Simple C question. <solved>

[docnascar]

Hi. I'm learning the basics of C and I can't figure this out.

Why does my program give me an answer of 2 if I use 4/2 (which is good). But if I use 2/4 my answer shows "0" and not "0.5" What am I missing in my code?

This works:

Code: [Select]

#include <stdio.h>
#include <math.h>

int main()
{
int answr;

answr = 4 / 2;
printf("The answer is : %d\n",answr);
getchar ();
return (0);
}


This shows "0"... Why   Thanks!!!

Code: [Select]

#include <stdio.h>
#include <math.h>

int main()
{
int answr;

answr = 2 / 4;
printf("The answer is : %d\n",answr);
getchar ();
return (0);
}


[Village Idiot]

Code: [Select]

int answr;
Change int to float   

An Integer is just that, an integer. A float carries the integer value -and- the numbers on the right hand side of the decimal point. For even more precision, use double.

 

[muungwana]

This is a classic example of why C language is seeing as being too easy to shoot yourself with and hence dangerous.

The compiler knows that information is going to be lost when a decimal number is assigned to an interger data type but it doesnt even show a warning. Other language will refuse to do such an assignment and will require a programmer to do explicit casting first.

[docnascar]

I tried double and float, but I still get an answer of 0 (or 0.00000 for this code) if I do 2/4. How do I get it to display 0.5? Thanks!

Code: [Select]

#include <stdio.h>
#include <math.h>

int main()
{
double answr;

answr = 2 / 4;
printf("The answer is : %2.5f \n",answr);

getchar ();
return (0);
}


[muungwana]

try with these lines

double answr;
answr = (double) 2 / 4;

[docnascar]

This is a classic example of why C language is seeing as being too easy to shoot yourself with and hence dangerous.

The compiler knows that information is going to be lost when a decimal number is assigned to an interger data type but it doesnt even show a warning. Other language will refuse to do such an assignment and will require a programmer to do explicit casting first.

I agree. For beginners like me, its not so straight forward and self explanatory.

try with these lines

double answr;
answr = (double) 2 / 4;

That worked. Thanks!

Can anyone explain why you would have to use (double) with your equation? To me, "answr" was expressed as double when its first called out, so I would not think that you would have to retell it that during the math part.

[docnascar]

Code: [Select]


float answr;
answr = (float) 2 / 4;
float works as well.

[muungwana]

"casting" is converting one data type into another.

in your example, you declared an object of type "double" named "answr" and then you want to put a result of 2 / 4.

Now, both "2" and "4" are integers and hence the computer decided that both are integers and hence a result should also be of type integer and hence "2 / 4" produced "0" since integer numbers do not have a decimal part and that part is ignored. The resulting number is then assigned to your variable and that is why the result was "0"

doing (double) 2 / 4 forced the calculation to be of type double and that is why it worked because the result will then be "0.5" and then "answr" will be assigned "0.5"

your code would have worked if you had any following combination (2.0 / 4), (2 / 4.0), (2.0 / 4.0). These numbers would have forced the compiler to see your numbers as not integers and a proper calculation and assignment would have taken place.

basically, your numbers made the computer assume you were doing an integer calculation and it went all the way with integers and that is why an explicit cast was required to force the computer to do the calculation with double data type in mind.

[docnascar]

Thanks. Great explanation. It makes sense.

[critter]

Passing values in declared variables can help to avoid a lot of this confusion.

Code: [Select]

#include <stdio.h>
#include <math.h>

main()
{
float answr, a, b;

a=2;
b=4;
answr=(a/b);
printf ("The answer is : %2.5f\n", answr);
return (0);
}


[docnascar]

Thanks Critter. Good example.

[The Chief]

This is a classic example of why C language is seeing as being too easy to shoot yourself with and hence dangerous.

I dispute that - C assumes the user at least knows what kind of data he will be manipulating, and uses strong (well, fairly strong) typing to prevent those almost undetectable bugs that arise from calling a function with the wrong parameters.

[muungwana]

This is a classic example of why C language is seeing as being too easy to shoot yourself with and hence dangerous.

I dispute that - C assumes the user at least knows what kind of data he will be manipulating, and uses strong (well, fairly strong) typing to prevent those almost undetectable bugs that arise from calling a function with the wrong parameters.

C complains when you call a function with a wrong parameter? i wrote this to test and this is what i got

code:

Code: [Select]

#include <stdio.h>
#include <math.h>

void output( char z )
{
    printf("The answer is : %d \n",z);   
}

int main()
{
char x = 'A' ;
int y = x ;
printf("The answer is : %d \n",y);
output( y ) ;
return (0);
}

result after compiling and running the program:

[ink@mtz ~]$ ./a.out
The answer is : 65
The answer is : 65

See the problem with C? i assigned a "char" datatype to an "int" data type and nobody complained, not even a warning. I then declared a function that took a type "char" but gave it a type "int" and again, no complain and no error from the compiler. I think this disputes your last statement.

[The Chief]

That worked. Thanks!

Can anyone explain why you would have to use (double) with your equation? To me, "answr" was expressed as double when its first called out, so I would not think that you would have to retell it that during the math part.

Because the divide operation was given two integers, so it provided an integer answer, without other direction.  As an alternative, you could have written either (or both) of the input numbers (the 2 and 4) with a '.0' after them.