Checking Multiple Values in if() statement

hey guys i keep going off and on with my programming resulting in my forgeting how to do most everything :p

ok so the program im working on is supposed to display all the ascii characters and their value I do not want to show the values that do not contain an ascii character (7,8,9,10,13,32) and i remember how to check the index for multiple values in a switch() but i think i remember doing it all in an if() statement but i cant remember what the correct syntax is for that.

i was trying this

Code:

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if(1 == 7 || 8 || 9 || 10 || 13 || 32) i++

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anyways ill post my code and if you can help me out that would be great Thanks

Code:

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#include <iostream>
using namespace std;
int main()
{
    cout<<"Here we go!\n";
    char var1 = 55;
    cout<<"\nvar1 is "<<var1<<endl;
    cout<<"the size of Var1 is: "<<sizeof(var1);
    for(unsigned char i =0; i<255; i++)
    {
          switch(i)  //how to check i for multiple values in if statement
          {
              case 7: i++;
              case 8: i++;
              case 9: i++;
              case 10: i++;
              case 13: i++;
              case 32: i++;
          }
          printf("%d %c",i, i); // How to display same results in c++ !c
          cout<<endl;
          }
    system("pause");
}//end main

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i also have another question, How would i get the same result the printf() statement has using cout?

Quote:

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Originally Posted by Zero_X

i was trying this

Code:

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if(1 == 7 || 8 || 9 || 10 || 13 || 32) i++

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Try this:

Code:

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if(1 == 7 || i == 8 || i == 9 || i == 10 || i == 13 || i == 32) i++

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or this

Code:

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          switch(i)  //how to check i for multiple values in if statement
          {
              case 7: case 8: case 9: case 10: case 13: case 32:
                i++;
          }

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awesome, i didnt know i could stack the case's like that. but now ive run into another problem im not sure if i will solve. for some reason im still getting 8 && 10 to show with no ascii character. i really have no idea what is going on with this.

Yeah, you've chosen to simplistically "increment past" the offending value. What's wrong with this logic?

You actually don't want to print the value, but to instead proceed with the loop to the next character, right? "Word it" so offending values don't print.

ahh yes I see now

Code:

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#include <iostream>
using namespace std;
int main()
{
    cout<<"Here we go!\n";
    char var1 = 55;
    cout<<"\nvar1 is "<<var1<<endl;
    cout<<"the size of Var1 is: "<<sizeof(var1);
    for(unsigned char i =0; i<255; i++)
    {
          if(i == 7 || i == 8 || i == 9 || i == 10 || i == 13 || i == 32)i++;
          else{printf("%d %c",i, i); // How to display same results in c++ !c
          cout<<endl;}
          }
    system("pause");
}//end main

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Thanks for the help Dave

It may be easier to see your question if it isn't in the code.

Code:

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  for ( unsigned char i =0; i<255; i++ )
  {
      if ( i == 7 || i == 8 || i == 9 || i == 10 || i == 13 || i == 32 )
      {
        /* skip */
      }
      else
      {
        // How to display same results in c++ !c
        cout << static_cast<int>(i) << ' ' << i << '\n';
      }
  }

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Or is otherwise perhaps highlighted.

I'm curious why you output past the 127th character.

Further, you could use <ctype.h> for a simpler implementation.

-Cheers