f[10] == *(f + sizeof(char) * 10); // if f is a char array
Watch out, if sizeof(char) is not 1 then you are in serious trouble. Doing + with a pointer increases it in increments of the size of the pointer, you don''t have to do it yourself.
difference between [] and *
Also, be very careful modifying those variables.
The first one could be a pointer to constant memory, whereas the second one could not.
char *temp = "Hello World.";
char blah[] = "Hello World.";
strtok(temp, " ,\t\n"); // Will probably crash.
strtok(blah, " ,\t\n"); // Won''t crash.
The first one could be a pointer to constant memory, whereas the second one could not.
char *temp = "Hello World.";
char blah[] = "Hello World.";
strtok(temp, " ,\t\n"); // Will probably crash.
strtok(blah, " ,\t\n"); // Won''t crash.
quote:Original post by C-Junkiequote:Original post by Raloth
Watch out, if sizeof(char) is not 1
It IS always 1. period.
... unless, say, you redefined char to be wchat_t, for say, internationalization ...
... or maybe some compilers that actually have internationalization support built in ...
he is trying to point out a bug in the code anyways. There shouldn''t be the sizeof statement in there at all.
aka: f[10] == *(f + 10) for any type.
but: f[10] == *(f + sizeof( type ) * 10) is incorrect.
... so no more random attacks for the hell of it ...
quote:Original post by MaulingMonkeyWhich, of course, you would NEVER do, because redefining keywords is disallowed by the standard...
... unless, say, you redefined char to be wchat_t, for say, internationalization ...
quote:... or maybe some compilers that actually have internationalization support built in ...Not that would change the size of char; this would break so much code, it''s not even funny.
quote:he is trying to point out a bug in the code anyways. There shouldn''t be the sizeof statement in there at all.
aka: f[10] == *(f + 10) for any type.
but: f[10] == *(f + sizeof( type ) * 10) is incorrect.
I''m pretty sure that SR was pointing exactly this out, implying that the compiler would automultiply by the type size. It would''ve been more clear with casts, tho:
f[10] == *(char*)((int)f + sizeof(char) * 10); // if f is a char array
quote:... so no more random attacks for the hell of it ...But it''s sooo much fun!
How appropriate. You fight like a cow.
quote:Original post by Sneftelquote:Original post by MaulingMonkeyWhich, of course, you would NEVER do, because redefining keywords is disallowed by the standard...
... unless, say, you redefined char to be wchat_t, for say, internationalization ...quote:... or maybe some compilers that actually have internationalization support built in ...Not that would change the size of char; this would break so much code, it''s not even funny.
Valid points, but my original point is it''s bad to assume
.quote:quote:he is trying to point out a bug in the code anyways. There shouldn''t be the sizeof statement in there at all.
aka: f[10] == *(f + 10) for any type.
but: f[10] == *(f + sizeof( type ) * 10) is incorrect.
I''m pretty sure that SR was pointing exactly this out, implying that the compiler would automultiply by the type size. It would''ve been more clear with casts, tho:
f[10] == *(char*)((int)f + sizeof(char) * 10); // if f is a char array
Not only more clear, but not in error for non char types, which would be the only reason for including the sizeof(char) item in the first place.
quote:quote:... so no more random attacks for the hell of it ...But it''s sooo much fun!
How appropriate. You fight like a cow.
* goes into a fury *
actually...yeah it IS fun
.
There was a post somewhere explaining that char doesn''t have to be 1 byte by the standards. It just happens that the major compilers make it 1, and all the characters you could want fit into 1 byte. But technically it can be greater than 1...
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