i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
but why the seemed same thing doesn't work on case 2?
thanks 10 1556
On Sep 13, 12:34 pm, goacr...@gmail. com wrote:
i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
but why the seemed same thing doesn't work on case 2?
thanks
I run the program in vc :)
On Sep 13, 9:34 am, goacr...@gmail. com wrote:
i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
but why the seemed same thing doesn't work on case 2?
thanks
This is becuase,
sizeof('h')[p] will return the size of the expression ('h')[p] which
will always be 1 as the p is pointer to char.
On Wed, 12 Sep 2007 21:34:01 -0700, go******@gmail. com wrote:
>i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
but why the seemed same thing doesn't work on case 2?
Compare these two expressions:
char n=(sizeof('h'))[p]; (1)
char n=sizeof(('h')[p]); (2)
Your second expression is being interpreted as (2), and what you
wanted was (1)
The problem lies in sizeof beˇing greedy when it refers to an object
and not to a type
Best regrads,
Zara
Zara wrote, On 13/09/07 07:00:
On Wed, 12 Sep 2007 21:34:01 -0700, go******@gmail. com wrote:
>i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
but why the seemed same thing doesn't work on case 2?
Compare these two expressions:
char n=(sizeof('h'))[p]; (1)
This is the same as
char n = (sizeof 'h')[p];
This will still not do what the OP expected in C although it will in
C++. This is because a character literal like 'h' has type int in C
(probably 4 on the OPs system, although it could be any other number)
whereas in C++ it would have type char.
char n=sizeof(('h')[p]); (2)
Your second expression is being interpreted as (2), and what you
wanted was (1)
The problem lies in sizeof beˇing greedy when it refers to an object
and not to a type
Also note that sizeof is an operator, NOT a function. This is why you
can do "sizeof expression" rather than "sizeof(express ion)" and is
probably also part of the OPs confusion.
--
Flash Gordon
On Sep 13, 2:00 pm, Zara <me_z...@dea.sp amcon.orgwrote:
On Wed, 12 Sep 2007 21:34:01 -0700, goacr...@gmail. com wrote:
i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
but why the seemed same thing doesn't work on case 2?
Compare these two expressions:
char n=(sizeof('h'))[p]; (1)
char n=sizeof(('h')[p]); (2)
Your second expression is being interpreted as (2), and what you
wanted was (1)
The problem lies in sizeof beˇing greedy when it refers to an object
and not to a type
Best regrads,
Zara
Thank you all.
It's clear now. :)
On Wed, 12 Sep 2007 21:34:01 -0700, goacross wrote:
i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
but why the seemed same thing doesn't work on case 2?
sizeof('h') is sizeof(int). If it is four, 4+"abc" will point past
the end of the string, so dereferencing it, even implicitly with
[], will cause UB. If sizeof(int) is 4, even pointer arithmetic
itself will cause UB, even if you don't dereference. If
sizeof(int) < 4, n will be 'b', 'c', or 0.
--
Army1987 (Replace "NOSPAM" with "email")
If you're sending e-mail from a Windows machine, turn off Microsoft's
stupid “Smart Quotes” feature. This is so you'll avoid sprinkling garbage
characters through your mail. -- Eric S. Raymond and Rick Moen
On Wed, 12 Sep 2007 21:34:01 -0700, go******@gmail. com wrote:
>i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
*(1+p) = *(p+1) = p[1] = 1[p].
Remove del for email
On Fri, 14 Sep 2007 12:35:41 +0200, Army1987 <ar******@NOSPA M.it>
wrote:
>On Wed, 12 Sep 2007 21:34:01 -0700, goacross wrote:
>i found something tricky this morning.
char *p="abc";
1. char m=1[p];// m='b'
2. char n=sizeof('h')[p]; // n=1;
I guess the reason of 1is,
1+p=p+1; so the same as p[1];
but why the seemed same thing doesn't work on case 2?
sizeof('h') is sizeof(int). If it is four, 4+"abc" will point past the end of the string, so dereferencing it, even implicitly with [], will cause UB. If sizeof(int) is 4, even pointer arithmetic itself will cause UB, even if you don't dereference. If sizeof(int) < 4, n will be 'b', 'c', or 0.
Except for variable length arrays (not germane to this thread), sizeof
never evaluates its operand. Consequently, nothing gets dereferenced.
The operand is analyzed only to determine its type, not whether it
actually exists.
Remove del for email
On Fri, 14 Sep 2007 14:18:21 -0700, Barry Schwarz wrote:
On Fri, 14 Sep 2007 12:35:41 +0200, Army1987 <ar******@NOSPA M.it>
wrote:
>>On Wed, 12 Sep 2007 21:34:01 -0700, goacross wrote:
>>2. char n=sizeof('h')[p]; // n=1;
sizeof('h') is sizeof(int). If it is four, 4+"abc" will point past the end of the string, so dereferencing it, even implicitly with [], will cause UB. If sizeof(int) is 4, even pointer arithmetic itself will cause UB, even if you don't dereference. If sizeof(int) < 4, n will be 'b', 'c', or 0.
Except for variable length arrays (not germane to this thread), sizeof
never evaluates its operand. Consequently, nothing gets dereferenced.
The operand is analyzed only to determine its type, not whether it
actually exists.
Yeah, damned syntax... sizeof('h')[p] means sizeof(('h')[p]), not
(sizeof('h'))[p] as I was thinking when I wrote that reply...
--
Army1987 (Replace "NOSPAM" with "email")
If you're sending e-mail from a Windows machine, turn off Microsoft's
stupid “Smart Quotes” feature. This is so you'll avoid sprinkling garbage
characters through your mail. -- Eric S. Raymond and Rick Moen This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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