Hey , I urgently need a program to calculate a Bodmas Expression In C
WaltP 2,905
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mridul.ahuja 4
Hey..i made a prog for bodmas expression..bt i dunnot know how to make it reads brackets..
plzz help..
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
char a[20];
int i,j,k,n,l,m;
int t,b=0,c=1,d;
clrscr();
printf("\n------m------\n\n For integers only..");
printf("\n\n To Find : ");
scanf("%s",&a);
n=strlen(a);
for(i=0;i<n;i++)
{
//j=i+1;
if((i==0)&&(a[0]!='('))
b=int(a[0])-48;
if((i==0)&&(a[0]=='-'))
{
b=0-int(a[1])-48+96;
i+=2;
}
g:
{
switch(a[i])
{
case '(' : j=i+1;
switch(a[j+1]) // 48 0
{ // 49 1
case '*' : c=(int(a[j])-48)*(int(a[j+2])-48);
break;
case '+' : c=(int(a[j])-48)+(int(a[j+2])-48);
printf(" %d \n",c);
break;
case '-' : c=(int(a[j])-48)-(int(a[j+2])-48);
break;
case '/' : c=(int(a[j])-48)/(int(a[j+2])-48);
break;
}
if(a[j-2]=='+')
{printf(" %d %d ",b,c); b=b+c; }
else if(a[j-2]=='-')
b=b-c;
else if(a[j-2]=='*')
b=b*c;
else if(a[j-2]=='/')
b=b/c;
else
b=c;
i=i+3;
break;
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
j=i;
switch(a[j+1])
{
case '*' : if(b==0)
b=1;
b=b*(int(a[j+2])-48);
break;
case '+' : b=b+(int(a[j+2])-48);
break;
case '-' : b=b-(int(a[j+2])-48);
break;
case '/' : b=b/(int(a[j+2])-48);
break;
}
i=i+2;
break;
case '+' :
if(a[i+1]!='(')
b=b+(int(a[i+1])-48);
else
{
goto g;
}
i=i+1;
break;
case '-' :
if(a[i+1]!='(')
b=b-(int(a[i+1])-48);
i=i+1;
break;
case '*' :
if(a[i-2]=='+')
{
c=int(a[i-1]-48)*int(a[i+1]-48);
b=b-int(a[i-1]-48)+c;
break;
}
else if(a[i-2]=='-')
{
c=a[i-1]*a[i+1];
b=b+a[i-1]+c;
break;
}
else
{ if(i==1)
{
b=(int(a[i-1])-48)*(int(a[i+1])-48);
break;
}
else
{
//b=b*(int(a[i+1])-48);
if(a[i+1]!='(')
b=b*(int(a[i+1])-48);
else if(a[i+1]=='(')
{i++; goto g;}
break;
}
}
case '/' :
if(a[i-2]=='+')
{
c=int(a[i-1]-48)/int(a[i+1]-48);
b=b-int(a[i-1]-48)+c;
break;
}
else if(a[i-2]=='-')
{
c=a[i-1]/a[i+1];
b=b+a[i-1]+c;
break;
}
else
{ if(i==1)
{
b=(int(a[i-1])-48)/(int(a[i+1])-48);
break;
}
else
{
//b=b*(int(a[i+1])-48);
b=b/(int(a[i+1])-48);
break;
}
}
case ')' : j=i;
switch(a[j+1])
{
case '*' : b=b*(int(a[j+2])-48);
break;
case '+' : b=b+(int(a[j+2])-48);
break;
case '-' : b=b-(int(a[j+2])-48);
break;
case '/' : b=b/(int(a[j+2])-48);
break;
}
i=i+2;
break;
}
}
}
printf("\n\n %d",b);
getch();
}zeroliken 79
Well your program does read brackets but it doesn't always output the right answer now does it?
You use a lot of unnecessary expressions in solving an equation
It's hard to narrow down which part of your code makes the problem
use comments on your codes for us to check properly what your doing
mridul.ahuja 4
(6+2) : it will read it...
but (6*2+4) : it wnt...
see the
case '('
part..
mridul.ahuja 4
Hey guys ,
plzz help modify my code..
WaltP 2,905
Hey guys ,
plzz help modify my code..
I don't see anywhere where you
use comments on your codes for us to check properly what your doing
If you are getting suggestions, I would suggest you don't ignore them. It just tells us all you want us to do is give you a working program.
Try outputting values at key points in your program to see what your program is actually doing. If the values aren't what you expect, you know what part of the code to look at.
mridul.ahuja 4
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
char a[20];
int i,j,k,n,l,m;
int t,b=0,c=1,d;
clrscr();
printf("\n------m------\n\n For integers only..");
printf("\n\n To Find : ");
scanf("%s",&a);
n=strlen(a);
a[x] - 48 : has been done to convert a number from character(ascii) to integer...
for(i=0;i<n;i++)
{
if((i==0)&&(a[0]!='('))
b=int(a[0])-48;
if((i==0)&&(a[0]=='-'))
{
b=0-int(a[1])-48+96;
i+=2;
}
g:
{
switch(a[i])
{
case '(' : j=i+1;
switch(a[j+1]) // Evaluating brackets eg. (4+6)
{ // 49 1
case '*' : c=(int(a[j])-48)*(int(a[j+2])-48); // multiplying in bracket
break;
case '+' : c=(int(a[j])-48)+(int(a[j+2])-48); // adding in bracket
printf(" %d \n",c);
break;
case '-' : c=(int(a[j])-48)-(int(a[j+2])-48); // subtracting in bracket
break;
case '/' : c=(int(a[j])-48)/(int(a[j+2])-48); // dividing in bracket
break;
}
// checking out pre bracket signs.. eg. 4*(7+3)
if(a[j-2]=='+')
{printf(" %d %d ",b,c); b=b+c; }
else if(a[j-2]=='-')
b=b-c;
else if(a[j-2]=='*')
b=b*c;
else if(a[j-2]=='/')
b=b/c;
else
b=c;
i=i+3;
break;
case '1': // in case the current element is a no. and not a sign
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
j=i;
switch(a[j+1])
{
case '*' : if(b==0)
b=1;
b=b*(int(a[j+2])-48);
break;
case '+' : b=b+(int(a[j+2])-48);
break;
case '-' : b=b-(int(a[j+2])-48);
break;
case '/' : b=b/(int(a[j+2])-48);
break;
}
i=i+2;
break;
case '+' : // if current element is a + sign..
if(a[i+1]!='(')
b=b+(int(a[i+1])-48);
else
{
goto g;
}
i=i+1;
break;
case '-' : // if current element is a - sign..
if(a[i+1]!='(')
b=b-(int(a[i+1])-48);
i=i+1;
break;
case '*' : // if current element is a * sign..
// in this case we'll take bodmas into account...
// eg. 4+2*3 will first be evaluated as 4+2 = 6 and then it will check
// the * sign after which it will subtract back 2 and then multiply
// 2 and 3 which will later be added to 4 to get the desired result..
if(a[i-2]=='+')
{
c=int(a[i-1]-48)*int(a[i+1]-48);
b=b-int(a[i-1]-48)+c;
break;
}
else if(a[i-2]=='-')
{
c=a[i-1]*a[i+1];
b=b+a[i-1]+c;
break;
}
else
{ if(i==1)
{
b=(int(a[i-1])-48)*(int(a[i+1])-48);
break;
}
else
{
//b=b*(int(a[i+1])-48);
if(a[i+1]!='(')
b=b*(int(a[i+1])-48);
else if(a[i+1]=='(')
{i++; goto g;}
break;
}
}
case '/' : // same as *
if(a[i-2]=='+')
{
c=int(a[i-1]-48)/int(a[i+1]-48);
b=b-int(a[i-1]-48)+c;
break;
}
else if(a[i-2]=='-')
{
c=a[i-1]/a[i+1];
b=b+a[i-1]+c;
break;
}
else
{ if(i==1)
{
b=(int(a[i-1])-48)/(int(a[i+1])-48);
break;
}
else
{
//b=b*(int(a[i+1])-48);
b=b/(int(a[i+1])-48);
break;
}
}
case ')' : j=i; // end of bracket..not working..
switch(a[j+1])
{
case '*' : b=b*(int(a[j+2])-48);
break;
case '+' : b=b+(int(a[j+2])-48);
break;
case '-' : b=b-(int(a[j+2])-48);
break;
case '/' : b=b/(int(a[j+2])-48);
break;
}
i=i+2;
break;
}
}
}
printf("\n\n %d",b);
getch();
}
