#include
int main() {
int i=10L;
long j=10;
int k=10;
printf("%d\n",sizeof(i));
printf("%d\n",sizeof(j));
printf("%d\n",sizeof(k));
printf("%d",sizeof((int)j));
return 0;
}
I am observing 4 bytes for all outputs, can anybody justify this.
Number of bytes per type is determined by the compiler. If you are running on a 64bit platform but are expecting 8 bytes (for 'long'), you need to set the appropriate compiler flag.
This is the shortcoming of the C family. There is still ambiguity what long short and normal int means in terms of storage size.
The first printf will print the size of an int (the type of the variable i; note that the assignment of the long value 10L to i does not change i's type; on systems where longs have more bits than ints, it would simply mean that the value would be truncated on assignment). On most modern systems, this would be 4 bytes.
The second printf prints the size of a long (the type of j). On 32-bit systems, this would be 4 bytes; on 64-bit Windows with Microsoft Visual C++, this would be 4 bytes; on 64-bit Linux (GNU C compiler), 8 bytes.
The third printf prints the size of an int again (the type of k), i.e., 4 bytes on most systems.
The fourth printf prints the size of an int (the type of the temporary value created when j is typecast to int before the application of the sizeof() operator), i.e., once again 4 bytes on most modern systems.
So on Windows with Visual C++ (32-bit, 64-bit) the program prints the number 4 four times; same thing on 32-bit Linux (GNU C), but on 64-bit Linux, it would print 4, 8, 4, 4. Other platforms/compilers may behave differently, and compiler flags may also affect the behavior.
(Clearing up some misconceptions posted here)
The C99 standards only gaurantee the following:
sizeof(short)
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