lib

《编译方法》课程设计内容２．《编译方法》课程设计内容

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\      End of run routine
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L1:
       HALT
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\      Print a number recursively
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L2:
       STI  R3,4(R1)			 \ Save the return address
       LDI  8(R1),R15			 \ Copy arg to R15
       LDA  10(R0),R5			 \ Constant 10
       DIV  R15,R5                       \ Non-zero if more than one digit
       BZE  L3				 \ Branch if only 1 digit
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       STI  R5,12(R1)			 \ Save on stack for later
       STI  R5,24(R1)			 \ Copy to new stack frame
       LDA  L2,R2			 \ Address to call
       STI  R1,16(R1)			 \ Save stack pointer
       LDA  16(R1),R1			 \ Increment stack pointer
       BAL  R2,R3			 \ Call recursively
       LDA  10(R0),R15			 \ Constant 10 in R15
       LDI  8(R1),R5			 \ Original number
       LDI  12(R1),R6			 \ Divided number
       MUL  R6,R15			 \ Multiply back by 10 in R15
       SUB  R5,R15			 \ Subtract from original for digit
L3:
       LDA  48(R0),R5			 \ ASCII '0' in R5
       ADD  R5,R15			 \ Add to get desired digit in R15
       TRAP				 \ Print the digit
       LDI  4(R1),R2			 \ Return address in R2
       LDI  0(R1),R1			 \ Restore stack pointer
       BAL  R2,R3			 \ Return
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\      Print a string
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L4:
       STI  R3,4(R1)			 \ Save the return address
       LDI  8(R1),R5			 \ String pointer in R5
       LDA  16777216(R0),R7		 \ 256 cubed for shifting 24 bits right
L5:
       LDI  0(R5),R6			 \ Next char in top byte  of R6
       LDR  R7,R15			 \ Constant 256 cubed
       DIV  R6,R15			 \ Char in R15
       BZE  L6				 \ EOS branch to exit
       TRAP				 \ Print the char
       LDA  1(R5),R5			 \ Step to next char
       BRA  L5				 \ Round again
L6:
       LDI  4(R1),R2			 \ Return address in R2
       LDI  0(R1),R1			 \ Restore stack pointer
       BAL  R2,R3			 \ Return
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