SizeOf operator in C : Why is the output 6?
#includeWhy is the output 6?#include char name[] = "aadit"; int main(void) { printf("%d\n",sizeof(name) ); }
Replies
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Kaustubh KatdareI 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. -
Jason EstibeiroThe 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. -
Shailaja TiwariYes it is the null character assumed to be appended at the end of the string ....
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Aadit KapoorThank you friends!
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#include
#include
#include
int main(void)
{
char names[][5] = {
"aadit",
"jobs",
"mark",
};
printf("%s\n",names[0]);
getch();
}
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