如何使用emu8086检查平衡括号
How to check balanced parentheses using emu8086
为了检查括号是否平衡，我用 c++ 语言编写了这个代码，但是我如何将这个代码转换成汇编 emu8086 程序。
#include <iostream> //main header file
#include <stack>
using namespace std;
void balance_parentheses();
int main()
{
int t;
cout << "Enter number of test cases:";
cin >> t;
for (int i = 0; i < t; i++) {
//calling of function for checking of brackets
balance_parentheses();
}
return 0;
}
void balance_parentheses()
{
stack<char> a;
string s;
cout << "Enter string may or may not containing parentheses:";
cin >> s;
int flag = 0; //flag is an arbitrary variable
for (int i = 0; i < s.length(); i++)
//for length of the string calculated by number of letters
{
if (s[i] == '{' || s[i] == '[' || s[i] == '(') {
//push function to enter terms in a stack
a.push(s[i]);
flag = 1;
}
if (!a.empty()) {
if (s[i] == '}') {
if (a.top() == '{')
// top of the stack
{
a.pop();
//pop function to delete terms from the top of array
continue;
}
else
break;
}
if (s[i] == ']') {
if (a.top() == '[') {
a.pop();
continue;
}
else
break;
}
if (s[i] == ')') {
if (a.top() == '(') {
a.pop();
continue;
}
else
break;
}
}
else {
break;
}
}
if ((a.empty()) && (flag == 1))
cout << "Balaned" << endl;
else
cout << "Not Balanced" << endl;
}
为简单起见，测试输入长度为 8 个固定值，在恰好取 8 个字符后，打印输出。
用于检查输入（示例如下）：
[[]({})] ------ 平衡
{({}[])} ------ 平衡
[]({)}[] ------ 不平衡
}()[(]){ ------ 不平衡
这里首先，@表示label开始，$表示procedure开始。如果汇编代码中存在多个标签和过程。然后可以轻松完成区分标签和程序。这只是我自己的编码方式。你可以忽略这个样式。
.model small ; declaring this code will be consists of one data segment and one code segment
.stack 100h ; stack is initializeed to offset address at 100h
.data ; Data segment
n_line db 0ah,0dh,"$" ; for new line
msg1 db 10,13,"Balanced$"
msg2 db 10,13,"Not Balanced$"
a1 dw 5bh ; '['
a2 dw 5dh ; ']'
b1 dw 7bh ; '{'
b2 dw 7dh ; '}'
c1 dw 28h ; '('
c2 dw 29h ; ')'
flag db 0
i db ?
.code ; Code segment
main proc
mov ax,@data ; copying starting address of data segment into ax register
mov ds,ax ; by copying ax into ds we are initializing data segment
mov cx,8 ; length is 8 fixed here
mov i,0d ;
xor ax,ax ; clearing ax
xor bx,bx ; clearing bx
xor dx,dx ; clearing dx
@input:
cmp i,8d
jge @input_end
mov ah,1 ; taking input
int 21h
mov ah,0 ; clearing previously stored 1 in ah
mov bp,sp ; checking stack is empty or not using bp. it can be done sp in this code
@if1: ; in this @if1 level: checking if current input is opening bracket or not
cmp ax,a1
je @push_it
cmp ax,b1
je @push_it
cmp ax,c1
je @push_it
jmp @if2
@push_it: ; if current input is opening bracket then push it
push ax
mov flag,1 ; and alter the default flag value
@if2:
cmp bp,100h ; checking if stack is empty or not
je @if2_end ; if not empty then
@inner_if1: ; in this @inner_if1 level, checking if top of the stack is co-responding closing bracket of '{' or not
cmp ax,b2 ; top==}
je @inner_if1_perform
jne @inner_if2
@inner_if1_perform:
pop bx ; top is popped and storing it to bx
cmp bx,b1
jne @inner_push1
je @inner_if2
@inner_push1:
push bx ; if not matched, then that popped value should be pushed
@inner_if2: ; in this @inner_if2 level, checking if top of the stack is co-responding closing bracket of '[' or not
cmp ax,a2 ; top==]
je @inner_if2_perform
jne @inner_if3
@inner_if2_perform:
pop bx ; top is popped and storing it to bx
cmp bx,a1
jne @inner_push2
je @inner_if3
@inner_push2:
push bx ; if not matched, then that popped value should be pushed
@inner_if3: ; in this @inner_if3 level, checking if top of the stack is co-responding closing bracket of '(' or not
cmp ax,c2 ; top== )
je @inner_if3_perform
jne @inner_if3_end
@inner_if3_perform:
pop bx ; top is popped and storing it to bx
cmp bx,c1
jne @inner_push3
je @inner_if3_end
@inner_push3:
push bx ; if not matched, then that popped value should be pushed
@inner_if3_end:
@if2_end:
inc i
jmp @input
@input_end: ; if ( (flag == 1) && (stack.empty()) )
cmp flag,1
je @next_level:
jne @print_msg2
@next_level:
cmp sp,100h ; cheking the stack pointer is returned to initial address or not means empty or not
je @print_msg1
jne @print_msg2
@print_msg1:
lea dx,msg1
mov ah,9
int 21h
@stop:
mov ah,4ch ; terminate the code
int 21h
main endp ; ending of main procedure
@print_msg2:
lea dx,msg2
mov ah,9
int 21h
jmp @stop
end main ; ending of code segment
; Input:
; [[]({})] ------ Balanced
; {({}[])} ------ Balanced
; []({)}[] ------ Not Balanced
; }()[(]){ ------ Not Balanced
为了检查括号是否平衡，我用 c++ 语言编写了这个代码，但是我如何将这个代码转换成汇编 emu8086 程序。
#include <iostream> //main header file
#include <stack>
using namespace std;
void balance_parentheses();
int main()
{
int t;
cout << "Enter number of test cases:";
cin >> t;
for (int i = 0; i < t; i++) {
//calling of function for checking of brackets
balance_parentheses();
}
return 0;
}
void balance_parentheses()
{
stack<char> a;
string s;
cout << "Enter string may or may not containing parentheses:";
cin >> s;
int flag = 0; //flag is an arbitrary variable
for (int i = 0; i < s.length(); i++)
//for length of the string calculated by number of letters
{
if (s[i] == '{' || s[i] == '[' || s[i] == '(') {
//push function to enter terms in a stack
a.push(s[i]);
flag = 1;
}
if (!a.empty()) {
if (s[i] == '}') {
if (a.top() == '{')
// top of the stack
{
a.pop();
//pop function to delete terms from the top of array
continue;
}
else
break;
}
if (s[i] == ']') {
if (a.top() == '[') {
a.pop();
continue;
}
else
break;
}
if (s[i] == ')') {
if (a.top() == '(') {
a.pop();
continue;
}
else
break;
}
}
else {
break;
}
}
if ((a.empty()) && (flag == 1))
cout << "Balaned" << endl;
else
cout << "Not Balanced" << endl;
}
为简单起见，测试输入长度为 8 个固定值，在恰好取 8 个字符后，打印输出。
用于检查输入（示例如下）：
[[]({})] ------ 平衡
{({}[])} ------ 平衡
[]({)}[] ------ 不平衡
}()[(]){ ------ 不平衡
这里首先，@表示label开始，$表示procedure开始。如果汇编代码中存在多个标签和过程。然后可以轻松完成区分标签和程序。这只是我自己的编码方式。你可以忽略这个样式。
.model small ; declaring this code will be consists of one data segment and one code segment
.stack 100h ; stack is initializeed to offset address at 100h
.data ; Data segment
n_line db 0ah,0dh,"$" ; for new line
msg1 db 10,13,"Balanced$"
msg2 db 10,13,"Not Balanced$"
a1 dw 5bh ; '['
a2 dw 5dh ; ']'
b1 dw 7bh ; '{'
b2 dw 7dh ; '}'
c1 dw 28h ; '('
c2 dw 29h ; ')'
flag db 0
i db ?
.code ; Code segment
main proc
mov ax,@data ; copying starting address of data segment into ax register
mov ds,ax ; by copying ax into ds we are initializing data segment
mov cx,8 ; length is 8 fixed here
mov i,0d ;
xor ax,ax ; clearing ax
xor bx,bx ; clearing bx
xor dx,dx ; clearing dx
@input:
cmp i,8d
jge @input_end
mov ah,1 ; taking input
int 21h
mov ah,0 ; clearing previously stored 1 in ah
mov bp,sp ; checking stack is empty or not using bp. it can be done sp in this code
@if1: ; in this @if1 level: checking if current input is opening bracket or not
cmp ax,a1
je @push_it
cmp ax,b1
je @push_it
cmp ax,c1
je @push_it
jmp @if2
@push_it: ; if current input is opening bracket then push it
push ax
mov flag,1 ; and alter the default flag value
@if2:
cmp bp,100h ; checking if stack is empty or not
je @if2_end ; if not empty then
@inner_if1: ; in this @inner_if1 level, checking if top of the stack is co-responding closing bracket of '{' or not
cmp ax,b2 ; top==}
je @inner_if1_perform
jne @inner_if2
@inner_if1_perform:
pop bx ; top is popped and storing it to bx
cmp bx,b1
jne @inner_push1
je @inner_if2
@inner_push1:
push bx ; if not matched, then that popped value should be pushed
@inner_if2: ; in this @inner_if2 level, checking if top of the stack is co-responding closing bracket of '[' or not
cmp ax,a2 ; top==]
je @inner_if2_perform
jne @inner_if3
@inner_if2_perform:
pop bx ; top is popped and storing it to bx
cmp bx,a1
jne @inner_push2
je @inner_if3
@inner_push2:
push bx ; if not matched, then that popped value should be pushed
@inner_if3: ; in this @inner_if3 level, checking if top of the stack is co-responding closing bracket of '(' or not
cmp ax,c2 ; top== )
je @inner_if3_perform
jne @inner_if3_end
@inner_if3_perform:
pop bx ; top is popped and storing it to bx
cmp bx,c1
jne @inner_push3
je @inner_if3_end
@inner_push3:
push bx ; if not matched, then that popped value should be pushed
@inner_if3_end:
@if2_end:
inc i
jmp @input
@input_end: ; if ( (flag == 1) && (stack.empty()) )
cmp flag,1
je @next_level:
jne @print_msg2
@next_level:
cmp sp,100h ; cheking the stack pointer is returned to initial address or not means empty or not
je @print_msg1
jne @print_msg2
@print_msg1:
lea dx,msg1
mov ah,9
int 21h
@stop:
mov ah,4ch ; terminate the code
int 21h
main endp ; ending of main procedure
@print_msg2:
lea dx,msg2
mov ah,9
int 21h
jmp @stop
end main ; ending of code segment
; Input:
; [[]({})] ------ Balanced
; {({}[])} ------ Balanced
; []({)}[] ------ Not Balanced
; }()[(]){ ------ Not Balanced