Cirrus Logic CS485 Manual de usuario descargar pdf (Pagina 16)

CS485G Spring 2015 16

1 int arith(int x, int y, int z)

2 {

3 int t1 = x+y;

4 int t2 = z+t1;

5 int t3 = x+4;

6 int t4 = y

48;

7 int t5 = t3 + t4;

8 int rval = t2

t5;

9 return rval;

10 }

7. Result of compilation

pushl %ebp # save base pointer

movl %esp,%ebp # new base pointer

movl 8(%ebp), %ecx # $c = x

movl 12(%ebp), %edx # $d = y

leal (%edx,%edx,2), %eax # $a = 3y

sall $4, %eax # $a = 48y [t4]

leal 4(%ecx,%eax), %eax # $a = x + 48y + 4 [t5]

addl %ecx, %edx # $d = x + y [t1]

addl 16(%ebp), %edx # $d = x + y + z [t2]

imull %edx, %eax # $a = (x+y+z)

(x+48y+4)

popl %ebp # restore base pointer

ret # return $a [rval]

8. Optimization converts multiple expressions to a single statement,

and a single expression might require multiple instructions.

9. The compiler generates the same code for (x+y+z)

(x+4+48

y).

10. Example

1 int logical(int x, int y)

2 {

3 int t1 = xˆy;

4 int t2 = t1 >> 17;

5 int mask = (1<<13) - 7; // 8185

6 int rval = t2 & mask;

7 return rval;

8 }

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Cirrus Logic CS485 Manual de usuario Pagina 16