[horsx]

Hi!
im trying to find the largest number from a selected row of an matrix. My code is ok until i insert negative numbers. I don't know what to write for solvinng this...

#include<iostream>
#include<cstdlib>
using namespace std;
int main ()
{

int rang=10,i,j, M[rang][rang],n,k,max=0;
char answer;

do{ 

do{cout<<"Insert the matrix rank(max. 10): "; cin>>n; }
while(n<1||n>rang);

cout<<"Insert elements: "<<endl;
for(i=0;i<n;i++)
for(j=0;j<n;j++){cout<<"M["<<i<<"]["<<j<<"]= "; cin>>M[i][j];}
cout<<endl;

for(int i=0; i<n;i++){ for(int j=0;j<n;j++) cout<<"\t"<<M[i][j]; cout<<"\n\n";}

do{ cout<<"Select a row: "; cin>>k;}
while(k<0||k>n);

for(j=0;j<n;j++) 
if (max<M[k-1][j]){max=M[k-1][j];}

cout<<"The largest number of the "<<k<<"  row is "<<max<<endl ;

cout << "Try again? (Y/N): "; 
cin >> answer;
cout << endl;
}
while(answer == 'Y' || answer == 'y');

}

[ishkabible]

'max' starts out at 0 so it will never be less than a negative value hence it will always say '0' if all the values are negative; you need to set 'max' to the first value in the row before the loop. that way 'max' starts out with a possible answer and not something invalid(like 0 in an all negative row).

while that is a problem you have several more.

1) you need an indentation style; i took the liberty of formatting your code(so I could read it)

#include<iostream>
#include<cstdlib>

using namespace std;

int main () {
	int rang=10,i,j, M[10][10],n,k,max=0;
	char answer;
	do{
		do {
			cout<<"Insert the matrix rank(max. 10): "; cin>>n;
		} while(n < 1 || n > rang);

		cout<<"Insert elements: "<<endl;
		for(i=0;i<n;i++) {
			for(j=0;j<n;j++) {
				cout<<"M["<<i<<"]["<<j<<"]= ";
				cin>>M[i][j];
			}
		}
		cout<<endl;

		for(int i=0; i<n;i++) {
			for(int j=0;j<n;j++) {
				cout<<"\t"<<M[i][j];
			}
			cout<<"\n\n";
		}

		do {
			cout<<"Select a row: ";
			cin>>k;
		} while(k<0||k>n);

		for(j=0;j<n;j++) {
			if (max<M[k-1][j]) {
				max=M[k-1][j];
			}
		}
		cout<<"The largest number of the "<<k<<"  row is "<<max<<endl ;
		cout << "Try again? (Y/N): ";
		cin >> answer;
		cout << endl;
	} while(answer == 'Y' || answer == 'y');
}

2) you can't create an array with a value not known at compile time.

int rang=10,i,j, M[rang][rang],n,k,max=0;

that is illegal in all standards of C++, only C99 and C11 offer variable length arrays.
instead it should be...

const int rang = 10;
int i, j, M[rang][rang], n, k, max=0;

then rang is known at compile time this way.

3) you redeclare(shadow) i and j a few times; just declare them were they are used(in for loops). shadowing variables leads to bad things.

This post has been edited by ishkabible: 14 February 2012 - 06:14 PM

[horsx]

Thanks for the answer and for solving the other problems.
So from what you are saying the code should be

max=M[k-1][j]
for(int j=0;j<n;j++) {
36
            if (max<M[k-1][j]) {
37
                max=M[k-1][j];
38
            }
39
        }
40
        cout<<"The largest number of the "<<k<<"  row is "<<max<<endl ;

I have another question. If i type a character when i insert the matrix rank or elements i'm stuck in a infinite loop. What should i do to avoid that?

[jimblumberg]

ishkabible, on 14 February 2012 - 07:09 PM, said:

2) you can't create an array with a value not known at compile time.

int rang=10,i,j, M[rang][rang],n,k,max=0;

that is illegal in all standards of C++, only C99 and C11 offer variable length arrays.

Variable Length Arrays are only guaranteed in C99, in C11 support is not required. See the following section of the C11 draft standard.

Quote

6.10.8.3 Conditional feature macros
The following macro names are conditionally defined by the implementation:
...
_ _STDC_NO_VLA_ _ The integer constant 1, intended to indicate that the
implementation does not support variable length arrays or variably
modified types.

Jim

[JackOfAllTrades]

Quote

I have another question. If i type a character when i insert the matrix rank or elements i'm stuck in a infinite loop. What should i do to avoid that?

FAQ Link

[ishkabible]

@Jim: good catch, I guess when I read about it I didn't look closely enough at it.