View Full Version : School assignment, arrays/matrices, help please :(

VanillaCone

November 6th, 2010, 02:47 AM

The assignment is to write several functions to do with matrices, The current function im having trouble with is the trace of a matrix, Which Is the sum Of numbers on the main diagonal.

so far i have

int trace (int **A, int n)

{

int i;

int j;

int sum;

for (i = 0; i < n; i++)

{

sum = 0;

for (j = 0; j < n; j++)

{

if (i = j)

sum += A[i][j];

}

}

return sum;

}

for simplicities Sake The matrix is a 4x4 and it is

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4.

when i run This program It tells me the trace of my matrix ( 1 + 2 + 3 +4) Is 9, when in reality it is 10, the program does not include the first value of the first row In its calculation and im not sure why, please help =(

worksofcraft

November 6th, 2010, 02:51 AM

The assignment is to write several functions to do with matrices, The current function im having trouble with is the trace of a matrix, Which Is the sum Of numbers on the main diagonal.

so far i have

int trace (int **A, int n)

{

int i;

int j;

int sum;

for (i = 0; i < n; i++)

{

sum = 0;

for (j = 0; j < n; j++)

{

if (i = j)

sum += A[i][j];

}

}

return sum;

}

for simplicities Sake The matrix is a 4x4 and it is

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4.

when i run This program It tells me the trace of my matrix ( 1 + 2 + 3 +4) Is 9, when in reality it is 10, the program does not include the first value of the first row In its calculation and im not sure why, please help =(

I think you want to move the sum = 0 outside BOTH of the loops, but I also think it would be much more efficient to just use one loop:

sum = 0;

for (i = 0; i < n; ++i) sum += A[i][i];

;)

p.s. and another thing to watch: the single = is an assignment so if (i = j) is actually setting i to be equal to j and then testing if it is zero. I suspect you wanted == which is the comparison operator

VanillaCone

November 6th, 2010, 02:54 AM

moving sum = 0; outside the loop did not solve the problem =(

worksofcraft

November 6th, 2010, 02:56 AM

moving sum = 0; outside the loop did not solve the problem =(

No I realized you need to change that = into == and just added a p.s. as you were typing :shock:

VanillaCone

November 6th, 2010, 02:59 AM

no doesnt work =(, It just refuses to Take that 1 Into the calculation

edit - maybe i should mention this is in C.

VanillaCone

November 6th, 2010, 03:13 AM

I fixed it but in a really stupid way

int trace (int **A, int n)

{

int i;

int j;

int sum;

for (i = 0; i < n; i++)

{

sum = 0;

for (j = 0; j < n; j++)

{

if (i == 0 && j == 0)

sum = sum + A[i][j];

else if (i = j)

sum += A[i][j];

}

}

return sum;

}

it now calculates properly But there HAS to be a less ugly solution to this, If anyone has one please let me know

worksofcraft

November 6th, 2010, 03:16 AM

I fixed it but in a really stupid way

int trace (int **A, int n)

{

int i;

int j;

int sum;

for (i = 0; i < n; i++)

{

sum = 0;

for (j = 0; j < n; j++)

{

if (i == 0 && j == 0)

sum = sum + A[i][j];

else if (i = j)

sum += A[i][j];

}

}

return sum;

}

it now calculates properly But there HAS to be a less ugly solution to this, If anyone has one please let me know

try this one:

// gcc trace.c

#include <stdio.h>

int trace (int **A, int n)

{

int i;

int j;

int sum;

sum = 0;

for (i = 0; i < n; i++)

{

for (j = 0; j < n; j++)

{

if (i == j)

sum += A[i][j];

}

}

return sum;

}

int main() {

int Row[] = { 1, 2, 3, 4 }; // a row of matrix data

int *A[] = { Row, Row, Row, Row }; // all 4 rows are the same

return !printf("sum = %d\n", trace(A, 4));

}

TiBaal89

November 6th, 2010, 03:24 AM

Were you specifically instructed to do the loops in that manner? It's sort of silly. You're running through all n^2 numbers when you're only interested in the diagonal. You could do something like (just typing this off the top of my head to give an idea):

sum = 0;

for (i=0; i<n; i++) {

sum += A[i][i];

}

VanillaCone

November 6th, 2010, 03:25 AM

i cant make my own Main function, Im forced To work with the one provided by the assignment ^.^

Tony Flury

November 6th, 2010, 09:17 AM

None of the solutions posted require you to re-write the main function.

The main bug in your original code is :

if (i = j)

Should be

if (i == j)

The fact you are using two loops in your function is a very odd design decision when one loop will work (as demonstrated by TiBaal89), but if you have to use two loops then you need to make sure you use the right comparison operators.

Interestingly in your fix where you added

if (i == 0 && j == 0)

sum = sum + A[i][j];

you used the correct operator in your added code.

Note : Confusion over the use and differences between "=" and "==" is very common, and is often the cause of bugs in C code. Many C compilers have options whereby they can warn against the use of "=" in if statements as although it is illegal it is easy to introduce bugs. The difference is :

(i == j) // Comparison operator ==

will produce TRUE when variable i is the same value as variable j, but neither i or j are changed.

(i = j) // Assignment operator =

will cause variable i to be set to the same value as variable j, and that expression takes the value of i - so if j is 0, i becomes 0 and the expression is 0 (which of course is the same as FALSE) - which is why your code did not work initially.

lisati

November 6th, 2010, 09:24 AM

Just a thought: what assumptions can we make about the start index of the array is passed to the function?

edit: Friendly reminder: the CoC expects that we don't do homework for people, but providing hints when someone is stuck.

Arndt

November 6th, 2010, 12:21 PM

Others have pointed out that you've written = when it should be ==. You can let the compiler help you with this by asking for warnings:

gcc -Wall -c code.c

code.c: In function ‘trace’:

code.c:16: warning: suggest parentheses around assignment used as truth value

Take the time to look at the manual page for gcc (or whatever compiler you're using); there may be many warnings that are useful for you (now or in the future), beyond the ones enabled by -Wall.

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