Discussion :: Library Functions - Find Output of Program (Q.No.2)
| Anusha said: (Mar 15, 2011) | |
| Can anyone please explain how it becomes 0 ? | |
| Archana said: (Mar 17, 2011) | |
| Please give me clarity how it became 0 ans may be 2, 3 also but you gave those values why? | |
| Sundar said: (Mar 17, 2011) | |
| @Anusha Use the casting operator (int) along with ceil(). This will convert the 'double' value to 'int' value. printf("%f, %d", floor(i), (int)ceil(i)); The output will be : 2.000000, 3 |
|
| Prakash said: (Jul 12, 2011) | |
| ceil(2.5)it returns 3, while converting the double to int it returns '0'. So, the output is '2.000000, 0'. |
|
| Swathi said: (Aug 30, 2011) | |
| Can anyone please explain me the procedure of double to int conversion ? | |
| Chandana said: (May 4, 2012) | |
| Conversion is confusing please explain the conversion of double to integer. | |
| Abc said: (May 6, 2012) | |
| Can anyone explain double to int conversion. | |
| Priyanka said: (Sep 3, 2012) | |
| Please explain the conversion of double to integer. | |
| Raghav said: (Oct 16, 2012) | |
| Double is 8 bytes and Int is 4 bytes. Hence specifying %d will yield in first 4 bytes from right which is Zero. | |
| Shubham said: (Mar 26, 2013) | |
| This is really confusing. Please somebody explain clearly. | |
| Djv said: (Aug 9, 2013) | |
| ceil(2.5) returns '3'. As double is 8 bytes- 3 is represented as : 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000011 So while converting double to int(which is of 4 bytes). The first four bytes will be taken into consideration. Hence returning the value '0' :). |
|
| Ramya said: (Aug 10, 2013) | |
| Ceil always return double value. But here return type is integer while converting double to integer 8 bytes is converted to 4 byte real value. So four bits from right to return zero. | |
| Ganesh Singh said: (Jan 18, 2014) | |
| I have got the o/p=0; but now I am confused with the concept used in the following case given below: int main() { double a=3; int b=a; printf("%d",b); } Please make me understood soon as possible. |
|
| Prince Rambade said: (Jun 8, 2014) | |
| Ok if i assume above explanation to be correct. Then pls explain how below program result in, Output : 0,0.00000000 ? int main() { float i = 2.5; printf("%d, %f", floor(i), ceil(i)); return 0; } ceil() returns a double & I have printed the value with %f format specifier but it results in 0.000000 how ? |
|
| Cvam Singh said: (Sep 18, 2014) | |
| Please specify that it will return from right or left. | |
| Amarjeet said: (Jun 12, 2016) | |
| It prints 0 because doubles priority is higher than an integer. That's why it prints 2.000000 and 0. |
|
| Swapnil Kudale said: (Nov 10, 2017) | |
| @Djv is correct. | |
| Rakhi Sinha said: (May 18, 2020) | |
| GCC compiler is giving some garbage value instead of 0. | |
Post your comments here:
Name *:
Email : (optional)
» Your comments will be displayed only after manual approval.