UNIX for Dummies Questions & Answers

This is what I have:

#include <stdio.h>

int main (void)
{
int integerVar;
int floatingVar;
int doubleVar;
int charVar;

integerVar = 100;
floatingVar = 331.79;
doubleVar = 8.44e+11;
charVar = 'W';

_Bool boolVar;

boolVar = 0;

printf ("integerVar = %i\n", integerVar);
printf ("floatingVar = %f\n", floatingVar);
printf ("doubleVar = %e\n", doubleVar);
printf ("doubleVar = %g\n", doubleVar);
printf ("charVar = %c\n", charVar);

printf ("boolVar = %i\n", boolVar);

return 0;
}

And this is what I got:

integerVar = 100
floatingVar = 0.000000
doubleVar = 6.674375e-308
doubleVar = 6.67437e-308
charVar = W
boolVar = 0

I don't know why the floatingVar and doubleVar's don't work. Does anybody know why this is?

I'm supposed to get:

integarVar = 100
floatingVar = 331.790009
doubleVar = 8.440000e+11
doubleVar = 8.44e+11
charVar = W
boolVar = 0

-- I'm using something I have never used before about a week ago, a program called Dev-C++ to write and compile and I'm using Command Prompt's bash to run it.

If anybody can figure out why I can't compile and run through Dev-C++, that would also be very helpful.

Code:

int integerVar;
int floatingVar;
int doubleVar;
int charVar;

Why have you declared them all as 'int's when they are not? Try:

Code:

int integerVar;
float floatingVar;
double doubleVar;
char charVar;

Thanks. :]

Now I feel stupid for not seeing that--and even worse for reading the original program wrong.

But that's what computers are for... to make us look stupid.