Turbo C++ -nan problem

[brianjay0914]

Heres what need to happen.
Enter two consecutive number , add them and show their average.


So The code is Quite correct
I can say

It worked before but if you try and try again .
and try to break it and it will

It'll show -NAN -NAN and -NAN


#include<stdio.h>
#include<conio.h>

float N1,N2,C,D,E;

main()
{
{
clrscr();
printf("Enter A Number:");
scanf("%f",&N1);
printf("Enter A Number Again:");
scanf("%f",&N2);
printf("%.2f and %.2f\n",N1,N2);
C=N2-N1;
D=N1+N2;
E=D*.50;

if(C!=1)
printf("Invalid");
{
if(C==1)
//printf("%f\n",D);
printf("%.2f",E);
}

}

getch();
return 0;
}




Tho it worked before . I want to know why it keeps happening when you try to break it . with same input or different.

what is -NAN how to prevent that from happening

 

[kanewolf]

NAN (or -NAN) is "not a number". Your scanf lines using the "%f" format will only accept floating point numbers. My guess (it has been a long time since I used Turbo C++) is that your float definition leaves the variables purposefully uninitialized. That is usually a compiler option. It is there to help developers.

 

[Alabalcho]

Apart from NAN issue, you could have trouble with the code comparing the value of N with number one. Remember, these are floating point values, and even if it prints as 1.000, it does mean that it is equal to one.

 

[kanewolf]

Apart from NAN issue, you could have trouble with the code comparing the value of N with number one. Remember, these are floating point values, and even if it prints as 1.000, it does mean that it is equal to one.

Good point. You should always add a ".0" to ensure you are doing a floating point comparison. As it is, you are comparing a float to an int. There will be a standardized conversion done, but it is better to be explicit in your compare.

 

[Alabalcho]

Apart from NAN issue, you could have trouble with the code comparing the value of N with number one. Remember, these are floating point values, and even if it prints as 1.000, it does mean that it is equal to one.

Good point. You should always add a ".0" to ensure you are doing a floating point comparison. As it is, you are comparing a float to an int. There will be a standardized conversion done, but it is better to be explicit in your compare.

Adding ".0" (or ".0f") won't help when comparing 1.000000 to 0.9999999. The "correct" way of comparing floating point numbers is

if (fabs(n1-n2) < eps)

where "eps" is quite small number for magnitude of the numbers.

 

[kanewolf]

Apart from NAN issue, you could have trouble with the code comparing the value of N with number one. Remember, these are floating point values, and even if it prints as 1.000, it does mean that it is equal to one.

Good point. You should always add a ".0" to ensure you are doing a floating point comparison. As it is, you are comparing a float to an int. There will be a standardized conversion done, but it is better to be explicit in your compare.

Adding ".0" (or ".0f") won't help when comparing 1.000000 to 0.9999999. The "correct" way of comparing floating point numbers is

if (fabs(n1-n2) < eps)

where "eps" is quite small number for magnitude of the numbers.

In the most generic sense, I would agree. In a practical sense, unnecessary. Human entered data in a program of this nature, that is unnecessarily complicated.