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# Array problem

Posted on 1998-12-14
Medium Priority
182 Views
My code
=======
#include<iostream.h>

void identity(int a[4][4], int m){
int i=m, j=m;
for(i=0;i<4;i++){
for(j=0;j<4;j++){
if(i==j)
a[i][j]=1;
else
a[i][j]=0;
return a[i][j];

}

}
}

void main(){
int a[4][4];
int m=4;
identity(a[4][4], m);
cout<a[i][j]
}

cout<<'\n';
}

My problem
==========

Write a function 'identity' which returns a 1 if the array argument passed to it represents an identity matrix, and returns 0 otherwise. An identity matrix is an m by m array of integers, where the values of the elements on the principle diagonal (row subscript == col subscript)equal 1, and all other elements are 0. for
example:
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
is an identity matrix of order 4 by 4,
1 0 0
0 1 0
0 0 1
is an identity matrix of order 3 by 3,and
1 1 5
0 1 0
0 0 1
is NOT an identity matrix. The function identity is passed two arguments: the array, and the order of the matrix ( i e the size of the two dimensions). Thus the prototype of the function would be written as :int identity (int a[][], int m);where a is the array and m represents the number of rows and columns. You may assume that the number of rows always equals the number of columns for the array.
0
Question by:muhbest
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LVL 10

Accepted Solution

RONSLOW earned 150 total points
ID: 1180179
#include <iostream.h>

/* no point in passing int m=4 if a[4][4] is hard coded */
/* the question you were given is not possible to answer */
/* exactly as specified.  but we cannot do much */

/* also the function should return an int not a void */
/* void identity(int a[4][4], int m) { */
int identity(int a[4][4], int m) {
/* no point initializing i and j as they are */
/* set in the loop */
/*      int i=m;      */
/*      int j=m;      */
int i;
/* you have hard-coded 4 in the for loops */
/*      for(i=0;i<4;i++){      */
/*            for(j=0;j<4;j++){      */
/* we should use 'm' instead */
for(i=0;i<m;i++){
/* declare j here where it is used */
/* this is nicer code */
int j;
for(j=0;j<m;j++){
/* this is CREATING an identity matrix */
/* NOT testing for one as was asked */
if(i==j)
a[i][j]=1;
else
a[i][j]=0;
/* this is not what should be returned */
return a[i][j];
}
}
/* no return value from here at all */
/* a good compiler will detect this error */
}

int main(int argc, char* argv[]) {
/* you haven't put any data into a[][] yet */
int a[4][4];
int m=4;
/* you haven't done anything with the */
/* return value from identity */
/* also the call is wrong .. you are passing */
/* the value of the [4][4] item of the array */
/* (which is outside it anyway) instead of the */
/* whole array */
/*      identity(a[4][4], m); */
int result = identity(a, m);
/* i and j aren't defined */
/* and you should be printing the result of */
/* the call to identity(...); */
/* BTW: if this is supposed to be 'C' code */
/* then you cannot use 'cout<<' etc */
/* cout<<a[i][j]; */
cout << result;
cout<<'\n';
return 0;
}

/* I suggest you think about the question */
/* a bit more and come back to me */

0

LVL 10

Expert Comment

ID: 1180180
PS: sorry about formatting .. that is what happens when I copy/paste source code into this silly edit box :-(

0

LVL 22

Expert Comment

ID: 1180181
RONSLOW, I think the assignment might expect a 1-D array that is to be treated as a 2D matrix, that is the only way to make it sizable at run-time. (That or a 1-D array of 1-D array pointers, but that really needs to be a class.)
0

LVL 10

Expert Comment

ID: 1180182
It could also be that the lecturer who posed the assignment stuffed up by saying "the prototype of the function would be written as :int identity (int a[][], int m)"

Anyway, as posed in the question, one cannot write such as function (it ain't valid C or C++).  If a was a flat array of m*m elements then that could be easily done.

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