Assembly MIPS：将C代码转换为程序集

我正在尝试将下面的C代码翻译成MIPS汇编语言，我有点理解其中的大部分内容，但是，我不知道第一行在汇编中的等价物是什么

int ary[3] = {2,3,4};

如果有人能看一看我的C到汇编的“翻译”并验证我是否在正确的轨道上，我将不胜感激

C代码

int ary[3] = {2,3,4};
int i=0;

//loop to double array values
for(i=0; i < 3; i++){
    ary[i] = ary[i]*2;
}

---

如果这是一个本地数组，您可以在堆栈上为它分配空间，然后从代码中初始化它。 C代码的asm翻译可能如下所示：

    addi $sp, $sp, -12     # allocate space for 3 words, $sp is now the address of the array
    addi $t0, $zero, 2
    sw $t0, ($sp)          # ary[0]=2
    addi $t0, $zero, 3
    sw $t0, 4($sp)         # ary[1]=3
    addi $t0, $zero, 4
    sw $t0, 8($sp)         # ary[2]=4

    addi $t0, $zero, 0     # initialize i=0

Start:
    sll $t1, $t0, 2        # i*4 for element size
    add $t1, $t1, $sp      # add base address of array, $t1 is now &ary[i]
    lw $t2, ($t1)          # load ary[i]
    sll $t2, $t2, 1        # mutiply by 2
    sw $t2, ($t1)          # store back to ary[i]
    addi $t0, $t0, 1       # i++
    addi $t1, $t0, -3      # check if i<3 by doing (i-3)<0
    bltz $t1, Start
    addi $sp, $sp, 12      # free the array

固定的asm版本是：

    addi $t1, $sp, 12      # end=&ary[3]
    addi $t0, $sp, 0       # ptr=ary

Start:
    lw $t2, ($t0)          # load ary[i]
    sll $t2, $t2, 1        # mutiply by 2
    sw $t2, ($t0)          # store back to ary[i]
    addi $t0, $t0, 4       # ptr++ (note it is incremented by 4 due to element size)
    bne $t0, $t1, Start    # ptr!=end

---

您的代码中有许多错误

addi $t1, baseAddress, 8 #cut off point to stop the loop; array[2]
addi $t1, $zero, $zero #initialize i=0

在第一行中，没有这样的指令，除非baseAddress是寄存器。第二行应该是

add

，而不是

addi

，因为$zero不是立即数

Start:
lw $t2, base(offset)
sll $t2, $t0, 1 #mutiply $t2 by 2  
sw $t2, base(offset)

上面的线路也有问题。您只需将一个单词加载到$t2，然后立即将另一个值存储到$t2，因此之前的加载没有意义

  #include <iostream>
  using namespace std;

  //prototypes

  int maxIs (int *x, int n);
  int minIs ( int *x, int n);
  void avgIs (int *x, int n, int *theAvg, int *theRem);

  int main(void)
  {
  int n = 8;
  int x[] = {1,2,3,4,5,6,7,8};
  int theMax, theMin, theAvg, theRem;

  theMax = maxIs(x,n);
  theMin = minIs(x,n);
  avgIs(x,n,&theAvg, &theRem);

  cout << "max = " << theMax << "\n";
  cout << "min = " << theMin << "\n";
  cout << "avg = " << theAvg << " " << theRem << "/" << n << "\n";
  cout << "Bye!\n";
  }

  //functions

 int maxIs (int *x, int n )
 {
int i;
int theMax = 0;
for (i=0; i<n; i++)
{
    if (x[i]>theMax) theMax =x[i];
}
return (theMax);
}

int minIs (int *x, int n )
{
int i;
int theMin = 0x7FFF;
for (i=0; i<n; i++)
{
    if (x[i]>theMin) theMin =x[i];
}
return (theMin);
}

void avgIs (int *x, int n, int *theAvg, int *theRem )
{
int i;
int theSum = 0;
for (i=0; i<n; i++)
{
    theSum += x[i];
}
*theAvg = theSum /n;
*theRem = theSum %n;
}

使用名称空间std； //原型 int-maxIs（int*x，int-n）； int mini（int*x，int n）； void avgIs（int*x，int*n，int*theAvg，int*theRem）； 内部主（空） { int n=8； int x[]={1,2,3,4,5,6,7,8}； int theMax、theMin、THAVG、theRem； 最大值=最大值（x，n）； theMin=mini（x，n）； avgIs（x、n、THAVG和theRem）；

---

您好，谢谢您的回复，我很感激。我想知道为什么lw和sw指令中没有结束语句。还有，为什么lw和sw指令中没有基址？不确定您所说的结束语句是什么，可能是一个

jr$ra

？因为您的C代码不是完整的函数或程序，我只是在一个simi中提供了asm代码至于缺少的基址：这是因为地址已经通过添加

$sp

（数组地址）计算出来，所以基址仅为0。如果需要，可以将其写出。请注意，基址不能是另一个寄存器，

lw$t2，$sp（$t1）

是非法的。您是否在错误的位置发布了此内容？这与问题无关，也无法回答问题。

Start:
lw $t2, base(offset)
sll $t2, $t0, 1 #mutiply $t2 by 2  
sw $t2, base(offset)

  #include <iostream>
  using namespace std;

  //prototypes

  int maxIs (int *x, int n);
  int minIs ( int *x, int n);
  void avgIs (int *x, int n, int *theAvg, int *theRem);

  int main(void)
  {
  int n = 8;
  int x[] = {1,2,3,4,5,6,7,8};
  int theMax, theMin, theAvg, theRem;

  theMax = maxIs(x,n);
  theMin = minIs(x,n);
  avgIs(x,n,&theAvg, &theRem);

  cout << "max = " << theMax << "\n";
  cout << "min = " << theMin << "\n";
  cout << "avg = " << theAvg << " " << theRem << "/" << n << "\n";
  cout << "Bye!\n";
  }

  //functions

 int maxIs (int *x, int n )
 {
int i;
int theMax = 0;
for (i=0; i<n; i++)
{
    if (x[i]>theMax) theMax =x[i];
}
return (theMax);
}

int minIs (int *x, int n )
{
int i;
int theMin = 0x7FFF;
for (i=0; i<n; i++)
{
    if (x[i]>theMin) theMin =x[i];
}
return (theMin);
}

void avgIs (int *x, int n, int *theAvg, int *theRem )
{
int i;
int theSum = 0;
for (i=0; i<n; i++)
{
    theSum += x[i];
}
*theAvg = theSum /n;
*theRem = theSum %n;
}