Number of bits in types
How correct is this?
char: 8 bit
short: 16 bit
long: 32 bit
long long: 64 bit
int: depending on your system (usually 32 bit today)
float: 32 bit (or depending on your system?)
double: 64 bit
This is only something I believe, but don't know. Is it the same in C++ as in C?
There are exactly CHAR_BIT bits in a byte/char, there are at least 8.
There are exactly sizeof(type) bytes in a type. [edit]Though whether or not there are padding bits is another matter.[/edit]
This is the same for both C and C++.
Everything else you showed may or may not be true -- other than besides the int, those will be the minimums. [edit]And I don't know that the larger floating types are required to be "bigger" or not.[/edit]
Last edited by Dave_Sinkula; 12-01-2006 at 09:01 PM. Reason: ...
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
Investigate limits.h -- it shows you some information like CHARBITS (8 bits per char)
This how you would check the memory information of your system. It depends really on you machine. But the following shows the information which was taken out from my machine
ssharish2005Code:#include<stdio.h> int main() { printf("These are the memory size information on my system\n"); printf("\n Int : %d bits\n", sizeof(int) * 8); printf(" Floart : %d bits\n", sizeof(float) * 8 ); printf(" Char : %d bits\n", sizeof(char) * 8); printf(" Double : %d bits\n", sizeof(double) * 8); printf(" Short : %d bits\n", sizeof(short) * 8); printf(" Long : %d bits\n", sizeof(long) * 8); getchar(); return 0; } /* my output These are the memory size information on my system Int : 32 bits Floart : 32 bits Char : 8 bits Double : 64 bits Short : 16 bits Long : 32 bits */
> printf("\n Int : %d bits\n", sizeof(int) * 8);
Read the other posts, 8 is an assumption.
Use CHAR_BIT
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
ohh sorry, thought that normally it would be 8. dint look at the Daves post. Updated code
Thanks SalemCode:#include<stdio.h> #include<limits.h> int main() { printf("These are the memory size information on my system\n"); printf("\n Int : %d bits\n", sizeof(int) * CHAR_BIT); printf(" Floart : %d bits\n", sizeof(float) * CHAR_BIT); printf(" Char : %d bits\n", sizeof(char) * CHAR_BIT); printf(" Double : %d bits\n", sizeof(double) * CHAR_BIT); printf(" Short : %d bits\n", sizeof(short) * CHAR_BIT); printf(" Long : %d bits\n", sizeof(long) * CHAR_BIT); getchar(); return 0; }
ssharish2005
For C89, the most portable way to handle size_t results in printf is to cast to unsigned long and use %lu. And again, I think only the char types are guaranteed to use all the bits as value bits.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
Thanks everyone. So a byte isn't necessary 8 bits, that's something new.
