How is output 0?

[jyoti300892]

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

void main()
{
printf("%d",(36 .0));
getch();
}
how output is 0

[Luuk]

no, it does not output '0', see below:

luuk@opensuse:~/tmp> cat test.c
#include<stdio.h>
#include<math.h>

void main()
{
        printf("%d\n",(36.0));
}
luuk@opensuse:~/tmp> make test
cc     test.c   -o test
luuk@opensuse:~/tmp> ./test
-733485528
luuk@opensuse:~/tmp> ./test
623260056
luuk@opensuse:~/tmp> ./test
694159224
luuk@opensuse:~/tmp> ./test
-1080378728
luuk@opensuse:~/tmp> ./test
2002711112
luuk@opensuse:~/tmp>

[donbock]

The "%d" argument to printf tells it to decode the next argument as an int. However, you passed a double. It should come as no surprise that the output is garbled.

[mHealth]

no, it does not output '0', see below:

luuk@opensuse:~/tmp> cat test.c
#include<stdio.h>
#include<math.h>

void main()
{
        printf("%d\n",(36.0));
}
luuk@opensuse:~/tmp> make test
cc     test.c   -o test
luuk@opensuse:~/tmp> ./test
-733485528
luuk@opensuse:~/tmp> ./test
623260056
luuk@opensuse:~/tmp> ./test
694159224
luuk@opensuse:~/tmp> ./test
-1080378728
luuk@opensuse:~/tmp> ./test
2002711112
luuk@opensuse:~/tmp>

Any idea, why is the output varying? Should it take
first 4 bytes and print?

[Luuk]

no idea why it's varying, but you should use this:

printf("%d",(int)(36.0));

or

printf("%4.0f",(36.0));

[meditation]

Because it is reading random memory location.

[donbock]

It is not reading random memory locations.
The displayed value varies because different compiler implementations use different encoding schemes for floating point numbers. You should never write code that takes advantage of implementation-specific characteristics such as floating point encoding scheme.

[aswal]

@donbock

What does "encoding schemes for floating point numbers" means?

[Luuk]

In simple English, it's the way how floating point numbers are stored in memory.

A human will 'store' the value of 2/10 as 'zero dot two' (or something)

A computer, or compiler, might do that differently.

[donbock]

In general, computer memory only contains integer numbers. However, we want the computer to handle other kinds of things (such as floating point number and strings). This is accomplished by encoding these other things as integer numbers (or sequences of integer numbers).

The example you are probably most familiar with is the ASCII encoding for representing printable characters as integer numbers (65 means 'A', 66 means 'B', etc). However, this is not the only way to encode characters (see EBCDIC).

[aswal]

@donbock

can you please list me the sources where i can learn more about this topic.

[Luuk]

@donbock:
"In general, computer memory only contains integer numbers"

In general, computer memory only contains binary (zero's and/or one's)
Claiming that a computer stores integer numbers is a simplification.

When a computer stored the next sequence of bits:
01000001
A human wants to read this as an integer, and starts calculation:
1*2^0+0*2^1+0*2 ^2+...+1*2^6 = 65
But if this person was counting in hexadecimal he would say:
41
(4*16+1=65; but also written in binary: '0100 0001')

[donbock]

@Luuk: yes, I was deliberately simplifying.
By the way, 01000001 (binary), 65 (decimal), and 41 (hexadecimal) are all integers. It doesn't matter that they use different bases.

@aswal: you can start by looking up character encoding and IEEE Floating Point in Wikipedia.

[Luuk]

@donbock: NO, 41 is not an integer (in this context)

It is a hexadecimal representation of the integer 65.