Java程序辅导

	 1	
2.3: Lab Exercise on translating nested control structures 
 
 
In this exercise, we’ll go through an example of how to translate nested control structures 
in C++ to MIPS. The thing to focus on is how mechanical this process is! It’s basically 
what a compiler does to translate your C++ or Java code automatically, to machine 
language or Java bytecodes. 
 
We start with a C++ program from a past quiz: 
 
 
#include  
 
using std::cin; 
using std::cout; 
 
int main() 
{ 
  int max, num = 4, i; 
 
  max = 100; 
 
  while (max > 42) { 
 
 
    for (i=1; i < num; i++) { 
 
      max = max / i; 
 
      if ((max > i) || (max < 2)) { 
 
        cout << max; 
 
      } 
    } 
    cin >> max; 
  } 
} 
 
	 2	
The first step is to identify the main blocks. It helps to draw boxes around the nested 
structures. For example, this helps me to see the structure of the code: 
 
  int max, num = 4, i; 
 
  max = 100; 
 
  while (max > 42) { 
 					
        
 
  } 
 
(It might also be helpful to draw more boxes, say around the if block. Whatever works for 
you…) 
   for (i=1; i < num; i++) { 
 
      max = max / i; 
 
      if ((max > i) || (max < 2)) { 
        cout << max; 
      } 
   } 
     
 
   cin >> max; 
	 3	
There are lots of ways to go about the translation. I find it easier to work at one level at a 
time. Let’s start with the outermost level, the while loop. Note that we can translate the 
while loop completely independently of what’s inside it! (Let’s make $s0 max, $s1 num, 
$s2 i.) 
 
        li   $s0, 100            #  max = 100; 
        li   $s1, 4              #  num = 4; 
 
        ble  $s0, 42, exit       #  while (max > 42) { 
 
while: 
 
                                 #    for (i=1; i < num; i++) { 
 
                                 #      max = max / i; 
 
                                 #      if ((max > i) || (max < 2)) { 
 
 
                                 #        cout << max; 
 
  
                                 #      } 
     
                                 #    } 
  
                                 #    cin >> max; 
 
 
        bgt  $s0, 42, loop       #  } 
 
	 4	
Next we deal with the body of the while loop. It contains two blocks: the for loop, and the 
cin statement. First we translate the for loop, once again leaving out the body of the for 
loop: 
 
        li   $s0, 100            #  max = 100; 
        li   $s1, 4              #  num = 4; 
 
        ble  $s0, 42, outw       #  while (max > 42) { 
 
while:  li   $s2, 1 
 
        bge  $s2, $s1, maxin    #    for (i=1; i < num; i++) { 
for: 
                                 #      max = max / i; 
 
                                 #      if ((max > i) || (max < 2)) { 
 
 
                                 #        cout << max; 
 
  
                                 #      } 
        add  $s2, $s2, 1 
        blt  $s2, $s1, for       #    } 
 maxin: 
                                 #    cin >> max; 
 
 
        bgt  $s0, 42, loop       #  } 
 
outw: 
 
	 5	
Again there are several ways to proceed, but let’s start filling in the body of the for loop. 
We translate the divide, and then the compound if: 
 
        li   $s0, 100            #  max = 100; 
        li   $s1, 4              #  num = 4; 
 
        ble  $s0, 42, outw       #  while (max > 42) { 
 
while:  li   $s2, 1 
 
        bge  $s2, $s1, maxin    #    for (i=1; i < num; i++) { 
for: 
        div  $s0, $s0, $s2      #      max = max / i; 
 
        bgt  $s0, $s2, out      #      if ((max > i) || (max < 2)) { 
        bge  $s0, 2, inci 
 
                                 #        cout << max; 
 
  
                                 #      } 
inci:   add  $s2, $s2, 1 
        blt  $s2, $s1, for       #    } 
 maxin: 
                                 #    cin >> max; 
 
 
        bgt  $s0, 42, loop       #  } 
 
outw: 
 
You should be able to handle the rest. The full solution is at: 
 http://unixlab.sfsu.edu/~whsu/csc256/QUIZ/s09q2blue  
 
In summary, keep in mind that we’re not writing code, we’re pretending to be a compiler, 
translating C++ code to MIPS assembly language. It’s a very mechanical procedure.