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c programming

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4 votes
 

What will be the output of the program in DOS (Compiler - Turbo C)?

#include<stdio.h>
double i;

int main()
{
    (int)(float)(char) i;
    printf("%d",sizeof(i));
    return 0;
}
  A.
4  
  B.
8
  C.
16
  D.
22
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2 Answers

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the answer is for 16bit platform of Turbo C

(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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answer is : (A)  4 

for the 64-bit system (the size of integer is 4)

first typecasting of i is done from double to char 

and then from char to float 

and then float to int

then the final size of the i will be 4 (the size of int )

answer may be 2(the size of int) if your system is of 32-bit configuration.

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