Why there is a difference between the number of bits given for a constant and a variable?

Question

The size of a constant float is 8 bytes while a variable float is just 4 bytes.

#include <stdio.h>
#define x 3.0
int main(){
    printf("%d", sizeof(x));
    return 0;
}

This also applies for a constant char (gives 4 bytes) while a char variable just gives 1 byte.

`3.0` is a `double`, `3.0f` is a `float`. https://ideone.com/w0ZBNs — Retired Ninja, Apr 15 '22 at 04:51
is this valid only for float and double ie can we transform a double constant to be an integer by adding a character? — TheKing, Apr 15 '22 at 04:59
C does not have constants of type `char`. `'x'` is an `int`. — chux - Reinstate Monica, Apr 15 '22 at 05:42
Note that the correct way to print a `size_t` value is `%zu` — the `z` modifier is for values of type `size_t`. Don't forget to add a newline at the end either: `printf("%zu\n", sizeof(x))` — Jonathan Leffler, Apr 15 '22 at 05:48
You can specify some variations on the type of an integer constant by adding `U` and `L` suffixes. `1` is an `int`; `1U` is an `unsigned int`; `1L` is a `long`; `1UL` is an `unsigned long`; `1LL` is a `long long` and `1ULL` is an `unsigned long long`. — Jonathan Leffler, Apr 15 '22 at 05:50

Rahul · Accepted Answer · 2022-04-15T08:10:52.210

I think this question has already been answered in a couple of previous posts. The basic idea is:

A) Consider this program in C:

#include <stdio.h>
#define x 3.0      /* without suffix, it'll treat it as a double */
#define y 3.0f     /* suffix 'f' tells compiler we want it as a float */

int main() {
    printf("%ld\n", sizeof(x)); /* outputs 8 */
    printf("%ld", sizeof(y)); /* outputs 4 */
    return 0;
}

Basically, double has more precision that float so it's more preferable in case of ambiguity. So, if you're declaring constant without suffix 'f', it'll treat it as a double.

B) Look into this one now:

#include <stdio.h>
#define x 'a'

int main() {
    char ch = 'b';
    printf("%ld\n", sizeof(x)); /* outputs 4 */
    printf("%ld", sizeof(ch)); /* outputs 1 since compiler knows what it's 
                                  exactly after we declared & initialized var 
                                  ch */
    return 0;
}

That constant value ('a') is converted to it's ASCII value (097) which is a number literal. Hence, it'll be treated as an integer. Please refer this link for more details: https://stackoverflow.com/questions/433895/why-are-c-character-literals-ints-instead-of-chars#:~:text=When%20the%20ANSI%20committee%20first,of%20achieving%20the%20same%20thing.

Why there is a difference between the number of bits given for a constant and a variable?

1 Answers1