C语言描述该程序流程：

#include <stdio.h>
int main()
{
	int arr1[]={11,33,23,542,12233,5443,267,456,234,453};
	int arr2[]={21,123,432,45,234,534,6517,678,879,1};
	int max=arr1[0];
	for(int i=1;i<10;i++)
	{
		if(arr1[i]>max)
		{
			max=arr1[i];
		}
	}
	printf("%d\n",max);
	max=arr2[0];
	for(int i=1;i<10;i++)
	{
		if(arr2[i]>max)
		{
			max=arr2[i];
		}
	}
	printf("%d",max);
	
}

汇编语言：

include irvine32.inc
.data
arr1 dword 11,33,23,542,12233,5443,267,456,234,453
arr2 dword 21,123,432,45,234,534,6517,678,879,1
num dword 9 ;需对比9次
.code
main proc
mov esi,offset arr1
mov eax,[esi] ;设第一个元素为最大值

findmax1:
cmp num,0 ;如果num为0，输出并开始找第二个数组的最大值
jz func
mov edx,[esi+4]
cmp eax,edx
jge nextnum1 ;如果该数不比eax大，就继续往下找
mov eax,edx

nextnum1:
add esi,4
dec num
jmp findmax1

func:
call writeint
mov esi,offset arr2
mov eax,[esi]
mov num,9

findmax2:
cmp num,0
jz output
mov edx,[esi+4]
cmp eax,edx
jge nextnum2
mov eax,edx

nextnum2:
add esi,4
dec num
jmp findmax2

output:
call writeint
main endp
end main

运行结果：