Size of operator

Question

#include<stdio.h>
int main(void) {
    printf("bytes occupied by '7' = %d\n", sizeof('7'));
    printf("bytes occupied by 7 = %d\n", sizeof(7));
    printf("bytes occupied by 7.0 = %d\n", sizeof(7.0));
      
	return 0;
}

Comments

bytes occupied by '7' = 4
bytes occupied by 7 = 4
bytes occupied by 7.0 = 8

Answer 1

Output:

bytes occupied by '7' = 4
bytes occupied by 7 = 4
bytes occupied by 7.0 = 8

The sizeof operator allows a program to determine how much memory is required to store values of a particular type. The syntax will be 

sizeof(type-name)

Now, $sizeof()$ may give different output according to machine

In a 32-bit gcc compiler

$$sizeof(char) \rightarrow 1 \\ sizeof(int) \rightarrow 4 \\  sizeof(float) \rightarrow 4 \\  sizeof(double) \rightarrow 8  $$

bytes occupied by '7' = 4

In C language, type of a character constant like 'x' [Here, '7'] is actually of type $int$, and thus $sizeof('a')$ [Here, $sizeof('7')$] is equal to $sizeof(int)$ which is $4$.

bytes occupied by 7 = 4

Now, $sizeof(7)$ is $sizeof(int)$ which is $4$

bytes occupied by 7.0 = 8

Now, $7.0$ is a $double$, not a $float$

because, 

if decimal floating-point constants suffixed by $f$ or $F$, then that indicates a constant of type $float$. 

If decimal floating point constants suffixed by $l$ or $L$ $\rightarrow$ indicates type $long \hspace{0.1cm} double$.

if decimal floating point constants left unsuffixed then that indicates a $double$ constant. 

So, $sizeof(7.0)$ indicates $sizeof(double)$ i.e. $8$ 

Comments

but $sizeof('c') \rightarrow $ will gives you  $4$ as this is treated like an integer in C, but if you implement sam thing in C++ you'll get $1$

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Then how to print size of single character directly using sizeof()?

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you've to give $sizeof(c)$ it'll print 1 on the screen