优先权 [英] precedence
问题描述
声明:
foo.b [0] = 7;
表示数据的数组成员的第一个元素结构体。我的
问题是,为什么在优先级图表中括号更高之后,首先是点运算符
之后是否先评估括号?谢谢
the statement:
foo.b[0] = 7;
indicates the first element of the array member of the data structure. My
question is, why aren''t the brackets evaluated first then the dot operator
after since the brackets are higher on the precedence chart? Thanks
推荐答案
" Andy White" <峰; br *********** @ msn.com> schrieb im Newsbeitrag
新闻:10 ************* @ corp.supernews.com ...
"Andy White" <br***********@msn.com> schrieb im Newsbeitrag
news:10*************@corp.supernews.com...
声明:
foo.b [0] = 7;
表示数据结构的数组成员的第一个元素。我的问题是,为什么在优先级图表中括号更高之后,首先不会对点
运算符进行括号评估?谢谢
the statement:
foo.b[0] = 7;
indicates the first element of the array member of the data
structure. My
question is, why aren''t the brackets evaluated first then the dot
operator
after since the brackets are higher on the precedence chart? Thanks
呃...首先评估大括号的内容:
long fkt(void)
{
printf(" fkt");
返回0;
}
class FOO
{
public:
long& B(){printf(" access B");返回b;}
长b;
};
FOO foo [100];
foo.B()[fkt()] = 7;
Er... The content of the braces _is_ evaluated first:
long fkt(void)
{
printf("fkt");
return 0;
}
class FOO
{
public:
long& B(){printf("access B "); return b;}
long b;
};
FOO foo[100];
foo.B()[fkt()] = 7;
在文章< 10 ************* @ corp。 supernews.com>,Andy White
< br *********** @ msn.com>写
In article <10*************@corp.supernews.com>, Andy White
<br***********@msn.com> writes
语句:
foo.b [0] = 7;
表示数据结构的数组成员的第一个元素。我的问题是,为什么在优先级图表上的括号更高之后,首先不会对括号运算符进行括号评估?谢谢
the statement:
foo.b[0] = 7;
indicates the first element of the array member of the data structure. My
question is, why aren''t the brackets evaluated first then the dot operator
after since the brackets are higher on the precedence chart? Thanks
因为你从名字开始,在这种情况下''foo''下标
运算符不适合foo,并且b不是自由变量,但是在foo的上下文中确定了一个
。
-
Francis Glassborow ACCU >
你可以做到!的作者参见 http:// www .spellen.org / youcandoit
对于项目构想和贡献: http://www.spellen.org/youcandoit/projects
Because you start from the name, in this case ''foo'' The subscript
operator is not appropriate to foo, and b isn''t a free variable but one
that is identified in the context of foo.
--
Francis Glassborow ACCU
Author of ''You Can Do It!'' see http://www.spellen.org/youcandoit
For project ideas and contributions: http://www.spellen.org/youcandoit/projects
" Andy White" <峰; br *********** @ msn.com>在消息中写道
news:10 ************* @ corp.supernews.com ...
"Andy White" <br***********@msn.com> wrote in message
news:10*************@corp.supernews.com...
声明:
foo.b [0] = 7;
表示数据结构的数组成员的第一个元素。我的问题是,为什么在优先级图表上的括号更高之后,首先不会对括号运算符进行括号评估?谢谢
the statement:
foo.b[0] = 7;
indicates the first element of the array member of the data structure. My
question is, why aren''t the brackets evaluated first then the dot operator
after since the brackets are higher on the precedence chart? Thanks
A''[]''运算符的优先级不高于''。''运算符。他们
具有相同的优先权。
问候,
Sumit。
A ''[]'' operator does not have higher precedence than a ''.'' operator. They
have the SAME precedence.
Regards,
Sumit.
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