Reputation: 733
I know some difference between char* and char[].
char x[] = "xxxx"
Is an array of chars;
char *y = "xxxx"
is a pointer to the literal (const) string;
And x[4]=='\0'
, and *(y+4) == '\0'
too.
So why sizeof(x)==5
and sizeof(y)==4
?
Upvotes: 4
Views: 7711
Reputation: 1
For char *x
, x is a pointer, which means you can change the pointed-to position by x++
, x+=2
, etc.
char x[]
is an array, which is a constant pointer so you cannot do x++
Upvotes: 0
Reputation: 7630
char x[] = "xxxx"
is an array of size 5 containing x x x x and \0.
char *y = "xxxx"
is a pointer to a string. It's length is 4 bytes, because that is the length of the pointer, not the string.
Upvotes: 10
Reputation: 37930
x
is really "xxxx\0"
. The nul terminator at the end of the string gives the array five bytes.
However, sizeof(y)
is asking for the size of a pointer, which happens to be four bytes in your case. What y
is pointing to is of no consequence to the sizeof()
.
Upvotes: 4
Reputation: 53037
The sizeof an array type is the size the array takes up. Same as sizeof("xxxx")
.
The sizeof a pointer type is the size the pointer itself takes up. Same as sizeof(char*)
.
Upvotes: 3