Reputation: 35
I was working on the following 2d-array program to output this result shown in picture:
I can't seem to get the min value for the result and get it displayed in array form. The code is below:
#include<stdio.h>
#define NUMROWS 2
#define NUMCOLS 3
//accessing elements of 2D array using pointers
int main(void){
const int table[NUMROWS][NUMCOLS]={{1,2,3},{5,6,7}};
int minvals[NUMROWS];
int i, j;
int *ptr = &table;
//accessing the elements of 2D array using ptr
printf("table values: min value\n");
for(int i=0;i<NUMROWS;i++){
for(int j=0;j<NUMCOLS;j++)
printf("%d ",*((ptr+i*NUMCOLS)+j));
printf("\n");
}
for(int i=0;i<NUMROWS;i++){
for(int j=0;j<NUMCOLS;j++)
printf("%d ",*((ptr+i*NUMCOLS)+j)<minvals[i]);
}
return 0;
}
Upvotes: 1
Views: 141
Reputation: 26345
The existence of minvals
would imply that you are expected to calculate the minimum value of each 'row' of table
before then moving on to printing. As it stands, had your program properly calculated the minimum values of each array, your printing would be rather out of order.
There's no need to do any tricky, manual pointer manipulation. Simple array subscription is much clearer.
Let's start simple and return to basics by looking at the way we find the minimum value in a one dimensional array, as it is the core of this problem.
To find the minimum value in an array we need a few things to start:
The array itself is obviously each subarray of table
, and the length in this case is known to be NUMCOLS
. Our initial value should either be INT_MAX
(or another type-appropriate maximum constant found <limits.h>
), such that every element in the array is equal to or less than our initial value, or a value from the array itself.
Often times we opt for the second option here, choosing the first element in the array as our initial value, and comparing it to the second and onward elements.
As such, finding the minimum value in a single 'row' would look like this
const int row[NUMCOLS] = { 9, 2, 5 };
int min = row[0];
for (int i = 1; i < NUMCOLS; i++)
if (row[i] < min)
min = row[i];
but since we want to find and record the minimum value of each 'row' in table
, we're going to use a nested loop. Instead of the min
variable from before, we store each value in the associated index of our minvals
array.
for (i = 0; i < NUMROWS; i++) {
minvals[i] = table[i][0];
for (j = 1; j < NUMCOLS; j++)
if (table[i][j] < minvals[i])
minvals[i] = table[i][j];
}
When it comes time to print, we're going to repeat our nested loop. Our inner loop prints each element of each 'row' of table
, and we end each iteration of the outer loop by printing the value found in minvals
with the same index of our 'row'.
for (i = 0; i < NUMROWS; i++) {
for (j = 0; j < NUMCOLS; j++)
printf("%6d", table[i][j]);
printf(":%6d\n", minvals[i]);
}
Here's a working example.
#include <stdio.h>
#define NUMROWS 2
#define NUMCOLS 3
int main(void) {
const int table[NUMROWS][NUMCOLS] = {
{ 9, 2, 5 },
{ 3, -4, -12 }
};
int minvals[NUMROWS];
int i, j;
for (i = 0; i < NUMROWS; i++) {
minvals[i] = table[i][0];
for (j = 1; j < NUMCOLS; j++)
if (table[i][j] < minvals[i])
minvals[i] = table[i][j];
}
puts("Table value: minimum values");
for (i = 0; i < NUMROWS; i++) {
for (j = 0; j < NUMCOLS; j++)
printf("%6d", table[i][j]);
printf(":%6d\n", minvals[i]);
}
}
A good further exercise for you would be to compose the logic of the inner loop for finding minimum values into a more generic function. Its function signature would look like
int min(int *array, size_t length);
allowing it to work on arrays of varying sizes. Then our outer loop could be as simple as:
for (i = 0; i < NUMROWS; i++)
minvals[i] = min(table[i], NUMCOLS);
Upvotes: 2
Reputation: 24726
The line
int *ptr = &table;
is wrong, because &table
is of type int (*)[2][3]
(i.e. a pointer to the entire table), whereas ptr
is a pointer to a single element. Also, your pointer is non-const
, so it cannot point be made to point into a const
array.
If you want ptr
to point to a single int
value, then you should declare it the following way:
const int *ptr = &table[0][0];
Also, you are reading the contents of the array minvals
, although that array contains uninitialized data. This does not make sense and causes undefined behavior.
Instead of doing complex pointer arithmetic with the expression
*((ptr+i*NUMCOLS)+j))
you can simply write the following:
table[i][j]
That way, you do not need the pointer ptr
and your code is simpler.
Upvotes: 1