内存中有10个分布在0至100内的正整数， 求小于60的数的个数num1，大于或等于60且小于80的数的个数num2，大于或等于80且小于100的数的个数num3
C语言描述该程序流程：
#include <stdio.h> int main() { int a[]={1, 20, 95, 32, 60, 70, 87, 56, 93, 99}; int num1=0,num2=0,num3=0; for(int i=0;i<10;i++) { if(a[i]<60) { num1++; } else if(a[i]<80) { num2++; } else { num3++; } } printf("%d\n%d\n%d",num1,num2,num3); }
汇编语言：
include irvine32.inc .data numbers dword 1, 20, 95, 32, 60, 70, 87, 56, 93, 99 num1 dword 0 num2 dword 0 num3 dword 0 count dword 10 ;需比较十次 .code main proc mov esi,offset numbers L1: cmp count,0 jz output mov ebx,[esi] cmp ebx,60 jl smallerthan60 cmp ebx,80 jl smallerthan80 inc num3 jmp nextnum smallerthan60: inc num1 jmp nextnum smallerthan80: inc num2 jmp nextnum nextnum: dec count add esi,4 jmp L1 output: mov eax,num1 call writeint call crlf mov eax,num2 call writeint call crlf mov eax,num3 call writeint main endp end main
运行结果：
