C program o/p

Question

main() {
float a=.5, b=.7;
if(b<.7)
  if(a<.5)
   printf("TELO");
  else
   printf("LTTE");
else
 printf("JKLF");

}

Comments

I think it is not about binary
It is the difference between float and double
https://stackoverflow.com/questions/2386772/what-is-the-difference-between-float-and-double
generally .7 is represented in double format
but in float there is different representation
right?

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Here when a and b are the float and they are compared with 0.5 and 0.7 which are double not float.

Note that 0.5f and 0.7f are float values.

Now, to deal with this we need the knowledge of Single (32 bit) and Double (64 bit) precision floating point number.

As we know that everything in the computer world is represented in binary.

So, (0.7)₁₀ will be converted to binary which will be (0.10110011001100.............)₂ infinite Since 0.7 is double so it will take only 64 bit and rest of the precision gets lost.

Coming to b=0.7 here b is float type so here 0.7 will be stored in b in the form (0.10110011001100...) only 32bit.

Now when if(b<0.7) will be performed b will be type promoted to double and the actual value which will be there in b is (0.10110011001100...) only 32bit and rest all 0's till 64 bit they are padded bits

now, when this comparison happen then obviously 0.7 is greater than b. Hence first if will be evaluated true.

Now, second "IF" 

here (0.5) when converted to binary it will be (0.100000....) till infinite, since 0.5 is double so it will take 64bit hence here 0.5 in 64bit precision will be (0.100000........) till 64 bit. Even here precision is lost but that precision is all 0.

Comming to a =0.5 here a is float type so here 0.5 will be stored in a in the form (0.100000....) only 32 bit.

Now when if(a<0.5) will be performed a will be type promoted to double and actual value which will be there in a is (0.100000...........) 64 bit. Now here 0.5 and a both have same value in 64 bit precision.

So, second if evaluetes to false and finally it will print "LTTE".

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Good  Jason

i also want to convey this

Answer 1

C provides three floating types , corresponding to different floating-point formats:

$$\text{float} \rightarrow \text{Single-precision floating point}\\ \text{double} \rightarrow \text{Double-precision floating point}\\ \text{long double} \rightarrow \text{Extended-precision floating point}$$

$float $ is suitable when the amount of precision isn't critical.

$double$ provides greater precisions.

$\text{long double}$ provides ultimate precision, it is rarely used.

Now, C standard doesnot provide how much precision $\text{float, double, long double}$ provides. So, Most computers follow IEEE Standard $754$.

Single precision (float) gives 23 bits of significant, 8 bits of exponent, and 1 sign bit.

Double precision (double) gives 52 bits of significant, 11 bits of exponent, and 1 sign bit.

 By default $float $ -ing constants are stored as double-precision numbers. When a C compiler finds $13.0$(assume) in a program, it arranges for the number to be stored in memory in the same format as a $double$ variable. This rule generally cause no problems, since $double$ values are converted automatically to $float$ when necessary.

main() {
float a=.5, b=.7;
if(b<.7)
  if(a<.5)
   printf("TELO");
  else
   printf("LTTE");
else
 printf("JKLF");

}

We're using $\color{blue}{float}$ to store $.5$ & $.7$. But $.5$ & $.7$ will be trated as  $\color{red}{double}$.

Now, .5 can be represented in binary form as $001111110000$ , which means $.5$, no error during conversion

but .7 can be represented in binary form as $00111111001100110011001100110011$, which actually means $0.699999988079071044921875$. So, there is an error during conversion which is $\approx -1.1920928955078125E-8$.

Now, as $.5$ & $.7$ both are considered as $double$, & be followed by $\color{blue}{\text{Double-precision format(64-bit)}}$ , then 

$$.5 \rightarrow 00111111000000000000000000000000\\ .7 \rightarrow 00111111001100110011001100110011$$

but as we stored $.5$ & $.7$ into  float $a$ & $b$ respectively. & we know, $float$ uses $\text{single precision format(32-bit)}$, so 

$$.5 \text{will be treated as 0.5.... (no change during conversion)}\\.7 \text{will be treated as 0.6999.... (changes during conversion)}$$

& $a's$ value is now $0.5$

$b's$ value is now $0.699...$

So, 

if(b<.7)

is true.

Now, 

if(a<.5)

this does not satisfy the condition.

So, we go to the $\color{red}{else}$ part & prints $\color{violet}{LTTE}$

Comments

@Subarna Das error during conversion is 1.19209289 e-8.

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@Subarna

u havenot mentioned double while representing 0.7 or 0.5

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Answer Updated

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@subarana das there will be no dangling else problem