sizeof

Question

#include <stdio.h> 
int main() 
{ 
	int a = 1; 
	char d[] = "ab"; 
	printf("%d", sizeof(a+d)); 
	return 0; 
} 

Explain the Output

 

Answer 1

First read  https://www.geeksforgeeks.org/sizeof-operator-c/

sizeof() operator returns size_t type, https://www.geeksforgeeks.org/size_t-data-type-c-language/

 

#include <stdio.h> 
int main() 
{ 
	int a = 1; 
	char d[] = "ab"; 
	printf("%ld", sizeof(d)); 
	return 0; 
} 

in general,  size of array = no.of elements* size of each element.

Note that, there is a null character present at end of the string by default. 

Output :- d is character array ==> no.of elements =  3 including null character, and each element takes 1 Bytes ==> 3 Bytes as output.

 

#include <stdio.h> 
int main() 
{ 
	int a = 1; 
	char d[] = "ab"; 
	printf("%ld", sizeof(a+d)); 
	return 0; 
} 

in this case integer is added to the character array ==> pointer + int ==> result type is pointer

Output :- size of pointer.

 

Now you may have doubt that, why sizeof(d) doesn't returns size of pointer ?

it is due to " type assigned to d during declaration. ", but in case of sizeof(d+a), type once again calculated, So it returns size of pointer ( Thanks to @arjun sir, for clarification. )

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IMPORTANT POINT :-  sizeof never evaluates the expression value, it just evaluates the type of expression only.

https://www.geeksforgeeks.org/anything-written-sizeof-never-executed-c/

https://www.geeksforgeeks.org/why-does-sizeofx-not-increment-x-in-c/

Comments

Thank you for explaining this.

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Thanks for the detailed explanation @Shaik Masthan. Thanks for the references.