hexadecimal calculator problem

In case of an odd number of numbers you should tell the user that he did something wrong.

But there's another problem:
On line 199 operators depends on sizeofinput, but it really depends on numofoperators

is there any way to make it calculate an odd number of numbers

You mean something like that: 1 + 2 + 3?

without taking precedence into account and the input is correct, it would look like this:

          for(int i = 0;i < numofoperators; i++)
          {
               switch(operators[i])
               {
                    case '+':
                         num[i+1] += num[i];
                    break;
                    case '-':
                         num[i+1] -= num[i];
                    break;
                    case '*':
                         num[i+1] *= num[i];
                    break;
                    case '/':
                         num[i+1] /= num[i];
                    break;
                    case '%':
                         num[i+1] %= num[i];
                    break;
               }
          }

          cout << "Answer in dec:" << num[numofoperators] <<endl;

thank you, its working now.

i tested it with division and subtraction last night and it keeps giving the wrong answer and i cant figure out how to get it working properly,it works fine with addition and multiplication,but with division and subtraction its just not giving the correct answer

Just for my education, what is the difference between a static cast like this:

static_cast<double>(j)

and an ordinary c cast like this:

(double) j

I read the documentation, but I don't see what the difference is.

ive got the calculator working, i changed the code in the loop a bit:

int temp = num[0];
          for(int i = 0,j = 0;i < numofoperators;i++)
          {
               switch(operators[i])
               {
                    case '+':
                         temp = num[i+1] += num[i];
                    break;
                    case '-':
                         temp = num[i] -= num[i+1];
                    break;
                    case '*':
                         temp = num[i+1] *= num[i];
                    break;
                    case '/':
                          temp = temp / num[i+1];
                    break;
                    case '%':
                          temp = temp % num[i+1];
                    break;
               }               
          }
          
          
          cout << "Answer in dec:" << temp <<endl;

@TheIdeasMan i dont know what the difference is.

ive got the calculator working, i changed the code in the loop a bit:

Unfortunately that's wrong too.

While the idea with temp is right since my previous approach didn't take the order into account.

The following gives the correct answer for 1 + 2 - 3 + 4 - 5 + 6:

int temp = num[0];
          for(int i = 0,j = 0;i < numofoperators;i++)
          {
               switch(operators[i])
               {
                    case '+':
                         temp += num[i+1];
                    break;
                    case '-':
                         temp -= num[i+1];
                    break;
                    case '*':
                         temp *= num[i+1];
                    break;
                    case '/':
                          temp /= num[i+1];
                    break;
                    case '%':
                          temp %= num[i+1];
                    break;
               }
          }


          cout << "Answer in dec:" << temp <<endl;

@TheIdeasMan
static_cast cannot cast away const and other 'attributes' while the c style cast doesn't care

@zoldri: I don't mean to hijack your thread, probably this is an easy question to ask coder777.

Also, with the calculator, do you check for division by zero?

@coder777

Thanks for your tip about the casting.

Would a (double) C cast to a const variable, be a compile error, meaning one would have to use const_cast? And if the variable was not a const, then one could just use (double).

Last edited on

@TheIdeasMan

I'm not completely sure what you mean. See

http://www.cplusplus.com/doc/tutorial/typecasting/

You can do this without a problem:

1
2

int x = 0;
const double d = x;

1
2

int x = 0;
const double d = (double)x;

1
2

const int x = 0;
const double d = x;

You usually cast from more to less precision:
long -> short // needs a cast (to make the warning go away) int -> double // don't need a cast

Ok, I have this last bit of clarification / question (I am worried because it is not my thread)

I had read the article before, but I couldn't see the difference when it is an ordinary type as opposed to base & derived classes.

What about this one:

1
2

const double d=3.14159265;
int i = static_cast<int>(d);

I am guessing that d remains the same, i equals 3. If not, it's compiler error, and would need a const_cast<int>?

and is this any different?

1
2

const double d=3.14159265;
int i = (int)d;

Last edited on

const_cast is only for pointer since it does matter if it points to a const or non value.

if const is on left of the assign operator (lvalue) it can be initialized only.

for numeric variables right of the assign operator (rvalue) it doesn't matter whether const or non const. So your examples are both ok. No difference.

For classes you should use dynamic_cast (with RTTI) since this is in fact save. It returns NULL if it's not the expected type (for pointer) or throws for references

Ok thanks coder777, I had better let the OP back to his own thread. Cheers

it didnt check for division by zero but it does now