GATE CSE 2019 | Question: 52

Question

Consider the following C program:

#include <stdio.h>
int main() {
    float sum = 0.0, j=1.0, i=2.0;
    while (i/j > 0.0625) {
        j=j+j;
        sum=sum+i/j;
        printf("%f\n", sum);
    }
    return 0;
}

The number of times the variable sum will be printed, when the above program is executed, is _________

Comments

5 will be answer

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Initially

i = 2.0, j = 1.0

J = 1 => i/j = 2/1  > 0.0625 // First print

J = 1+1 => i/j = 2/2 > 0.0625 //second print

J = 2+2 => i/j = 2/4  > 0.0625// Third print

J = 4+4 => i/j = 2/8 > 0.0625 // fourth print

J = 8+8 => i/j = 2/16 > 0.0625 // Fifth print

J = 16+16 => i/j = 2/32 > 0.0625 (False)

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$\mathbf{5}$ times.

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Use normal calculator and look

2/1 -------------------1

2/(1+1)---------------2

2/(2+2)-------------3

2/(4+4)--------------4

2/(8+8)--------------5

2/(16+16) while condition fails as 0.0625 is not greater than 0.0625

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i want to know how float divide by divde is store as integer by compiler ; give me proper explanation

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@Satish_kr_Verma-CSE- 

there is none integer data type variable used in above code :) 

even it is explicitly mentioned all variables are float data type,

float sum = 0.0, j=1.0, i=2.0;

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@Satish_kr_Verma-CSE-  

When dividing two floating-point numbers (such as float or double) in C, the result of the division is also a floating-point number. If the operation involves one operand as a float (or double), the other operand is automatically promoted to float (or double), ensuring that the result remains a floating-point type. This automatic promotion preserves the decimal part of the result, maintaining precision.

However, if this floating-point result is stored in an integer variable, the fractional portion is discarded, and only the integer part is retained. This process is known as type casting or type conversion, and in C, it occurs automatically when a float or double result is assigned to an int variable. Additionally, explicit type casting can be used to manually convert a floating-point result to an integer if needed. For instance, (int)(a / b) would store only the integer part of the division result by truncating the decimal portion.

In the given code, only float data types are used, meaning that both the operands and the result of the division remain in float form, preserving the decimal part throughout the calculation. No integer data types are involved, so the result stays a floating-point value. However, if desired, this result could be stored in an integer by applying implicit or explicit type conversion, which would truncate the decimal and keep only the integer portion.

Answer 1

$i = 2.0, j=1.0$

while $\left( \frac{i}{j} > 0.0625\right)$

$j = 1$
$ \frac{i}{j} = \frac{2}{1}  > 0.0625$
$j=j+j, 1^{\text{st}}$ PRINT

$j=2$
$ \frac{i}{j} = \frac{2}{2}  > 0.0625$
$j=j+j, 2^{\text{nd}}$ PRINT

$j=4$
$ \frac{i}{j} = \frac{2}{4}  > 0.0625$
$j=j+j, 3^{\text{rd}}$ PRINT

$j=8$
$ \frac{i}{j} = \frac{2}{8}  > 0.0625$
$j=j+j, 4^{\text{th}}$ PRINT

$j=16$
$ \frac{i}{j} = \frac{2}{16}  > 0.0625$
$j=j+j, 5^{\text{th}}$ PRINT

$j=32$
$ \frac{i}{j} = \frac{2}{32}  = 0.0625$
$\text{Break}$

Total $5$ times sum will be printed.

Comments

Typo  $2/32$

Answer 2

5 times while loop will iterate..

Answer 3

while loop condition is :-

$\frac{2}{2^{k}}> 0.0625$

this condition will be true for k=0,1,2,3,4 but fails for k=5

so it'll print sum 5 times.

 

Answer 4

 

 

sum = 0.0
i = 2.0
j = 1.0

Iteration	Current j Value	Condition Check: 2/j > 1/16	Is it True?	Action Inside Loop
1st	1.0	2/1 = 2 > 0.0625	Yes	j becomes 2.0, prints sum (1st time)
2nd	2.0	2/2 = 1 > 0.0625	Yes	j becomes 4.0, prints sum (2nd time)
3rd	4.0	2/4 = 0.5 > 0.0625	Yes	j becomes 8.0, prints sum (3rd time)
4th	8.0	2/8 = 0.25 > 0.0625	Yes	j becomes 16.0, prints sum (4th time)
5th	16.0	2/16 = 0.125 > 0.0625	Yes	j becomes 32.0, prints sum (5th time)
6th	32.0	2/32 = 0.0625 > 0.0625	No	Loop terminates immediately

Final Answer: 5

Answer 5

So answer is 5.

Answer 6

Firstly, for simplicity convert 0.0625 into 1/16.

0.0625 = 625/10000 = 1/16

The loop executes while i/j > 1/16. Here, i always remains 2 and j is doubled (j = j + j) after every iteration. Hence, the value of i/j is halved in each iteration. Since the question only asks how many times sum is printed, we only need to check the loop condition.

Initially: i = 2, j = 1

(1) 2/1 > 1/16 → True → Print

↓

(2) 2/2 > 1/16 → True → Print

↓

(3) 2/4 > 1/16 → True → Print

↓

(4) 2/8 > 1/16 → True → Print

↓

(5) 2/16 > 1/16 → True → Print

↓

2/32 > 1/16 → False

Loop terminates.

Hence, sum is printed 5 times.