Answer Posted / hasan ali mirza

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int preceed(char c);
void push(char c);
char pop();
char stk[30];
int tos=-1;
int main()
{
int i,j=0,n,u,v;
char infix[30],postfix[30],c;
printf("Enter the infix expression:");
scanf("%s",&infix);
n=strlen(infix);
for(i=0;i<n;i++)
{
if((infix[i]>='a'&&infix[i]<='z')||(infix[i]>='A'&&infix[i]<='Z'))
{
postfix[j]=infix[i];
j++;
}
else if(infix[i]=='^'||infix[i]=='*'||infix[i]=='/'||infix[i]=='+'||infix[i]=='-')
{
u=preceed(stk[tos]);
v=preceed(infix[i]);
while(u>=v&&stk[tos]!='(')
{
postfix[j]=pop();
j++;
u=preceed(stk[tos]);
}
push(infix[i]);
}
else if(infix[i]=='(')
{
push(infix[i]);
}
else if(infix[i]==')')
{
c=pop();
while(c!='(')
{
postfix[j]=c;
j++;
c=pop();
}
}
else
{
printf("Equation has error\n");
exit(0);
}
}
while(tos!=-1)
{
postfix[j]=pop();
j++;
}
postfix[j]=='\0';
printf("The equation in postfix:%s",postfix);
}
void push(char c)
{
tos++;
stk[tos]=c;
}
char pop()
{
char val;
val=stk[tos];
tos--;
return(val);
}

int preceed(char c)
{
int v;
switch(c)
{
case '^':v=3;
break;
case '*':
case '/':v=2;
break;
case '+':
case '-':v=1;
break;
default:v=0;
break;
}
return(v);
}

Draw the diagram showing the function stack, illustrating the variables that were pushed on the stack at the point when function f2 has been introduced . type def struct { double x,double y} point; } main( int argc, char *arg[3]) { double a; int b,c; f1(a,b); } f1(double x, int y) {point p; stack int n; f2(p,x,y) } f2(point p, double angle) { int i,j,k,int max;

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