write a program for size of a data type without using
sizeof() operator?
Answer Posted / p sahana upadhya
#include<stdio.h>
#include<math.h>
main()
{
float size;
printf("The Size of Integer Data Type is %d
Bytes\n",(sizeof(int)));
size=pow(2,(sizeof(int)*8));
printf("The Range of Integer Data Type is -%.0f to
%.0f\n\n",(size/2),((size/2)-1));
printf("The Size of Long Integer Data Type is %d
Bytes\n",(sizeof(long int)));
size=pow(2,(sizeof(long int)*8));
printf("The Range of Long Integer Data Type is -%.0f
to %.0f\n\n",(size/2),((size/2)-1));
printf("The Size of Unsigned Long Data Type is %d
Bytes\n",(sizeof(long int)));
size=pow(2,(sizeof(long int)*8));
printf("The Range of Unsigned Long Data Type is 0 to
%.0f\n\n",size);
printf("The Size of Float Data Type is %d Bytes\n",
(sizeof(float)));
size=pow(2,(sizeof(float)*8));
printf("The Range of Float Data Type is -%.0f to
%.0f\n\n",(size/2),((size/2)-1));
printf("The Size of Double Data Type is %d Bytes\n",
(sizeof(double)));
size=pow(2,(sizeof(double)*8));
printf("The Range of Double Data Type is -%.0f to
%.0f\n\n",(size/2),((size/2)-1));
printf("The Size of Long Double Data Type is %d
Bytes\n",(sizeof(long double)));
size=pow(2,(sizeof(long double)*8));
printf("The Range of Long Double Data Type is -%.0f
to %.0f\n\n",(size/2),((size/2)-1));
printf("The Size of Char Data Type is %d Bytes\n",
(sizeof(char)));
size=pow(2,(sizeof(char)*8));
printf("The Range of Character Data Type is -%.0f to
%.0f\n\n",(size/2),((size/2)-1));
printf("The Size of Unsigned Char Data Type is %d
Bytes\n",(sizeof(char)));
size=pow(2,(sizeof(char)*8));
printf("The Range of Unsigned Char Data Type is 0 to
%.0f\n\n",size);
getch();
return 0;
}
This program may seems long, but it definitely works!!!!!
| Is This Answer Correct ? | 3 Yes | 3 No |
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