]
        addi a3, a3, 4    # ++Z
        blt  t0, s1, loop # if i < W, go to loop
        ret               # else, return to main

现在，使用数据实现是64位整数（a、X、Y和Z为64位长），但仍使用RV32IM。我的方法是，由于RV32IM中的一个寄存器只能保存32位值，因此我必须使用两个寄存器保存高位和低位32位，但如何在RV32IM汇编中实现？