#define STR "test1"
Why does this take 6 bytes?
sizeof(STR) = 6
views:
207answers:
7@DigitalRoss: that is what I was thinking....thanks.
Tommy
2010-09-30 22:29:34
+5
A:
It has nothing to do with #define. A character array would be the same size:
const char str[] = { "test1" };
sizeof (str) == 6
The reason this string is 6 bytes long is that strings in C have a terminating NUL character to mark the end.
Ben Voigt
2010-09-30 22:29:40
That's not an array, it's a pointer to a string literal, and `sizeof(str)` will return the size of the pointer type for your machine.
Carl Norum
2010-09-30 22:31:12
What this meant to say is that `const char str[] = "test1";` is the same as `const char str[] = {'t','e','s','t','1','\0'};`
aschepler
2010-09-30 22:33:40
Umm, yeah. I feel really horrible, especially since I thought about the difference between sizeof array and sizeof pointer before typing the answer, and then my fingers hit the wrong keys anyways.
Ben Voigt
2010-09-30 22:38:00
+2
A:
Strings in C are arrays of chars, with a null terminator i.e. they end with the \0. The common alternative is Pascal-style strings, where the string stores the array of chars without the null terminator, and stores the length of the string somewhere instead.
Michael Williamson
2010-09-30 22:29:59
Speaking of Pascal strings in the context of C and C++ and string literals, is terribly misleading.
Mark Ransom
2010-09-30 22:39:37
+5
A:
a #define just does a text replacement before compiling.
#define STR "test1"
sizeof(STR);
is actually seen by the compiler as
sizeof("test1");
now why is that 6 and not 5? because there's a null terminator at the end of the string.
miked
2010-09-30 22:33:42
+1
A:
What the others said ... BUT
In C, preprocessing tokens take no space. It depends on how you use them
#define STR "test1"
char x[] = STR; /* 6 bytes */
char *y = STR; /* sizeof (char*) bytes (plus possibly 6 bytes) */
int ch = STR[3]; /* 1 byte (or sizeof (int), depending on how you look at it) */
if (ch == STR[1]) /* 1 byte (or sizeof (int) or no bytes or ...) */
printf("==>" STR "<==") /* 5 bytes ??? */
pmg
2010-10-01 00:23:06
A:
The latest C compiler has a feature to guess if the person writing the program is in a learning phase and give answers which make them search wider and deeper, and thus enrich their knowledge.
After programming for some time, depending of your learning, you might see the value go down to 5. ;-)
JK.. as someone else said, it symbolically nothing at the end which ironically takes a byte.
Rajan
2010-10-01 00:38:34