Thread: i am a bit lost with this one

Be aware that the program is finished when you reach a "return" statement in main().

The printf documentation may seem confusing if you have never done any programming before. Anyway, you cannot include the variable in quotes (with C); it will just literally print "i".

Code:

#include <stdio.h>  /* needed for printf */

int main() {
     char example[]="hello world";
     int x=5;

     printf("%s, x is %d\n", example, x);
     return 0;
}

The green part is the "template". The underlined bits are "conversion specifiers". For each conversion specifier in the template, you need a matching argument. In this example, there are two (in blue), one is a "string" (a char pointer) and the other one (important to you) is a "digit", which would be an integer.

\n is a newline, try it without that to see the difference.

Code:

#include <stdio.h>
int main ()
{
    int x;
    int y;
    int i =1;                //#c
    int power;            //change to int power = 1; as per #d of assignment
   
  
    
    scanf(" %d", &x);   //#a
    scanf(" %d", &y);   //#b
    
    i= power * x;       //put this inside while loop, before the ++i;
   

    while (i<=y)      //#f
    { i++;                //#g

       }
       
       return 0;               //move this line to the very end of the program
       
       ++i;                       //remove this line
       
       printf("int power is i");  //#h I'm not sure if this should go inside the while loop or not
                                            // I think it should go inside, however.      
       }

is this right now

Let's see where we are:

a. input integer variable x with scanf
b. input integer variable y with scanf
c. initialise integer variable i to 1
d. initialise integer variable power to 1
e. while i <= y
f. multiply power by x and assign the result to power
g. increment i by 1
h. output the integer variable power with printf.

So, a few things to be fixed, yet.

Edit:

The console window your program is running in will close immediately after running, in many compilers. It's the default action of a
console window in Windows.

To keep the wiindow open long enough to see it properly, you might need to add this line of code:

getchar();

Note that each getchar() call will only remove *one* char from the keyboard buffer. If you have more than one, it will need some further help:

Code:

while((c = getchar()) != '\n');   //removes all chars in the keyboard buffer including the first newline char.
getchar();                                //now getchar() by itself will wait for the next "enter" key before proceeding

return 0;