I"m beginning to learn C .The book I follow tells that when any variable has been declared as type int , in 32 bit computer the maximum integer that can be stored is 2147483647 while on 16
bit it is 32767.I'd initialized a variable to 32768 which I could print ,however when I tried to print 2147483647 it gave -1(while I tried to print it on to the screen). So could any one tell me how many bits my compute uses?
Ask your system:
[edit]Code:#include <stdio.h> #include <limits.h> int main(void) { printf("INT_MAX = %d\n", INT_MAX); printf("INT_MIN = %d\n", INT_MIN); return 0; }Code:#include <stdio.h> #include <limits.h> int main(void) { printf("CHAR_BIT = %d\n", CHAR_BIT); printf("sizeof(int) = %d\n", (int)sizeof(int)); printf("An int occupies %d bits.\n", (int)(CHAR_BIT * sizeof(int))); return 0; } /* my output CHAR_BIT = 8 sizeof(int) = 4 An int occupies 32 bits. */
Last edited by Dave_Sinkula; 12-15-2005 at 10:04 AM.
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.*
Or, alternatively:This will give you a direct result for any modern machine where there are 8 bits per smallest addressable memory unit larger than a single bit, which is usually a byte.Code:#include <stdio.h> int main(void) { printf("There are %d bits in an int on this machine\n", sizeof(int) * 8); return 0; }
[edit]It looks as though I was beaten to the punch, and Dave even accounts for different sizes of char. Well done, Dave![/edit]
Last edited by filker0; 12-15-2005 at 10:09 AM. Reason: Beaten to the punch
Insert obnoxious but pithy remark here
