divw (Divide Word) Instruction

Assembler Language Reference

divw (Divide Word) Instruction

Purpose

Divides the contents of a general-purpose register by the contents of another general-purpose register and stores the result in a third general-purpose register.

Note: The divw instruction is supported only in the PowerPC architecture.

Syntax

PowerPC
divw	RT,RA,RB
divw.	RT,RA,RB
divwo	RT,RA,RB
divwo.	RT,RA,RB

Description

The divw instruction divides the contents of general-purpose register (GPR) RA by the contents of GPR RB, and stores the result in the target GPR RT. The dividend, divisor, and quotient are interpreted as signed integers.

For the case of -2**31 / -1, and all other cases that cause overflow, the content of GPR RT is undefined.

The divw instruction has four syntax forms. Each syntax form has a different effect on Condition Register Field 0 and the Fixed-Point Exception Register.

Syntax Form	Overflow Exception (OE)	Fixed-Point Exception Register	Record Bit (Rc)	Condition Register Field 0
divw	0	None	0	None
divw.	0	None	1	LT,GT,EQ,SO
divwo	1	SO, OV	0	None
divwo.	1	SO, OV	1	LT,GT,EQ,SO

The four syntax forms of the divw instruction never affect the Carry bit (CA) in the Fixed-Point Exception Register. If the syntax form sets the Overflow Exception (OE) bit to 1, the instruction affects the Summary Overflow (SO) and Overflow (OV) bits in the Fixed-Point Exception Register. If the syntax form sets the Record (Rc) bit to 1, the instruction affects the Less Than (LT) zero, Greater Than (GT) zero, Equal To (EQ) zero, and Summary Overflow (SO) bits in Condition Register Field 0.

Parameters

RT	Specifies target general-purpose register where result of operation is stored.
RA	Specifies source general-purpose register for dividend.
RB	Specifies source general-purpose register for divisor.

Examples

The following code divides the contents of GPR 4 by the contents of GPR 6 and stores the result in GPR 4:

# Assume GPR 4 contains 0x0000 0000.
# Assume GPR 6 contains 0x0000 0002.
divw 4,4,6
# GPR 4 now contains 0x0000 0000.

The following code divides the contents of GPR 4 by the contents of GPR 6, stores the result in GPR 4 and sets Condition Register Field 0 to reflect the result of the operation:
```
# Assume GPR 4 contains 0x0000 0002.
# Assume GPR 6 contains 0x0000 0002.
divw. 4,4,6
# GPR 4 now contains 0x0000 0001.
```
The following code divides the contents of GPR 4 by the contents of GPR 6, places the result in GPR 4, and sets the Summary Overflow and Overflow bits in the Fixed-Point Exception Register to reflect the result of the operation:
```
# Assume GPR 4 contains 0x0000 0001.
# Assume GPR 6 contains 0x0000 0000.
divwo 4,4,6
# GPR 4 now contains an undefined quantity.
```
The following code divides the contents of GPR 4 by the contents of GPR 6, places the result in GPR 4, and sets the Summary Overflow and Overflow bits in the Fixed-Point Exception Register and Condition Register Field 0 to reflect the result of the operation:
```
# Assume GPR 4 contains 0x8000 0000.
# Assume GPR 6 contains 0xFFFF FFFF.
divwo. 4,4,6
# GPR 4 now contains undefined quantity.
```

Related Information

Fixed-Point Processor .

Fixed-Point Arithmetic Instructions .