gcc：这有什么不合法的？ [英] gcc: What's illegal about this ?
问题描述
我需要一个线索。我得到一个增量无效左值来自gcc
4.1.2在这一行...
sum + = *（（unsigned short *）iphp）++;
....但我不知道问题是什么。 gcc 3.4.3也没有打扰它。
机器细节：
$ uname -a
Linux tmxnw-fc6 2.6.22.9-61.fc6＃1 SMP Thu Sep 27 18:48:03 EDT 2007
i686 i686 i386 GNU / Linux
$ gcc --version
gcc（GCC）4.1.2 20070626（Red Hat 4.1.2-13）
I need a clue. I''m getting an "invalid lvalue in increment" from gcc
4.1.2 on this line...
sum += *((unsigned short *)iphp)++ ;
....but I don''t see what the problem is. gcc 3.4.3 isn''t bother by
it, either.
Machine specifics:
$ uname -a
Linux tmxnw-fc6 2.6.22.9-61.fc6 #1 SMP Thu Sep 27 18:48:03 EDT 2007
i686 i686 i386 GNU/Linux
$ gcc --version
gcc (GCC) 4.1.2 20070626 (Red Hat 4.1.2-13)
推荐答案
uname -a
Linux tmxnw-fc6 2.6.22.9-61.fc6＃1 SMP Thu Sep 27 18:48:03 EDT 2007
i686 i686 i386 GNU / Linux
uname -a
Linux tmxnw-fc6 2.6.22.9-61.fc6 #1 SMP Thu Sep 27 18:48:03 EDT 2007
i686 i686 i386 GNU/Linux
gcc --version
gcc（GCC） 4.1.2 20070626（Red Hat 4.1.2-13）
gcc --version
gcc (GCC) 4.1.2 20070626 (Red Hat 4.1.2-13)
2008年3月14日星期五17:49:16 +0000，Richard Eich写道：
On Fri, 14 Mar 2008 17:49:16 +0000, Richard Eich wrote:
我需要一个线索。我得到一个增量无效左值来自gcc
4.1.2在这一行...
sum + = *（（unsigned short *）iphp）++;
......但我不知道问题是什么。 gcc 3.4.3并不打扰它，
。
I need a clue. I''m getting an "invalid lvalue in increment" from gcc
4.1.2 on this line...
sum += *((unsigned short *)iphp)++ ;
...but I don''t see what the problem is. gcc 3.4.3 isn''t bother by it,
either.
（无符号短*）iphp不是左值。你想在哪里存储
（unsigned short *）iphp + 1？据推测，您希望将其存储在iphp中。如果你想要那个，那就写下来吧。
sum + = *（（unsigned short *）iphp）;
iphp =（（unsigned short *）iphp）+ 1;
你可以认为它类似于
int i;
（ i + 1）= -1;
（双倍）i = 1.5;
(unsigned short *)iphp is not an lvalue. Where do you want to store
(unsigned short *)iphp + 1? Presumably, you want to store it in iphp. If
you want that, write that.
sum += *((unsigned short *)iphp);
iphp = ((unsigned short *)iphp) + 1;
You can consider it similar to
int i;
(i + 1) = -1;
(double) i = 1.5;
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