Cirrus Logic CS485 Manuel d'utilisateur télécharger pdf (Page 20)

CS485G Spring 2015 20

absDiff:

pushl %ebp # save base pointer

movl %esp,%ebp # new base pointer

movl 8(%ebp), %edx # d = x

movl 12(%ebp), %eax # a = y

cmpl %eax, %edx # x <> y?

jle .L6 # jump if x <= y

subl %eax, %edx # d = x - y

movl %edx, %eax # a = x - y

jmp .L7 # goto exitpoint

.L6: # elsepoint

subl %edx, %eax # a = y - x

.L7: # exitpoint

popl %ebp # restore %ebp

ret # return

11. C can sometimes use a single conditional expression to han-

dle such cases:

1 val = x>y ? x-y : y-x;

21 do while loops

1. C code to count how many 1’s in a parameter

1 int countOnes(unsigned x) {

2 int result = 0;

3 do {

4 result += x & 0x1;

5 x >>= 1;

6 } while (x);

7 return result;

8 }

2. Same thing, represented with goto

1 2 ... 15 16 17 18 19 20 21 22 23 24 25 ... 66 67

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Cirrus Logic CS485 Manuel d'utilisateur Page 20