[C] How do I get around this warning message?

[Ravager]

When I attempt to compile in GCC on Linux, I get the following warning:

ISO C90 does not support the `%lf' printf format

The code in question is:

  Quote

void do_the_calculation(double a, double b, double c, double calculation, double root1, double root2)

{

  if (a != 0)

  {

printf("\nYou entered the numbers %lf, %lf, and %lf\n", a, b, c);

calculation = pow(b,2)-4*a*c;

printf("The discriminant of the quadratic equation is: %lf\n\n ", calculation);

root1 = (-b+(pow(b,2)-4*a*c))/(2*a);

root2 = (-b-(pow(b,2)-4*a*c))/(2*a);

printf("Roots are: %lf and %lf\n\n", root1, root2);

  }

  else

printf("Non-real answer will result if a = 0");

}

How can I get around this?

Obviously I want to printf a double. My lab instructor is picky and requires us to fix all warning messages. The odd part is that this error message only occurs on the Linux machines at my university, but compiling this program within Windows results in no errors.

Is this something Linux-specific, or is there actually something not right with my code? Please note this GCC compiler was modified to be VERY sensitive.

Thanks.

[kjordan2001]

It must be somewhat old too to stick by ISO C90. %lf was put in the spec for ISO C99.

[Ravager]

  kjordan2001 said:

It must be somewhat old too to stick by ISO C90.  %lf was put in the spec for ISO C99.

586566156[/snapback]

Hooray for the obsolete.

PS: Good to see you around here, still. I'll be posting here a lot more often. ;)

[digweed]

Ravager, do you go to U of C and in Engineering? I had a problem similar to that in my lab today.