buf [n] ++是左值吗？ [英] Is buf[n]++ a lvalue?
问题描述
以下代码给出了错误C2105：'''''''需要l值。
char buf [20];
buf [n ] ++ =''A''; // buf [n] ++ isn''ta lvalue
为什么buf [n] ++不是左右？这是根据C标准还是
这样的原因？
如果buf [n] ++不是左值，那么为什么以下代码有效吗？
void func（char * a，char * b）
{
while（* a ++ = * b ++）{}
}
Stub< st ** @ asof.com>这样说：
char buf [20];
buf [n] ++ =''A''; // buf [n] ++ isn''ta左值
为什么buf [n] ++不是左值？这是根据C标准还是存在某种原因？
如果buf [n] ++不是左值，那么为什么以下代码有效？
while（* a ++ = * b ++）{}
因为* a ++与*（a ++）相同。 a是一个指针，所以
增加它是好的。然而，buf [n]是一个你无法增加的角色。
-
Christopher Benson-Manica |我*应该*知道我在说什么 - 如果我
ataru（at）cyberspace.org |不，我需要知道。火焰欢迎。
Stub写道：以下代码给出了错误C2105：''''''''需要l值。
char buf [20];
buf [n] ++ =''A''; // buf [n] ++ isn''ta lvalue
为什么buf [n] ++不是左右吗？这是根据C标准还是存在某种原因？
为什么你认为这两个选项是相互排斥的？
;-)。 C标准*中的事情是有原因的！
然而，最有可能的表达方式是：
buf [ n ++] =''A'';
如果buf [n] ++不是左值，那么为什么以下代码有效？
void func（char * a，char * b）
{while /（* a ++ = * b ++）{}
}
因为这里*指针*被解除引用（* a是左值，
是吗？），执行赋值（产生表达式的值）
最后每个指针都递增。
HTH，
- g
-
Artie Gold - 德克萨斯州奥斯汀
哦，对于常规老垃圾邮件的美好时光。
Christopher Benson-Manica写道：
但是，buf [n]是一个你无法增量的角色。
是的，你可以。
#include< stdio.h>
int main（无效）
{
char buf [] =" 3.1415926" ;;
int n = 0;
buf [n] ++;
printf（"％s \ nn"，buf）;
返回0;
}
-
Richard Heathfield： bi ****@eton.powernet.co.uk
Usenet是一个奇怪的地方。 - Dennis M Ritchie，1999年7月29日。
C FAQ： http://www.eskimo.com/~scs/C-faq/top.html
K& R答案，C书等：< a rel =nofollowhref =http://users.powernet.co.uk/etontarget =_ blank> http://users.powernet.co.uk/eton
The following code gives error C2105: ''++'' needs l-value.
char buf[20];
buf[n]++ = ''A'';//buf[n]++ isn''t a lvalue
Why buf[n]++ isn''t a lvalue? Is this according to C Standard or is there
some reason for this?
If buf[n]++ isn''t a lvalue, then why the following code works?
void func(char *a, char *b)
{
while(*a++=*b++){}
}
Stub <st**@asof.com> spoke thus:
char buf[20];
buf[n]++ = ''A'';//buf[n]++ isn''t a lvalue Why buf[n]++ isn''t a lvalue? Is this according to C Standard or is there
some reason for this? If buf[n]++ isn''t a lvalue, then why the following code works? while(*a++=*b++){}
Because *a++ is the same as *(a++). a is a pointer and so it''s fine to
increment it. buf[n], however, is a character, which you cannot increment.
--
Christopher Benson-Manica | I *should* know what I''m talking about - if I
ataru(at)cyberspace.org | don''t, I need to know. Flames welcome.
Stub wrote:The following code gives error C2105: ''++'' needs l-value.
char buf[20];
buf[n]++ = ''A'';//buf[n]++ isn''t a lvalue
Why buf[n]++ isn''t a lvalue? Is this according to C Standard or is there
some reason for this?
Why would you think that these two options are mutually exclusive?
;-). Things in the C standard *are* there for a reason!
Most likely, however, the expression you want is:
buf[n++] = ''A'';
If buf[n]++ isn''t a lvalue, then why the following code works?
void func(char *a, char *b)
{
while(*a++=*b++){}
}
Because here the *pointers* are dereferenced (*a is an lvalue,
yes?), the assignment is performed (yielding the expression''s value)
and finally each pointer is incremented.
HTH,
--ag
--
Artie Gold -- Austin, Texas
Oh, for the good old days of regular old SPAM.
Christopher Benson-Manica wrote:
buf[n], however, is a character, which you cannot
increment.
Yes, you can.
#include <stdio.h>
int main(void)
{
char buf[] = "3.1415926";
int n = 0;
buf[n]++;
printf("%s\n", buf);
return 0;
}
--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
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