printf((char *) i); runtime error? (i as integer)

Question

In a for loop, I am trying to use printf to print the current i value. This line: printf((char *) i); is giving me runtime errors. Why is this?!

Below is a quick fizzbuzz solution that is doing this:

void FizzBuzz()
{
    for (int i = 0; i < 20; i++)
    {
        printf((char *)i);
        if ((i % 3 == 0) && (i % 5 == 0))
        {
            printf("FizzBuzz \n");
        }
        else if (i % 3 == 0)
        {
            printf("Fizz \n");
        }
        else if (i % 5 == 0)
        {
            printf("Buzz \n");
        }
        else 
        {
            printf("%d\n", i);
        }
    }
}

Answer 1

Because that's not how printf works. You want this instead:

printf("%d\n", i);

Or even better,

std::cout << i;

Answer 2

If you are using C++, you should use cout instead of printf:

#include <iostream>
using namespace std;

int main() {
        int i = 42;
        cout << "The answer is: " << i << endl;
}

Answer 3

By that statement what you are telling is :"there is a sting starting at location i, display it " Surely that is not what you intended. Use format string %d to print a integer

Answer 4

The way printf works is that it takes a string like this:

printf( "the number is: " );

If you then want an integer in the last section of the string you need to use an escape character and then pass the int in as another paramater:

printf( "the number is %d", i );

There is more information here on the escape characters you can use.

You will also need to include:

#include <stdio.h>

EDIT

Sorry my terminology was wrong, % followed by a sequence is called a conversion specification not an escape characer.

HTH.

Answer 5

The first argument of printf() is a C-style null-terminated string. It's meant to be used as a format (thus the "f") with % formatting sequences to print the remaining arguments.

By using printf((char *) i); you are instructing the computer to print every byte starting at the address that i points to, until it encounters a null. Unfortunately, given that i is usually used for counters, it probably points to very low memory. Most modern operating systems prohibit access to such memory from user-space programs to prevent bugs from creating security holes, and will send signals to offending programs. The default response to such a signal, unless you trap it, is to kill the process.

To instead display the number stored in i, use printf("%d\n", i);. To display the value of i as a character, try putc((char)i); or printf("%c\n", i);. If i really is a pointer to a character, try putc((char)(*i));.