1 / 12

Branching & Shifting

Branching & Shifting. CS 235 Computer Organization & Assembly Language. Two’s Complement Branching Conditions. When the numbers are signed (two’s complement), the branching conditions introduced earlier are appropriate. bl branch < 0 (N xor V) = 1

Download Presentation

Branching & Shifting

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Branching & Shifting CS 235 Computer Organization & Assembly Language

  2. Two’s ComplementBranching Conditions When the numbers are signed (two’s complement), the branching conditions introduced earlier are appropriate. bl branch < 0 (N xor V) = 1 ble branch <= 0 Z or (N xor V)=1 be branch == 0 Z = 1 bne branch != 0 Z = 0 bge branch >= 0 (N xor V) = 0 bg branch > 0 Z or (N xor V)=0

  3. How are Conditions Set? Signed numbers • Z is set when all bits of the result are 0 • N is set when the msb is 1 • V is set • 1) when the register is not long enough to hold the result • 2) on subtraction, d = m – s, when the sign of d is the same as the sign of s and the sign of m is different from s For example, most_negative – any_positive_number (-8) – (2) = 10, not representable in 4 bits

  4. Conditions with Unsigned Arithmetic • No N flag!!! • Condition codes set differently • On addition, a carry out of the msb indicates overflow and the carry bit C is set • On subtraction. We imagine an extra bit to the left of the number. By forming the two’s complement of the subtrahend, and adding, we would expect a carry out of the most significant bit to be added to the imaginary bit to indicate a positive result. So the C bit is set if there is no carry out of the msb. • Result: the test for unsigned overflow is a test of the carry bit C.

  5. 4 Bit Examples 1) 12 1100 09 1001 21 10101 On addition, the carry bit set indicates overflow.

  6. 4 Bit Examples 2) 12 1100 - 09 0111 = (10000 –1 - 1001) +1 03 10011 and the carry indicates a correct result

  7. 4 Bit Examples 3) 09 1001 - 12 0100 = (10000 –1 - 1100) +1 21 01101 The carry bit is not set, so the result is incorrect

  8. Unsigned Branching Conditions When the numbers are unsigned, the branching conditions are different. blu branch < 0 C = 1 bleu branch <= 0 Z or C =1 be branch == 0 Z = 1 bne branch != 0 Z = 0 bgeu branch >= 0 C = 0 bgu branch > 0 Z and C =0

  9. Condition Code Testsbased directly upon condition codes bneg branch on negative N = 1 bpos branch on positive N = 0 bz branch on zero Z = 1 bnz branch on not zero Z = 0 bvs branch overflow set V = 1 bvc branch no overflow V = 0 bcs branch carry set C = 1 bcc branch carry clear C = 0

  10. Shifting • Three shifts • Shift right logical srl rs1, rs2_or_im, rdest • Shift right arithmetic sra rs1, rs2_or_im, rdest • Shift left logical sll rs1, rs2_or_im, rdest The rs1 is shifted the numbers of bits in the second operand and the result is stored in the destination

  11. Shift is a Multiply by 2 • A fast multiply • Used instead of .mul • Consider %o0 * 11 1110 = 10112 = 8 + 2 + 1 sll %o0, 3, %o1 ! 8*x sll %o0, 1, %o2 ! 2*x add %o0, %o1, %o3 ! x + 8*x add %o2, %o3, %o3 ! 8*x + 2*x + x ! = 11*x

  12. SPARC Boolean Opcodes Recall Boolean Operations (02:38) andcc regs1, reg_or_imm, regdest andncc regs1, reg_or_imm, regdest xorcc regs1, reg_or_imm, regdest orcc regs1, reg_or_imm, regdest xnorcc regs1, reg_or_imm, regdest orncc regs1, reg_or_imm, regdest

More Related