【发布时间】：2009-12-22 10:53:48
【问题描述】：
我是汇编语言的新手，我必须实现一个函数，在我的例子中是 sin(x)，它可以从 C 源文件中调用。 我必须制作 2 个单独的文件：*.c 和 *.s 在 ubuntu 上通过 gcc 编译时一切正常，但是当程序执行时它给我错误“分段错误”...
要编译，我输入：
gcc -c -o sinc.o sinx.c -g3
gcc -c -o sins.o sinx.s -g3
gcc -o sinx sinc.o sins.o -g3
当我启动程序时：
./sinx
打印出来：
.......insert x:
我把 x 放进去，然后：
segmentation fault
然后它停止了。 这是两个文件：
/*-----------------------------------Sin[x]----------------------------------------*/
extern float Sin(float x); //extern assembly function
#include <stdio.h> //necessary libraries
int main() // main function
{
float x; //allocate variable 'x' as a float
float sine; //allocate variable 'sine' as a float
printf("Calculate Sin[x], with 'x' in radians\n"); //purpose of the program
printf("Insert 'x': "); //user prompt
scanf("%f",&x); //get input value
sine=Sin(x); //calculate sin(x)
printf("Sin[%f]=%f\n",x,sine); //print sin(x)
return 0; //successful exit
}
/*----------------------------------------------------------------------------------*/
和
#---------------------------Sin[x]-----------------------------#
.data
.text
.globl Sin #global function visible by main function
#function sumplus
sumplus:
pushl %ebp
movl %esp,%ebp
fld1 #load 1 in ST(0)
movl 12(%ebp),%ecx #ecx=i
power:
fmul %st(0),%st(1) #ST(0)=ST(0)*ST(1),ST(1)=x
loop power #at the end ST(0)=x^i
movl 12(%ebp),%ecx #ecx=i
xorl %edx,%edx #reset edx used in mul
movl $1,%eax #set accumulator
factorial:
mul %ecx #eax=eax*ecx
loop factorial #at the end eax=i!
pushl %eax
fild -4(%ebp) #ST(0)=i!,ST(1)=x^i,ST(2)=x
popl %eax
fdiv %st(1),%st(0) #ST(1)=ST(1)/ST(0)=(x^i)/i!
fld 8(%ebp) #load partial result in ST(0)
fadd %st(0),%st(2) #ST(0)=updated partial result
fstp 8(%ebp) #store partial result into the stack
fcomip %st(0),%st(0) #pop i!
fcomip %st(0),%st(0) #pop x^i/i!,ST(0)=x
leave
ret
#function summinus
summinus:
pushl %ebp
movl %esp,%ebp
fld1 #load 1 in ST(0)
movl 12(%ebp),%ecx #ecx=i
power2:
fmul %st(0),%st(1) #ST(0)=ST(0)*ST(1),ST(1)=x
loop power2 #at the end ST(0)=x^i
movl 12(%ebp),%ecx #ecx=i
xorl %edx,%edx #reset edx used in mul
movl $1,%eax #set accumulator
factorial2:
mul %ecx #eax=eax*ecx
loop factorial2 #at the end eax=i!
pushl %eax
fild -4(%ebp) #ST(0)=i!,ST(1)=x^i,ST(2)=x
popl %eax
fdiv %st(1),%st(0) #ST(1)=ST(1)/ST(0)=(x^i)/i!
fld 8(%ebp) #load partial result in ST(0)
fsub %st(0),%st(2) #ST(0)=updated partial result
fstp 8(%ebp) #store partial result into the stack
fcomip %st(0),%st(0) #pop i!
fcomip %st(0),%st(0) #pop x^i/i!,ST(0)=x
leave
ret
#function develop
develop:
pushl %ebp
movl %esp,%ebp
sub $8,%esp #allocate room for local variables
movl $9,-4(%ebp) #store development-order,only odd values
fldz #load 0.0 in ST(0)
fstp -8(%ebp) #store ST(0) and pop to collect results
Cycle:
movl -4(%ebp),%eax #eax=i,development-order
xorl %edx,%edx #reset edx because of div
sub $1,%eax #i-1
movl $2,%ecx #divisor
div %ecx #eax=(i-1)/2,edx=0
div %ecx #eax=((i-1)/2)/2,remainder edx=0 or edx!=0
movl %edx,%ecx #ecx=remainder
jecxz Sumplus #n even,(-1)^n=+1
Summinus:
call summinus #n odd,(-1)^n=-1
jmp Restore #if sum- occured skip sum+
Sumplus:
call sumplus
Restore:
movl -4(%ebp),%ecx #restore counter
sub $2,-4(%ebp) #update order
loop Cycle #decrement ecx
fcomip %st(0),%st(0) #pop x
fld -8(%ebp) #load final result in ST(0)
leave
ret
#function sin
Sin:
pushl %ebp
movl %esp,%ebp
fld 8(%ebp) #push onto the stack 'x' value
fldpi #load pi into ST(0),x in ST(1)
ControlPi:
fcomi %st(1) #compare pi and x
jae LoadNPi #stop ControlPi, x is less than pi
fsub %st(1),%st(0) #store (x-pi) in ST(1)
jmp ControlPi #return to control
LoadNPi:
fimul -1 #(-pi) in ST(0),x in ST(1)
ControlPi2:
fcomi %st(1) #compare -pi and x
jbe PopStack #stop ControlPi2, x is greater than -pi
fadd %st(1),%st(0) #store (x+pi) in ST(1)
jmp ControlPi2 #return to control
PopStack:
fcomip %st,%st(0) #compare -pi to -pi then pop register stack
call develop #call develop function
leave #restore ebp
ret #return
那么，错误在哪里？我怎么解决这个问题？ 谢谢。
【问题讨论】：
标签： c gcc assembly x86 segmentation-fault