SizeOf operator in C : Why is the output 6?
#include <stdio.h>
#include <string.h>
char name[] = "aadit";
int main(void)
{
printf("%d\n",sizeof(name) );
}
Why is the output 6?
0
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Administrator • Mar 18, 2014
I guess because it includes the null at the end; the terminating \0. By the way the correct way to get the length of string would be strlen() and not sizeof(); because sizeof will return the length of the array, not the string.
C/C++ experts may offer better help.
PS: Been away from programming since eternity.Are you sure? This action cannot be undone. -
Member • Mar 18, 2014
The sizeof() function provides the size of the data type in bytes. The character array consist of a terminating character '\0' (hence name[5]='\0') (as told by #-Link-Snipped-# ) and since each char value occupies a space of one byte, hence the result is 6 ... 6 bytes.
And again as #-Link-Snipped-# said, if you want the length of the string use strlen() function. Then you would get the answer as 5.Are you sure? This action cannot be undone. -
Member • Mar 18, 2014
Yes it is the null character assumed to be appended at the end of the string ....Are you sure? This action cannot be undone. -
Member • Mar 18, 2014
Thank you friends!Are you sure? This action cannot be undone.