Discussion :: Complicated Declarations - Find Output of Program (Q.No.2)
| Raja said: (Jul 8, 2010) | |
| The option does not matches with the answer of the online compiler which is giving the answer as 4. Why? |
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| Sundar said: (Jul 8, 2010) | |
| Hi Raja, C is a machine dependent language. Typically we assume that the C program is compiled on DOS or Windows platform with Turbo C. The above program will give '2' as output in Turbo C. But, this online compiler runs on a 64-bit platform (Linux server - GCC compiler). So it produces 4 as output. Just try the below code in in Turbo C and Online Compiler, you will understand it better: int a; printf("%x", &a); Have a nice day! |
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| Ankit Anand said: (Jul 9, 2010) | |
| Actually its not about C language, its about compiler we are using, I am using gcc 4:4.4.31-ubuntu1. It gives sizeof(int) = 4, but turbo compiler on windows gives sizeof(int) = 2. and Yes questions based on size of leads to confusion...due to their compiler dependency. |
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| Sasikala [Chennai] said: (Jul 9, 2010) | |
| Hi Ankit Anand and Sundar, Thanks for your informations. While answering the questions in written test, which one should I follow? TurboC or GCC? Windows or Linux? Thanks in advance. |
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| Himanshu said: (Sep 12, 2011) | |
| What the meaning of these three declaration... which is consider by compiler... int, char, float? | |
| Suresh said: (Nov 18, 2011) | |
| I gets cast to a char, then to a float and finally to an int whose size is then used in sizeof. In Turbo C int is 2 bytes. | |
| Sasi said: (Dec 1, 2011) | |
| Can you please explain (int) (float) (char) i this statement? | |
| Kapilsuman said: (Jan 2, 2012) | |
| (int)(float)(char) i; it runs like that typecast i into char first which is of 1 byte. : 1 byte<-(char) i than in float : 4 byte <- char variable i(size is 1 byte) than in int : 2 byte<- float variable i(size is 4 byte) so finally i is of 2 byte integer variable |
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| Shri said: (Feb 3, 2012) | |
| Very good explanation kapilsuman. Kudos to you. | |
| Kaleem said: (Mar 18, 2012) | |
| That right way to explain.... weldn kapilsuman | |
| Aakash Gupta said: (Apr 23, 2013) | |
| I'm getting following errors in Borland c++: (int)(float)(char) i; //Error: undefined symbol 'i' printf("%d", sizeof((int)(float)(char)i)); //ERROR: Not an allowed type & I'm getting following errors in GCC compiler: In function 'main': Line 6: error: 'i' undeclared (first use in this function) Line 6: error: (Each undeclared identifier is reported only once Line 6: error: for each function it appears in.) Please Help with this? |
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| Tinku said: (May 10, 2013) | |
| Can you please explain that variable "i" will be declared as which datatype and why? (int)(float)(char) i; @Aakash: For me the code is running on Linux. I suggest you to try code on some other C version. Hope this helps! |
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| Kranthi Kumar G said: (Jul 9, 2013) | |
| type casting will like this. char size(1) will converted to float size(4). float size(4) will then converted to size int(2). So answer is 2. |
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| Archana said: (Nov 7, 2013) | |
| But look at the declaration of I globally at the top. What does it signify in the pgm? | |
| Shiv said: (Oct 2, 2014) | |
| double i; int main() { (int)(float)(char) i; printf("%d", sizeof(i)); return 0; } Output:8 Why is this? Can anyone give me a valid solution. |
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| Naresh said: (Feb 7, 2016) | |
| Is it possible to convert float (4 bytes) to int (2 bytes)? | |
| Pavan Kumar said: (Jun 17, 2016) | |
| What is typecasting? | |
| Jangum said: (Sep 7, 2016) | |
| Typecasting: Converting one datatype into another datatype. int x; char v; v = (int) x; // x (int) is converted into char datatype and it's value stored into char variable v. |
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| Juel Khan said: (May 12, 2020) | |
| Data type placed in first position in sizeof operator will works only. So, the output is 4 for 64-bit machine and 2 for 32-bit machine. Thanks to all. | |
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