where's the mistake?

[Totti10]

hi,

can anyone tell me why this does not compile?

Code: Select all

#include <stdio.h>
#pragma argsused
#pragma hdrstop
#include<iostream>
#include<conio>
using namespace std;
int main(void)
{
    int i = 0;

    for(i = 3; i < 7; i++)
    {
    cout << "%c" << (char)i, i);
    }
    getch();
    return(0);
}


[Zen]

Dont know about pragma, but try something like this maybe..

Code: Select all

#include <stdio.h>
#pragma argsused
#pragma hdrstop
#include<iostream>
#include<conio>
using namespace std;
int main(void)
{
    int i = 0;

    for(i = 3; i < 7; i++)
    {
    cout <<(char)i<<" - "<<i<<endl;
    }
    getch();
    return(0);
}

[icujc]

#include<conio>

should be

#include<conio.h>

[Jimbo]

Code: Select all

    cout << "%c" << (char)i, i);


its this line thats the problem. notice how you have one '(' and two ')'. also, you are mixing up cout with printf(). cout uses the '<<' operator to ouput variables to a string, but it lacks printf()'s capability to format any variables outputted.

[raimo]

You should decide whether to use C++ or C.
If C++, then replace <stdio.h> with <cstdio>, throw those #pragmas into rubbish can and learn how to use iostream classes.
std::cout << static_cast<char>(i); is probably what you want to say on that one line.

[c0de]

Code: Select all

#include <iostream>

using namespace std;

int main(void)
{
  int i = 0;
  for(int i = 0; i <= 6; i++)
    {
       cout << i << "\n";
    }
}


wouldn't that give you your desired output? I know it does not convert the int to a char but it still output 1 2 3 4 5 etc...
eh?