error in solution of ANTI-BLOT SYSTEM(spoj) - c++

Question

hello we have to convert an expression like "43 + machula0 = 163"
as 43 + 120 = 163.means we have to scan the expression and then find the missing part.my program runs good without putting the while loop but it shows the error when i run it with the loop.
#include<iostream>
#include<cstring>
#include<cctype>
#include<cstdlib>
using namespace std;
void convert(char *a,int size)
{
int i=0;
char c =a[i];
int f1,f2,f3,n1,n2,s=0; //f1,f2,f3 are flags to check machula
f1=0;f2=0;f3=0;n1=0;n2=0;
while(c!='+')//to store no. before plus in n1
{
c=a[i];
if(isalpha(c))//to check whether the character is alphabet or not
{
f1=1;
}
else if(isdigit(c))//to check whether the character is digit
{
int a = c-'0';
n1=n1*10 + a;
}
i++;
}
while(c!='=')//to store no. before plus in n2
{
c=a[i];
if(isalpha(c))
{
f2=1;
}
else if(isdigit(c))
{
int k = c-'0';
n2=n2*10 + k;
}
i++;
}
while(i!=size)//to store no. before plus in s
{
c=a[i];
if(isalpha(c))
{
f3=1;
}
else if(isdigit(c))
{
int h = c-'0';
s=s*10 + h;
}
i++;
}
if(f3==1)
{
s=n1+n2;
cout<<n1<<" + "<<n2<<" = "<<s<<endl;
}
else
{
if(f1==1)
{
n1=s-n2;
cout<<n1<<" + "<<n2<<" = "<<s<<endl;
}
else if(f2==1)
{
n2=s-n1;
cout<<n1<<" + "<<n2<<" = "<<s<<endl;
}
else
{
cout<<n1<<" + "<<n2<<" = "<<s<<endl;
}
}}
int main()
{
int t;
cin>>t;
while(t--)
{
char *a;
a= new char[10000];
cin.getline(a,10000);
int size = strlen(a);
convert(a,size);
delete []a;
}
return 0;
}

Answer 1

On second thought I do have some useful input. If you are typing this in by hand, there is a attendance to type in the input like this:
1
43 + machula0 = 163" as 43 + 120 = 163
and that leads to problems
int main()
{
int t;
cin>>t; // gets up to the end of the number typed.
// It does not get the carriage return from hitting enter to trigger the input.
while(t--)
{
char *a; // Not relevant to the problem, but this is an awesomely bad idea.
// Use a string
a= new char[10000];
cin.getline(a,10000); // returns instantly with an empty string because of the
// left-over enter used above to get the number.
int size = strlen(a);
convert(a,size); // most of convert does not check size, so instantly out of
// array bounds and crash
delete []a;
}
return 0;
}
It will be valid for input like this:
1 43 + machula0 = 163" as 43 + 120 = 163
Better solution:
void convert(string &a)
{
for (char c: a)
{
//parse logic goes here
}
// output logic goes here
}
int main()
{
int t;
cin >> t;
cin.get();
while (t--)
{
string line;
if (getline(cin, line))
{
convert(line);
}
}
return 0;
}
Unfortunately the convert function needs a lot more work. Most obviously the nearly complete lack of testing for the end of the string.