错误：函数被视为微观 [英] error: function is taken as an micro
问题描述
大家好，
i使用一个名为minor的函数来查找
矩阵元素的次要元素，
i声明为：
float minor（float A [] [COL]，int m，int n，int i，int j）; //找到未成年人
在A [] []
并在文件中稍后定义为：
float minor（float A [] [COL]，int m，int n，int i，int j）
{
if（m == 0 || n == 0）//单个coloumn或单行
返回1.0;
float temp [ROW] [COL]; //用于查找mod的临时矩阵（将有一行
和\
//小于父矩阵
int p，p1，q，q1; //索引
for（p = 0，p1 = 0; p <= m&& p1< =（m-1）; p ++，p1 ++）{
if（p == i）{
p1--;
继续; //忽略行i
}
for（q = 0，q1 = 0; q< = n&& q1< =（n-1）; q ++，q1 ++）{
if（q == j）{
q1 - ;
继续; //忽略coloumn j
}
temp [p1] [q1] = A [p] [q];
}
}
float x = mod（temp，m-1，n-1）; //返回temp的决定因素
返回x;
}
当我使用gnu g ++编译它时它给出了以下错误：
solve_equation.cpp:13:51：错误：宏次要传递了5个参数，但是
只需要1
solve_equation.cpp：68：49：错误：宏次要传递了5个参数，但
仅需1个
solve_equation.cpp：76：51：错误：宏次要传递了5个参数，但是
只需要1
solve_equation.cpp：121：51：错误：宏次要通过了5个参数，
但只需要1
solve_equation.cpp：76：错误：无效的函数声明
因为看起来编译器正在服用函数minor（）作为宏.....
但为什么???
以及函数声明如何无效？
i在fedora c8上使用gnu g ++ ........
谢谢
mohan gupta
hello everyone,
i use a function called minor to find the minor of an element of a
matrix ,
i declare it as :
float minor(float A[ ][COL],int m,int n,int i,int j);//find the minor
in A[][]
and define it later in the file as :
float minor(float A[][COL],int m,int n,int i,int j)
{
if(m==0 ||n==0)//single coloumn or single row
return 1.0;
float temp[ROW][COL];//temp matrix used to find mod (will have one row
and \
//less than the parent matrix
int p,p1,q,q1;//indexes
for(p=0,p1=0;p<=m&& p1<=(m-1);p++,p1++){
if(p==i){
p1--;
continue;//ignore row i
}
for(q=0,q1=0;q<=n && q1<=(n-1);q++,q1++){
if(q==j){
q1--;
continue;//ignore coloumn j
}
temp[p1][q1]=A[p][q];
}
}
float x= mod(temp,m-1,n-1);//return the determinant of temp
return x;
}
when i compile it using gnu g++ it gives the following error :
solve_equation.cpp:13:51: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:68:49: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:76:51: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:121:51: error: macro "minor" passed 5 arguments,
but takes just 1
solve_equation.cpp:76: error: invalid function declaration
as it seems the compiler is taking function minor() as macro .....
but why???
and how is the function declaration invalid??
i use gnu g++ on fedora c8........
thank you
mohan gupta
推荐答案
8月21日，12：19 * pm，mohi< mohangupt ... @ gmail.comwrote：
On Aug 21, 12:19*pm, mohi <mohangupt...@gmail.comwrote:
因为看起来编译器正在将函数minor（）作为宏.....
但为什么???
以及函数声明如何无效？
as it seems the compiler is taking function minor() as macro .....
but why???
and how is the function declaration invalid??
这不是一个标准的宏;它是您的代码特有的东西，
您的系统或第三方库。你可以在你的
标题中#undef：
#ifdef minor
#undef minor
#endif
干杯！ --M
That''s not a standard macro; it''s something specific to your code,
your system, or third-party libraries. You can #undef it in your
header:
#ifdef minor
# undef minor
#endif
Cheers! --M
2008年8月21日星期四09:19:40 -0700，mohi写道：
On Thu, 21 Aug 2008 09:19:40 -0700, mohi wrote:
大家好，
i使用一个名为minor的函数来查找
矩阵元素的次要元素，
i声明它as：
float minor（float A [] [COL]，int m，int n，int i，int j）; //在
$ b中找到未成年人$ b A [] []
hello everyone,
i use a function called minor to find the minor of an element of a
matrix ,
i declare it as :
float minor(float A[ ][COL],int m,int n,int i,int j);//find the minor in
A[][]
[...]
[...]
当我使用gnu编译它时g ++它给出以下错误：
solve_equation.cpp:13:51：错误：宏次要传递了5个参数，但是
只需要1
solve_equation.cpp：68：49：错误：宏次要传递了5个参数，但
仅需1个
solve_equation.cpp：76：51：错误：宏次要传递了5个参数，但是
只需要1
solve_equation.cpp：121：51：错误：宏次要传递了5个参数，但是
只需要1
solve_equation.cpp：76：错误：函数声明无效
似乎编译器将函数minor（）作为宏.....但是
为什么???
when i compile it using gnu g++ it gives the following error :
solve_equation.cpp:13:51: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:68:49: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:76:51: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:121:51: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:76: error: invalid function declaration
as it seems the compiler is taking function minor() as macro ..... but
why???
也许真的*是*在某处定义了一个名为minor的宏？尝试
给你的函数一个不同的名字，看它是否编译。或插入
一些代码如：
#ifdef minor
#error macro minor已定义！
$ b在你的函数声明之前$ b #endif
。
Maybe there really *is* a macro called `minor'' defined somewhere? Try
giving your function a different name and see if it compiles. Or insert
some code like:
#ifdef minor
#error macro minor already defined!
#endif
before your function declaration.
以及函数声明如何无效？
and how is the function declaration invalid??
如果在
声明中定义了一个宏次要，它将无效，因为预处理器将替换
中的'minor''宏在编译器看到之前你的函数声明。
-
Lionel B
8月21日，9：37 pm，Lionel B< m ... @ privacy.netwrote：
On Aug 21, 9:37 pm, Lionel B <m...@privacy.netwrote:
2008年8月21日星期四09:19:40 -0700，mohi写道：
On Thu, 21 Aug 2008 09:19:40 -0700, mohi wrote:
大家好，
hello everyone,
i使用一个名为minor的函数来查找
矩阵元素的次要元素，
i将其声明为：
i use a function called minor to find the minor of an element of a
matrix ,
i declare it as :
float minor（float A [] [COL]，int m，int n，int i，int j）; //在
中找到未成年人A [] []
float minor(float A[ ][COL],int m,int n,int i,int j);//find the minor in
A[][]
[...]
[...]
当我编译它时g gnu g ++它给出了以下错误：
when i compile it using gnu g++ it gives the following error :
solve_equation.cpp：13：51：error：macro" minor"传递了5个参数，但是
只需要1
solve_equation.cpp：68：49：错误：宏次要传递了5个参数，但
仅需1个
solve_equation.cpp：76：51：错误：宏次要传递了5个参数，但是
只需要1
solve_equation.cpp：121：51：错误：宏次要传递了5个参数，但
仅需1
solve_equation.cpp：76：错误：无效函数声明
solve_equation.cpp:13:51: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:68:49: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:76:51: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:121:51: error: macro "minor" passed 5 arguments, but
takes just 1
solve_equation.cpp:76: error: invalid function declaration
因为看起来编译器正在将函数minor（）作为宏.....但是
为什么???
as it seems the compiler is taking function minor() as macro ..... but
why???
也许真的*是*在某处定义了一个名为minor的宏？尝试
给你的函数一个不同的名字，看它是否编译。或插入
一些代码如：
#ifdef minor
#error macro minor已定义！
$ b在你的函数声明之前$ b #endif
。
Maybe there really *is* a macro called `minor'' defined somewhere? Try
giving your function a different name and see if it compiles. Or insert
some code like:
#ifdef minor
#error macro minor already defined!
#endif
before your function declaration.
以及函数声明如何无效？
and how is the function declaration invalid??
如果在
声明中定义了一个宏次要，它将无效，因为预处理器将替换
中的'minor''宏在编译器看到之前你的函数声明。
-
Lionel B
It will be invalid if there is a macro `minor'' defined at the point of
declaration, since the preprocessor will substitute the `minor'' macro in
your function declaration before the compiler sees it.
--
Lionel B
谢谢你们......
这是未成年人被宣布为宏的真正问题
某处:::
但我从来没有把它定义为微型和下级我使用任何第三个
派对库，
起始包括和我的代码的函数声明如下：
#include< iostream>
#include< cmath>
#include< cstdlib> ;
使用命名空间std;
#define ROW 20
#define COL 20
#ifdef minor
#undef minor
#endif
void get_array（float [ROW] [COL]， int *，int *）; //获取数组中的输入
void inverse（float [] [COL]，int，int）; //反转arary
void multiply（float [] [COL]，int，int，float [ ] [COL]，int，int，float
result [] [COL]）; //乘以两个数组
float minor（float A [] [COL]， int m，int n，int i，int j）; //在A []中找到次要的
[]
float mod（float A [] [COL] ，int m，int n）;
void transpose（float temp [] [COL]，int m，int n，float A [] [COL]）; // find
中的临界温度$
void显示（浮动A [] [COL]，int m，int n）; //显示数组A [] []
void divide（float array [] [COL]，int m，int n，float value）; //划分
矩阵数组
// value
（其中我现在未定义次要）,,,,但我知道所有标准
libarray变量用下划线声明
说" __micro（我知道它不是一个坚硬而快速的规则）所以
哪里可以宣布????它可以在cmath ????
谢谢
mohan
thank you guys ...
that was the real problem that minor was declared as macro
somewhere:::
but i never defined it as an micro and nether am i using any third
party library ,
the starting includes and function declarations of my code are as:
#include<iostream>
#include<cmath>
#include<cstdlib>
using namespace std;
#define ROW 20
#define COL 20
#ifdef minor
#undef minor
#endif
void get_array(float [ROW][COL],int* ,int* );//get the input in array
void inverse(float [][COL],int ,int);//inverse the arary
void multiply(float [][COL],int,int,float [][COL],int,int,float
result[][COL] );//multiply two array
float minor(float A[][COL],int m,int n,int i,int j);//find the minor
in A[][]
float mod(float A[][COL],int m,int n);
void transpose(float temp[][COL],int m,int n,float A[][COL]);//find
trnpos of temp in A
void display(float A[][COL],int m,int n);//display array A[][]
void divide(float array[][COL],int m,int n,float value);//divide
matrix array by
//value
(where i now undefined minor) ,,,,but as i know all the standard
libarray variable are declared with an underscore
say" __micro "(i know that its not an hard and fast rule though) so
where could it have been declared ????can it be in cmath????
thank you
mohan
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