Good People,
I'm new to assembly, I'm having trouble dividing. When I compile the program at the command line, this error appears Floating point exception .
.section .data
.global op1
.global op2
.global resultado
.section .text
.global Conta
Account:
pushl %ebp
movl %esp, %ebp
movl op1, %edx
movl op2, %eax
divl %edx, %eax
movl %eax, resultado
movl %ebp, %esp
popl %ebp
ret
The instruction div , divides the value (unsigned integer) stored in the pair of registers edx:eax (dividend - 64 bits) by the target operator (can be a register or memory location - 32 bits), and stores the quotient in eax and the rest in edx .
As you are storing the value of op1 in edx , and dividing edx:eax by edx (which is part of the dividend), the result of the division is probably greater than 32 bits and this generates the error Floating point exception (or divide error as Intel's manual ).
edx and use a register other than edx and eax to store the divisor (if the division you are trying to do is 32 bits even and not 64 ).
Here is an example of your changed code, using the ebx register to store the divisor:
Conta:
push %ebp
movl %esp, %ebp
movl op2, %ebx # armazena o divisor em ebx
xorl %edx, %edx # zera edx (parte alta do dividendo)
movl op1, %eax # armazena o dividendo (parte baixa) em eax
divl %ebx, %eax # divide edx:eax por ebx
movl %eax, resultado
movl %ebp, %esp
pop %ebp
ret
Example run: 13-by-3 split:
Conta () at teste.a:30
30 movl op2, %ebx
(gdb)
31 xorl %edx, %edx
(gdb)
32 movl op1, %eax
(gdb)
33 divl %ebx, %eax
(gdb) info registers
eax 0xd 13 <----- parte "baixa" do dividendo
ecx 0x0 0
edx 0x0 0 <----- parte "alta" do dividendo
ebx 0x3 3 <----- divisor
...
(gdb) s
34 movl %eax, resultado
(gdb) info registers
eax 0x4 4 <----- quociente
ecx 0x0 0
edx 0x1 1 <----- resto
ebx 0x3 3
...