Control Structures and I/O in SAL

Tom Kelliher, CS 220

Sept. 17, 1997

Announcements

Advice on using HLL and converting to SAL.

Assignment

Friday will be a lab session to write and run a SAL program. Here's the description:

Write a SAL program to read an integer, n, then read n integers and output the sum of the positive integers read.

Control Structures

1. Unconditional branch: b target

2. Conditional branches with implicit comparison to 0: bltz, bgtz, blez, bgez, beqz, bnez. Example: beqz divisor, error

3. Conditional branches which compare two operands: blt, bgt, ble, bge, beq, bne. Example: beq i, j, endif

Examples

1. if (i >= 0)
      j = 1;
   else
      j = -1;
   

2. if (i <= -2 || i >= 2)
      i *= i;
   else if (i == 0)
      i = i / j;
   else
      j = i;
   

3. for (i = 1, sum = 0; i < 1025; i *= 2)
      sum += i;
   

4. while (i < 1000)
   {
      if (i % j == 0)
         break;
   
      i += k;
   }
   

I/O

1. get input --- Input operation. input can be a .word, .float, or a .byte. Entire line read for the former two.

2. put output --- Output operation. output can be a .word, .float, or a .byte. No newline added. (So, how do we get a newline?)

3. puts string --- Output a string.