ISO C++ forbids comparison between pointer and integer

dmckay · 10-15-2009, 03:21 PM

Hi there,

I'm new to programming (just started it at University) and I'm creating a simple exercise program which is giving me the error in the topic, here's the code

Code:

#include <stdio.h>


const float lowrate = 2;
const float highrate = 5;
const float taxrate = 20;

float accountbalance, interestpaid, newbalance, tax, total;
char taxed;

main()
{
      printf("\n\n\t\tPlease input your account balance");
      scanf("%d",&accountbalance);
      printf("\n\n\t\tIs the account to be taxed?");
      scanf("%c",&taxed);
      getchar();
      
      /*Calculate the interest */
      if(accountbalance < 1000)
      {
                        interestpaid = accountbalance * (lowrate / 100);
                        newbalance = accountbalance + interestpaid;
                        
      }
      interestpaid = accountbalance * (highrate / 100);
      newbalance = accountbalance + interestpaid;
      if(taxed == "y" )
      {
               tax = interestpaid * (taxrate / 100);
               newbalance = newbalance - tax;
              }        
      /*Display the outcomes*/
      
      printf("\n\n\t\tInterest Added = %d", interestpaid);
      printf("\n\n\t\tTotal Account Balance = %d", newbalance);
      getchar();
}

It is saying there is something wrong with the line
if(taxed == "y" )

tuxdev · 10-15-2009, 03:27 PM

This is actually C, although C++ can compile it.

As for your problem, "y" is a const char * that points to a 2-element array containing the character 'y' and 0(null, '\0'). 'y' is the char literal you want.

dmckay · 10-15-2009, 03:31 PM

So I change "char taxed;" to "const char taxed;" ?

It's still giving me the same compile error but with the following added

uninitialized const `taxed'

dmckay · 10-15-2009, 03:35 PM

Seem to have fixed it by changing the line to :-

if( &taxed == "y" )

johnsfine · 10-15-2009, 03:38 PM

Quote:

Originally Posted by tuxdev

'y' is the char literal you want.

Quote:

Originally Posted by dmckay

Seem to have fixed it by changing the line to :-

if( &taxed == "y" )

Try reading the answer you were given.

Change to

Code:

      if( taxed == 'y' )

(&taxed == "y") will compile, but it doesn't have the correct meaning. (taxed == 'y') will compile and has the correct meaning.

nadroj · 10-15-2009, 03:39 PM

Quote:

Originally Posted by dmckay

Code:

char taxed;

It is saying there is something wrong with the line
if(taxed == "y" )

you have to compare characters using single quotes, so this code should instead be

Code:

if (taxed == 'y')

dmckay · 10-15-2009, 03:45 PM

Thanks a lot for the help guys

nadroj · 10-15-2009, 03:46 PM

Quote:

Originally Posted by dmckay

Seem to have fixed it by changing the line to :-

if( &taxed == "y" )

by "fixed" you mean that the compiler doesnt complain anymore? since youre just starting, be aware that if the compiler compiles your code without errors or warnings, it does not mean your code is correct! it simply means that your code has correct syntax and semantics for the language the compiler is for (i.e., C++ in your case).

also, if youre using an IDE, look around for how to turn on more compiler options, i.e. to be "pedantic" (more strict) and show all warnings, as some might not be shown. if you use the command line to compile your programs, i.e. "g++", add the options "-Wall -pedantic" to get more information about your code from the compiler.

basically, since you got rid of the semantic error, it doesnt mean it "works" (does what you think it does).

tuxdev · 10-15-2009, 04:09 PM

Also -Wextra and -Weffc++. I also use -ansi for portability.