[SOLVED] simple if-else with output beyond wits!!

mq15 · 06-26-2022, 02:47 AM

Hello there.

Please consider this simple C program;

Code:

#include <stdio.h>
int main() {
    float a = .5, b = .7;
    if (b < .7)
       if (a < .5)
          printf("TELO");
       else
          printf("LTTE");
    else
       printf("JKLF");
    return 0;
}

Here is the output:

Quote:

LTTE

The output that I expect is JKLF as the condition b < .7 is false; therefore, the control should not enter the inner if and should reach its own or else printing JKLF.

QUESTION#1: How does the control reach LTTE?

However, if the float data type is changed to int, constants .5 and .7 to 5 and 7 respectively, then the output yields what I expect, i.e., JKLF

QUESTION#2: How does the control reach JKLF after the aforementioned change?

Thanking you in anticipation.

rtmistler · 06-26-2022, 06:02 AM

Probably something to do with the fact that their floating point.

I'd look at it in the debugger where (a) you can see the exact values of those variables, need to use -O0, and (b) you can see the disassembly of the compare lines.

I wonder if the constants are treated as ints and rounded up.

I'd try the same experiment with variables in place of the constants.

EdGr · 06-26-2022, 06:17 AM

.5 and .7 are double precision. The "if" statements are comparing float against double.
Ed

Ser Olmy · 06-26-2022, 07:53 AM

Why don't you change the values to something like .5000001 and .7000001 respectively, and see what that does?

And how about modifying the printf statements to print the values as well? I'm sure that will tell you something.

rtmistler · 06-26-2022, 08:28 AM

Whether or not you've run into this before, this is a detail to remember, that floats and doubles are 'special'. Not that there aren't any conventions, there absolutely are, but one has to learn them. Myself, the few times I work with them, I learn what I need. Certain types of scientific programming sometimes require a lot of this.

teckk · 06-26-2022, 08:28 AM

As others have said.

test1.c

Code:

#include <stdio.h>

int main() 
{
    double a = .5, b = .7;
    
    if (b < .7)
        if (a < .5)
            printf("TELO");
        else
            printf("LTTE");
    else
        printf("JKLF");
        
    printf("\n");
        
    return 0;
}

//gcc test1.c -o test1

Code:

./test1
JKLF

mq15 · 06-26-2022, 09:16 AM

Quote:

Originally Posted by Ser Olmy

...
And how about modifying the printf statements to print the values as well? I'm sure that will tell you something.

Output with float;
a = 0.500000 b = 0.700000 LTTE

Output with int;
a = 5 b = 7 JKLF

mq15 · 06-26-2022, 09:19 AM

Quote:

Originally Posted by EdGr

.5 and .7 are double precision. The "if" statements are comparing float against double.
Ed

Would you please elaborate?

Ser Olmy · 06-26-2022, 09:45 AM

Quote:

Originally Posted by mq15

Would you please elaborate?

I'm not EdGr, but I'll give it a shot.

Consider this program:

Code:

#include <stdio.h>
int main() {
    float a = .7;
    double c = .7;
    if (a < c) {
      printf("What the...?\n");
    }
    return 0;
}

Run it, watch the result, and then recompile with both values changed to .5.

What's going on here, is that some decimal values do not map directly to binary, so different approximations are used when such values are stored as double or float respectively. Conversion between types will amplify these inaccuracies.

Constants in if statements are not treated as the type they are being compared to. Instead, they default to double. As a result, the statement if (a < .7) in your original program ends up becoming "if the result of typecasting a to double is less than the double approximation of the decimal number .7", which is probably not what you wanted. Or what anyone would want, really.

teckk · 06-26-2022, 09:52 AM

https://www.tutorialspoint.com/float-and-double-in-c
https://www.c-programming-simple-steps.com/c-float.html

Did you try this?

test2.c

Code:

#include <stdio.h>

int main() 
{
    float a = 1.5;
    float b = 1.7;
    
    if ( b < 1.7 )
        if ( a < 1.5 )
            printf("TELO");
        else
            printf("LTTE");
    else
        printf("JKLF");
        
    printf("\n");
        
    return 0;
}

//gcc test2.c -o test2

Code:

./test2
JKLF

EdGr · 06-26-2022, 12:14 PM

Quote:

Originally Posted by mq15

Would you please elaborate?

The precision is different. You generally don't want to mix float and double.

You can make the compiler see the constants as float by appending an 'F', i.e. .5F and .7F
Ed

mq15 · 06-26-2022, 11:57 PM

Quote:

Originally Posted by Ser Olmy

I'm not EdGr, but I'll give it a shot.

Consider this program:

Code:

#include <stdio.h>
int main() {
    float a = .7;
    double c = .7;
    if (a < c) {
      printf("What the...?\n");
    }
    return 0;
}

Run it, watch the result, and then recompile with both values changed to .5.

What's going on here, is that some decimal values do not map directly to binary, so different approximations are used when such values are stored as double or float respectively. Conversion between types will amplify these inaccuracies.

Constants in if statements are not treated as the type they are being compared to. Instead, they default to double. As a result, the statement if (a < .7) in your original program ends up becoming "if the result of typecasting a to double is less than the double approximation of the decimal number .7", which is probably not what you wanted. Or what anyone would want, really.

The output of your code is:

Quote:

What the...?

Which clearly means, float a = .7 is considered smaller than double c = .7.

But the output of the code after changing .7 to .5 simply returns. This clearly means that float a = .5 is considered greater than double c = .5

This is strange and unintelligible to me.

EdGr · 06-27-2022, 01:55 AM

The different behavior is due to .5 being expressed exactly in binary floating-point while .7 is being expressed as an approximation.
Ed

NevemTeve · 06-27-2022, 02:56 AM

When in doubt, just print the difference

Code:

printf("%g-%g=%g\n", b, 0.7, b-0.7);

mq15 · 06-28-2022, 11:03 PM

Thanks to everyone who participated. It literally helped. However, EdGr concluded it here;

Quote:

Originally Posted by EdGr

The different behavior is due to .5 being expressed exactly in binary floating-point while .7 is being expressed as an approximation.
Ed

And NevemTeve provided a simple means for the proof;

Quote:

Originally Posted by NevemTeve

When in doubt, just print the difference

Code:

printf("%g-%g=%g\n", b, 0.7, b-0.7);

So, please observe the code and the output;

Code:

#include <stdio.h>
int main() {
    float  b = .7, a = 0.5;
    printf("%g -%g = %g\n", a, 0.5, a-0.5);
    printf("%g -%g = %g\n", b, 0.7, b-0.7);
    return 0;
}

Quote:

0.5 -0.5 = 0
0.7 -0.7 = -1.19209e-08

Happy Coding!