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![4. Design, Develop and Implement a Program in C for converting an Infix Expression
to Postfix Expression. Program should support for both parenthesized and free
parenthesized expressions with the operators: +, -, *, /, %(Remainder), ^(Power)
and alphanumeric operands.
#include<stdio.h>
#include<conio.h>
inttop=-1,stack[20],pos=0,length;
charsymbol,temp,infix[20],postfix[20];
void push(char symbol)
{
stack[++top]=symbol;
}
int pop()
{
return(stack[top--]);
}
int preced(char symbol)
{
int p;
switch(symbol)
17](https://image.slidesharecdn.com/program4-240502104105-59d98cef/85/program-4-doc-in-computer-science-and-engineering-1-320.jpg)
![{
case'^':p=3;
break;
case'/':
case'*':
case'%':p=2;
break;
case'+':
case'-':p=1;
break;
case')':
case'(':p=0;
break;
case'#':p=-1;
break;
}
return(p);
}
void ifix_pfix(char infix[],char postfix[])
{
int i=0;
push('#');
while(i<strlen(infix))
{
symbol=infix[i];
switch(symbol)
{
case'(': push(symbol);
break;
case')': temp=pop();
while (temp!='(')
{
}
break;
case'^':
case'+':
case'-':
case'*':
case'/':
postfix[pos++]=temp;
temp=pop();
case'%': while(preced(stack[top])>=preced(symbol))
{
18](https://image.slidesharecdn.com/program4-240502104105-59d98cef/85/program-4-doc-in-computer-science-and-engineering-2-320.jpg)
![postfix[pos++]=pop();
}
push(
symb
ol);
break
;
default:
postfix[pos++]
=symbol;
break;
}
i++;
}
while((temp=pop())!='#')
{
postfix[pos++]=temp;
}
}
void main()
{
clrscr();
printf("Enter the Infix expression---
>"); scanf("%s",infix);
ifix_pfix(infix,postfix);
printf("Postfix expression is--->
%sn",postfix); getch();
}](https://image.slidesharecdn.com/program4-240502104105-59d98cef/85/program-4-doc-in-computer-science-and-engineering-3-320.jpg)
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Please need help on C++ language.Infix to Postfix) Write a program.pdf
Please need help on C++ language.Infix to Postfix) Write a program.pdfpristiegee Please need help on C++ language.
Infix to Postfix) Write a program that converts an infix expression into an equivalent postfix
expression. The rules to convert an infix expression into an equivalent postfix expression are as
follows:
Initialize pfx to an empty expression and also initialize the stack.
Get the next symbol, sym, from infx.
If sym is an operand, append sym to pfx.
If sym is (, push sym into the stack.
If sym is ), pop and append all of the symbols from the stack until the most recent left
parentheses. Pop and discard the left parentheses.
If sym is an operator:
Pop and append all of the operators from the stack to pfx that are above the most recent left
parentheses and have precedence greater than or equal to sym.
Push sym onto the stack.
After processing infx, some operators might be left in the stack. Pop and append to pfx
everything from the stack.
In this program, you will consider the following (binary) arithmetic operators: +, -, *, and /.
You may assume that the expressions you will process are error free. Design a class that stores
the infix and postfix strings. The class must include the following operations:
getInfix: Stores the infix expression.
showInfix: Outputs the infix expression.
showPostfix: Outputs the postfix expression.
convertToPostfix: Converts the infix expression into a postfix expression. The resulting postfix
expression is stored in pfx.
precedence: Determines the precedence between two operators. If the first operator is of higher
or equal precedence than the second operator, it returns the value true; otherwise, it returns the
value false.
A + B - C;
(A + B ) * C;
(A + B) * (C - D);
A + ((B + C) * (E - F) - G) / (H - I);
A + B * (C + D ) - E / F * G + H;
Infix Expression: A+B-C;
Postfix Expression: AB+C-
Infix Expression: (A+B)*C;
Postfix Expression: AB+C*
Infix Expression: (A+B)*(C-D);
Postfix Expression: AB+CD-*
Infix Expression: A+((B+C)*(E-F)-G)/(H-I);
Postfix Expression: ABC+EF-*G-HI-/+
Infix Expression: A+B*(C+D)-E/F*G+H;
Postfix Expression: ABCD+*+EF/G*-H+
PLEASE PROVIDE FOLLOWING.
A UML diagram for your class.
The header file for your class.
The implementation file for your class.
The source code for your test program.
Solution
#include
using namespace std;
const int MAX = 50 ;
class InfixToPostfix
{
private :
char target[MAX], stack[MAX] ;
char *s, *t ;
int top ;
public :
InfixToPostfix( ) ;
void getInfix ( char *str ) ;
void showInfix () ;
void push ( char c ) ;
char pop( ) ;
void convertToPostfix( ) ;
int precedence ( char c ) ;
void showPostfix( ) ;
void Delete();
} ;
InfixToPostfix :: InfixToPostfix( )
{
top = -1 ;
strcpy ( target, \"\" ) ;
strcpy ( stack, \"\" ) ;
t = target ;
s = \"\" ;
}
void InfixToPostfix :: getInfix ( char *str )
{
s = str ;
}
void InfixToPostfix :: showInfix ( )
{
cout<<\"Infix Expression :\"<= precedence ( *s ) )
{
*t = opr ;
t++ ;
opr = pop( ) ;
}
push ( opr ) ;
push ( *s ) ;
}
else
push ( *s ) ;
s++ ;
}
if ( *s == \')\' )
{
opr = pop( ) ;
while ( ( opr ) != \.
Write a program to convert a given INFIX into POSTFIX. Make sure .pdf
Write a program to convert a given INFIX into POSTFIX. Make sure .pdfFOREVERPRODUCTCHD Write a program to convert a given INFIX into POSTFIX. Make sure the program checks for \"(
)\", and check the balances of parentheses.
Write a program to convert a given INFIX into POSTFIX. Make sure the program checks for \"(
)\", and check the balances of parentheses.
Solution
#include
#include
#include
#define size 100
using namespace std;
char s[size];
int top=-1;
push(char z)
{ /*PUSH Function*/
s[++top]=z;
}
char pop()
{ /*POP Function*/
return(s[top--]);
}
int pr(char z)
{
switch(z)
{
case \'#\': return 0;
case \'(\': return 1;
case \'+\':
case \'-\': return 2;
case \'*\':
case \'/\': return 3;
}
}
int main()
{
char in[100],post[100],c,z;
int i=0,k=0;
printf(\"\ \ Enter your Infix Expression : \");
scanf(\"%s\",in);
push(\'#\');
while( (c=in[i++]) != \'\\0\')
{
if( c == \'(\') push(c);
else
if(isalnum(c)) post[k++]=c;
else
if( c == \')\')
{
while( s[top] != \'(\')
post[k++]=pop();
z=pop(); /* Remove ( */
}
else
{ /* Operator */
while( pr(s[top]) >= pr(c) )
post[k++]=pop();
push(c);
}
}
while( s[top] != \'#\') /* Popping from stack */
post[k++]=pop();
post[k]=\'\\0\';
printf(\"\ \ Infix Expression: %s \ \ Postfix Expression: %s\ \",in,post);
}.
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I need a complete working code for a solution to this problem using .pdf
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Ans:
#include
using namespace std;
class InfixToPostfix
{
private:char *p;//for infix expression
char *q;//for postfix expression
char *stack; //stack array
int top; //top
public: InfixToPostfix();//constructor
void getInfix();//ro read infix expression
void showInfix();//to display infix expression
void showPostfix();//to display postfix expression
void push(char ele);//to push elements into the stack
char pop();//to pop elements from stack
void convertToPostfix();//to convert infix to postfix
int precedence(char);//to find precedence of an operator
~InfixToPostfix();//destructor
};
InfixToPostfix::InfixToPostfix(void)
{
top=-1;
p=new char[100];
q=new char[100];
stack=new char[10];
}
void InfixToPostfix::getInfix(){
cout<<\"enter infix expression:\";
cin>>p;
}
void InfixToPostfix::showInfix(){
cout<<\"\ Given Infix Expression: \"<=precedence(*(p+i)) && top!=-1)
{
ch=pop();
q[j]=ch;j++;
}//end of while
push(*(p+i));//push next symbol into the stack
} //end of else
}//end of for loop
while(top!=-1)
{
ch=pop();
q[j]=ch;j++;
}
q[j]=\'\\0\';
}//end of convertToPostfix()
void InfixToPostfix::showPostfix() //to display postfix expression
{
cout<<\"Postfix Expression: \"<.
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quotient = quotient / 2;
}
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for(j = i -1 ;j> 0;j--)
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return 0;
}
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巴利亚多利德大学毕业证书学校原版文凭补办UVa成绩单办本科成绩单xule9cv6nd <P>专业制作海外大学文凭西班牙文凭巴利亚多利德大学成绩单？【q微1954292140】Buy Universidad de Valladolid Diploma购买美国毕业证，购买英国毕业证，购买澳洲毕业证，购买加拿大毕业证，以及德国毕业证，购买法国毕业证（q微1954292140）购买荷兰毕业证、购买瑞士毕业证、购买日本毕业证、购买韩国毕业证、购买新西兰毕业证、购买新加坡毕业证、购买西班牙毕业证、购买马来西亚毕业证等。包括了本科毕业证，硕士毕业证。【q微1954292140】巴利亚多利德大学毕业证办理，巴利亚多利德大学文凭办理，巴利亚多利德大学成绩单办理和真实留信认证、留服认证、巴利亚多利德大学学历认证。学院文凭定制，巴利亚多利德大学原版文凭补办，扫描件文凭定做，100%文凭复刻。<BR>主营项目：<BR>1、真实教育部国外学历学位认证《西班牙毕业文凭证书快速办理巴利亚多利德大学官方办理文凭》【q微1954292140】《论文没过巴利亚多利德大学正式成绩单》，教育部存档，教育部留服网站100%可查.<BR>2、办理UVa毕业证，改成绩单《UVa毕业证明办理巴利亚多利德大学毕业证办理》【Q/WeChat：1954292140】Buy Universidad de Valladolid Certificates《正式成绩单论文没过》，巴利亚多利德大学Offer、在读证明、学生卡、信封、证明信等全套材料，从防伪到印刷，从水印到钢印烫金，高精仿度跟学校原版100%相同.<BR>3、真实使馆认证（即留学人员回国证明），使馆存档可通过大使馆查询确认.<BR>4、留信网认证，国家专业人才认证中心颁发入库证书，留信网存档可查.<BR>《巴利亚多利德大学学位证定制西班牙毕业证书办理UVa国外文凭购买》【q微1954292140】学位证1:1完美还原海外各大学毕业材料上的工艺：水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠。文字图案浮雕、激光镭射、紫外荧光、温感、复印防伪等防伪工艺。<BR>【q微1954292140】办理巴利亚多利德大学毕业证(UVa毕业证书)毕业证成绩单购买【q微1954292140】巴利亚多利德大学offer/学位证、留信官方学历认证（永久存档真实可查）采用学校原版纸张、特殊工艺完全按照原版一比一制作</P>
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Career Planning After Class XII: Your Roadmap to Success
Career Planning After Class XII: Your Roadmap to SuccessDr. Radhika Sharma Title:
Career Planning After Class XII: Your Roadmap to Success
Description:
Choosing the right career after Class XII is one of the most important decisions in a student's life. This presentation provides a comprehensive guide to career planning, covering various streams such as Science, Commerce, and Humanities. It highlights emerging fields, top courses, leading colleges, and tips on how students can make informed career choices based on their interests, skills, and aspirations.
Whether you're a student, parent, or educator, this PPT serves as a valuable resource for career awareness and decision-making.
Topics Covered:
Importance of Career Planning
Traditional vs. Emerging Career Options
Courses and Colleges after XII
Career Options Stream-wise (Science, Commerce, Arts)
Vocational and Professional Pathways
Tips for Smart Career Decision-Making
Future Trends in Careers
Uploaded to help students build a clear, confident, and successful future.
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program 4.doc in computer science and engineering
- 1. 4. Design, Develop and Implement a Program in C for converting an Infix Expression to Postfix Expression. Program should support for both parenthesized and free parenthesized expressions with the operators: +, -, *, /, %(Remainder), ^(Power) and alphanumeric operands. #include<stdio.h> #include<conio.h> inttop=-1,stack[20],pos=0,length; charsymbol,temp,infix[20],postfix[20]; void push(char symbol) { stack[++top]=symbol; } int pop() { return(stack[top--]); } int preced(char symbol) { int p; switch(symbol) 17
- 2. { case'^':p=3; break; case'/': case'*': case'%':p=2; break; case'+': case'-':p=1; break; case')': case'(':p=0; break; case'#':p=-1; break; } return(p); } void ifix_pfix(char infix[],char postfix[]) { int i=0; push('#'); while(i<strlen(infix)) { symbol=infix[i]; switch(symbol) { case'(': push(symbol); break; case')': temp=pop(); while (temp!='(') { } break; case'^': case'+': case'-': case'*': case'/': postfix[pos++]=temp; temp=pop(); case'%': while(preced(stack[top])>=preced(symbol)) { 18
- 3. postfix[pos++]=pop(); } push( symb ol); break ; default: postfix[pos++] =symbol; break; } i++; } while((temp=pop())!='#') { postfix[pos++]=temp; } } void main() { clrscr(); printf("Enter the Infix expression--- >"); scanf("%s",infix); ifix_pfix(infix,postfix); printf("Postfix expression is---> %sn",postfix); getch(); }