[英]Recursively check parenthesis balance of an equation in C
I am trying to write a program in C that recursively checks the balance of an equations parenthesis. 我试图用C编写一个程序,该程序递归检查方程式括号的平衡。 Ie for every open there is a closing.
即,每次打开都有一个关闭。
This is what I currently have, but I can't get it to ignore characters and spaces. 这是我目前拥有的,但是我无法忽略字符和空格。
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#define MAX_SIZE 100
struct Stack{
int top;
char arr[MAX_SIZE];
} st;
void init(){
st.top = -1;
}
bool isEmpty(){
if(st.top == -1){
return true;
}else{
return false;
}
}
bool isFull(){
if(st.top == MAX_SIZE-1){
return true;
}else{
return false;
}
}
void push(char item){
if(isFull()){
printf("Stack is full");
exit(0);
}else{
st.top++;
st.arr[st.top] = item;
}
}
void pop(){
if(isEmpty()){
printf("Stack is empty");
exit(0);
}else{
st.top--;
}
}
char gettop(){
return st.arr[st.top];
}
bool ArePair(char opening,char closing)
{
if(opening == '(' && closing == ')') return true;
else if(opening == '{' && closing == '}') return true;
else if(opening == '[' && closing == ']') return true;
return false;
}
void main()
{
char in_expr[MAX_SIZE],a,temp;
int length=0,i,j,count;
init();
printf("Enter an expression to check:");
scanf("%s", in_expr);
length = strlen(in_expr);
for(i=0;i<length;i++){
if (in_expr[i] != '(' && in_expr[i] != ')' && in_expr[i] != '{' && in_expr[i] != '}' && in_expr[i] != '[' && in_expr[i] != ']') {
i++;
}
else if(in_expr[i] == '(' || in_expr[i] == '{' || in_expr[i] == '['){
push(in_expr[i]);
a = in_expr[i];
printf("%c",a);
}
else if(in_expr[i] == ')' || in_expr[i] == '}' || in_expr[i] == ']'){
// a = st.arr[st.top];
a = in_expr[i];
printf("%c", a);
}
if(isEmpty() || !ArePair(gettop(),in_expr[i])){
printf("\nInvalid expression - Not Balanced!\n");
exit(0);
}
else{
pop();
}
i++;
// }
}
if(isEmpty()){
printf("\nValid expression - Perfectly Balanced!\n");
}else{
printf("\nInvalid expression - Not Balanced!\n");
}
}
Any help is greatly appreciated and if you need more details, don't hesitate to ask! 非常感谢您的帮助,如果您需要更多详细信息,请随时询问! If this is a duplicate I apologize, I am still searching for another solution through other posts here.
如果这很抱歉,我仍在通过此处的其他帖子搜索其他解决方案。
A few bugs: 一些错误:
Don't do i++
except in the for
clause. 除了
for
子句,不要执行i++
。 You're [potentially] skipping past chars that you want to see. 您[潜在地]跳过了想要查看的字符。
Also, don't check for isEmpty
or ArePair
unless the current char is a closer. 另外, 除非当前字符更近,否则不要检查
isEmpty
或ArePair
。
Here's a corrected version [please pardon the gratuitous style cleanup]. 这是一个更正的版本[请原谅免费的样式清理]。 Note that there may be other bugs, but I think this will get you closer:
请注意,可能还有其他错误,但是我认为这将使您更加接近:
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#define MAX_SIZE 100
struct Stack {
int top;
char arr[MAX_SIZE];
} st;
void
init()
{
st.top = -1;
}
bool
isEmpty()
{
if (st.top == -1) {
return true;
}
else {
return false;
}
}
bool
isFull()
{
if (st.top == MAX_SIZE - 1) {
return true;
}
else {
return false;
}
}
void
push(char item)
{
if (isFull()) {
printf("Stack is full");
exit(0);
}
else {
st.top++;
st.arr[st.top] = item;
}
}
void
pop()
{
if (isEmpty()) {
printf("Stack is empty");
exit(0);
}
else {
st.top--;
}
}
char
gettop()
{
return st.arr[st.top];
}
bool
ArePair(char opening, char closing)
{
if (opening == '(' && closing == ')')
return true;
else if (opening == '{' && closing == '}')
return true;
else if (opening == '[' && closing == ']')
return true;
return false;
}
void
main()
{
char in_expr[MAX_SIZE],
a,
temp;
int length = 0,
i,
j,
count;
init();
#if 0
printf("Enter an expression to check:");
scanf("%s", in_expr);
#else
strcpy(in_expr,"( ( a + b ) * (c + d) ) * ( ( w + x ) * (y + z) )");
//strcpy(in_expr,"( ( a + b ) * (c + d) ) * ( ( w + x ) * y + z) )");
#endif
length = strlen(in_expr);
for (i = 0; i < length; i++) {
int chr = in_expr[i];
printf("input: '%c'\n",chr);
if (chr != '(' && chr != ')' && chr != '{' && chr != '}' && chr != '[' && chr != ']') {
//i++;
continue;
}
if (chr == '(' || chr == '{' || chr == '[') {
push(chr);
a = chr;
printf("%c", a);
continue;
}
if (chr == ')' || chr == '}' || chr == ']') {
a = chr;
printf("%c", a);
}
if (isEmpty() || !ArePair(gettop(), chr)) {
printf("\nInvalid expression - Not Balanced!\n");
exit(0);
}
else {
pop();
}
//i++;
}
if (isEmpty()) {
printf("\nValid expression - Perfectly Balanced!\n");
}
else {
printf("\nInvalid expression - Not Balanced!\n");
}
}
Here's a version that uses a switch
statement that may be easier to read: 这是使用
switch
语句的版本,可能更易于阅读:
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#define MAX_SIZE 100
struct Stack {
int top;
char arr[MAX_SIZE];
} st;
void
init()
{
st.top = -1;
}
bool
isEmpty()
{
if (st.top == -1) {
return true;
}
else {
return false;
}
}
bool
isFull()
{
if (st.top == MAX_SIZE - 1) {
return true;
}
else {
return false;
}
}
void
push(char item)
{
if (isFull()) {
printf("Stack is full");
exit(0);
}
else {
st.top++;
st.arr[st.top] = item;
}
}
void
pop()
{
if (isEmpty()) {
printf("Stack is empty");
exit(0);
}
else {
st.top--;
}
}
char
gettop()
{
return st.arr[st.top];
}
bool
ArePair(char opening, char closing)
{
if (opening == '(' && closing == ')')
return true;
else if (opening == '{' && closing == '}')
return true;
else if (opening == '[' && closing == ']')
return true;
return false;
}
void
main()
{
char in_expr[MAX_SIZE],
a,
temp;
int length = 0,
i,
j,
count;
init();
#if 0
printf("Enter an expression to check:");
scanf("%s", in_expr);
#else
strcpy(in_expr,"( ( a + b ) * (c + d) ) * ( ( w + x ) * (y + z) )");
//strcpy(in_expr,"( ( a + b ) * (c + d) ) * ( ( w + x ) * y + z) )");
#endif
length = strlen(in_expr);
for (i = 0; i < length; i++) {
int chr = in_expr[i];
printf("input: '%c'\n",chr);
switch (chr) {
case '(':
case '{':
case '[':
push(chr);
a = chr;
printf("%c", a);
break;
}
case ')':
case '}':
case ']':
a = chr;
printf("%c", a);
if (isEmpty() || !ArePair(gettop(), chr)) {
printf("\nInvalid expression - Not Balanced!\n");
exit(0);
}
pop();
break;
}
}
if (isEmpty()) {
printf("\nValid expression - Perfectly Balanced!\n");
}
else {
printf("\nInvalid expression - Not Balanced!\n");
}
}
Here's a version that adds a self test function that allows multiple unit tests. 这是一个添加了自测功能的版本,该功能允许进行多个单元测试。 It's what I've done when I've written expression parsers in the past.
这是我过去编写表达式解析器时所做的事情。
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#define MAX_SIZE 100
struct Stack {
int top;
char arr[MAX_SIZE];
} st;
void
init()
{
st.top = -1;
}
bool
isEmpty()
{
if (st.top == -1) {
return true;
}
else {
return false;
}
}
bool
isFull()
{
if (st.top == MAX_SIZE - 1) {
return true;
}
else {
return false;
}
}
void
push(char item)
{
if (isFull()) {
printf("Stack is full");
exit(0);
}
else {
st.top++;
st.arr[st.top] = item;
}
}
void
pop()
{
if (isEmpty()) {
printf("Stack is empty");
exit(0);
}
else {
st.top--;
}
}
char
gettop()
{
return st.arr[st.top];
}
bool
ArePair(char opening, char closing)
{
if (opening == '(' && closing == ')')
return true;
else if (opening == '{' && closing == '}')
return true;
else if (opening == '[' && closing == ']')
return true;
return false;
}
// RETURNS: 1=balanced, 0=unbalanced
int
test_expr(const char *in_expr)
{
int i;
int length;
//printf("\ntest_expr: '%s'\n",in_expr);
init();
length = strlen(in_expr);
for (i = 0; i < length; i++) {
int chr = in_expr[i];
//printf("input: '%c'\n",chr);
switch (chr) {
case '(':
case '{':
case '[':
push(chr);
//printf("%c", chr);
break;
case ')':
case '}':
case ']':
//printf("%c", chr);
if (isEmpty() || !ArePair(gettop(), chr)) {
printf("Invalid expression - Not Balanced!\n");
return 0;
exit(0);
}
pop();
break;
}
}
if (isEmpty()) {
printf("Valid expression - Perfectly Balanced!\n");
return 1;
}
else {
printf("Invalid expression - Not Balanced!\n");
return 0;
}
}
void
testone(int expected,const char *expr)
{
int isvalid;
printf("\ntestone: '%s'\n",expr);
isvalid = test_expr(expr);
printf("%s\n", (isvalid == expected) ? "PASS" : "FAIL");
}
void
testall(void)
{
testone(1,"( ( a + b ) * (c + d) ) * ( ( w + x ) * (y + z) )");
testone(0," ( a + b ) * (c + d) ) * ( ( w + x ) * (y + z) )");
testone(0,"( ( a + b ) * (c + d) ) * ( ( w + x ) * y + z) )");
testone(0,"( ( a + b ) * (c + d) ) * ( ( w + x ) * (y + z) ");
testone(0,"( ( a + b ) * (c + d) ) * ( ( w + x ) * (y + z )");
testone(0,"( ( a + b ) * (c + d) ] * ( ( w + x ) * (y + z) )");
}
int
main(void)
{
#if 0
char in_expr[MAX_SIZE];
printf("Enter an expression to check:");
scanf("%s", in_expr);
test_expr(in_expr);
#else
testall();
#endif
return 0;
}
You can take a simpler approach with an incremental state that stores the pending blocks, for example a char
array with the opening characters. 您可以采用具有存储待处理块的增量状态的更简单方法,例如,带有开头字符的
char
数组。
Here is the code: 这是代码:
#include <stdio.h>
#define MAX_EXPR 256
#define MAX_LEVEL 32
enum { EXPR_BALANCED, EXPR_TOO_DEEP, EXPR_PAREN_MISMATCH, EXPR_UNCLOSED_PAREN };
int check_balance(const char *expr) {
char state[MAX_LEVEL];
const char *p;
int level = 0;
state[level++] = 0; // force mismatch on state[0]
for (p = expr; *p; p++) {
switch (*p) {
case '(':
case '{':
case '[':
if (level >= MAX_LEVEL)
return EXPR_TOO_DEEP;
state[level++] = *p;
break;
case ')':
if (state[--level] != '(')
return EXPR_PAREN_MISMATCH;
break;
case '}':
if (state[--level] != '{')
return EXPR_PAREN_MISMATCH;
break;
case ']':
if (state[--level] != '[')
return EXPR_PAREN_MISMATCH;
break;
}
}
if (level != 1)
return EXPR_UNCLOSED_PAREN;
return EXPR_BALANCED;
}
int main() {
char expr[MAX_EXPR];
printf("Enter expressions to check, one per line: ");
while (fgets(expr, sizeof expr, stdin) {
switch (check_balance(expr)) {
case EXPR_TOO_DEEP:
printf("too many nesting levels\n");
break;
case EXPR_PAREN_MISMATCH:
printf("closing character mismatch\n");
break;
case EXPR_UNCLOSED_PAREN:
printf("closing character missing\n");
break;
case EXPR_BALANCED:
printf("expression is balanced\n");
break;
}
}
return 0;
}
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