Lecture 3

 

• Compile Two C Assignments
  
  • C is one level above assembler
    
  • Change c program segment into MIPS assembly language:
    
    • 
      a = b + c;
      
    • d = a – e;
      
  • Solution
    
    • Add a,b,c
      
    • Sub d,a,e
      

Data Path Memory: MIPS Registers

 

• Register = STORAGE LOCATION (ONE WORD LONG)
  
• 1 word long = 32 bits
  
• Numbered in reverse direction
  
• 32 registers
  
  • named (each register has a unique specific name)
    
  • each register has 32 bits
    
    • why? We can't have too much memory because this memory is very expensive, and we don't want too few because we want to do interesting things. (broad answer)
      
  • each register has a specific name
    
  • there are temporary registers and save registers
    
    • 7 S registers
      
      • Addressed like this: $t0
        
    • 10 T registers
      
      • Addressed like this: $s0
        

Main Memory as an Array of Words

 

• A 'word' is 4 bytes
  
  • A byte is 8 bits
    
• Big Endian is a 32 bit word
  
• The array structure G is imbedded in word memory array at Byte 16(base address)
  
• Word address of G[3] is base address + offset
  
• Base address = 16
  
• Offset = relative word * bytes/word = 3*4=12
  

Address of G[3] = 16+3*4 = 28