Programming

Question

#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
	char *p="good";
	char a[]="good";
	printf("%d%d%d",sizeof(p),sizeof(*p),strlen(p));
	printf("%d%d",sizeof(a),strlen(a));
	getch();
}

What will be output? Pls Explain

Answer 1

sizeof(p) will print 8 as it returns the size of the char pointer which can be 4 or 8 on 32bit and 64 bit machine respectively.

sizeof(*p) will return 1 since the size of 1 char("g" in this case) is 1 byte.

strlen(p) will return 4, which is quite obvious.

sizeof(a) will return 5 since it is a char array trailing with the terminator symbol

And finally, strlen(a) will return 4.

Comments

but the size of pointer is always 2.sizeof(*p) should be 2 noo why it is 1 ??

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very nice explanation

Answer 2

Considering the system to be 32 bit, means every instruction will be of 32 bit.

Now,

Char *p = 'good'; here p will be a pointer variable having the base address of string 'good'.

Char a[] =' good'; here also a is just a mnemonic having the base address of string 'good'.

Eg- lets assume base address is 'x'. So a[2] = (a+2) = (base address 'a' + 2*sizeof(char)).

Now, sizeof(p) = sizeof(a) = 4(32 bits)

         Sizeof(*p) = 4,  since *p will be pointing to string's first character and its length is 4.

        Strlen(a) = strlen(p) = 32, since length of a or p will be equal to 32 bits.