I had searched so much on the forum, but I didn't find a solution for my problem.
How do I print a matrix that is given like this matrix[2][2]={{1,2},{3,4}} and give it back like:
1 2
3 4
#include <stdio.h>
int main(){
int i,j;
int x [2][2]={{1,2},{3,4}};
for(i=0; i<2; i++) {
for(j=0; j<2; j++) {
printf(" %d", x[i][j]);
}
printf("\n");
}
}
This program support not only 2x2 matrix but also int matrix[2][3]={{1,2,3},{4,5,6}}; or other size.
#include <stdio.h>
int main(void) {
int matrix[2][2]={{1,2},{3,4}};
size_t i, j;
for (i = 0; i < sizeof(matrix)/sizeof(matrix[0]); i++) {
for (j = 0; j < sizeof(matrix[i])/sizeof(matrix[i][0]); j++) {
if (j > 0) putchar(' ');
printf("%d", matrix[i][j]);
}
putchar('\n');
}
return 0;
}
Simple nested for loop:
for (int i = 0; i < 2; i++){
for(int j = 0; j < 2; j++){
printf("%d\t", matrix[i][j]);
}
printf("\n");
}
for (int i = 0; i < 2; i++){
for(int j = 0; j < 2; j++){
printf("%d\t", matrix[i][j]); // tab-separated. Did you want a space?
}
if (i < 2 - 1) printf("\n"); // newline except at the end
}
is one way.
Related
here below i have given multiplication of matrices in c language using for loop but can any help me make a more simplified version or can any help me make it using while loop
i want a simplified version
i want a code in while loop
:) just learning
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a[10][10], b[10][10], mul[10][10], r, c, i, j, k;
system("cls");
printf("enter the number of row=");
scanf("%d", &r);
printf("enter the number of column=");
scanf("%d", &c);
printf("enter the first matrix element=\n");
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
scanf("%d", &a[i][j]);
}
}
printf("enter the second matrix element=\n");
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
scanf("%d", &b[i][j]);
}
}
printf("multiply of the matrix=\n");
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
mul[i][j] = 0;
for (k = 0; k < c; k++)
{
mul[i][j] += a[i][k] * b[k][j];
}
}
}
//for printing result
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
printf("%d\t", mul[i][j]);
}
printf("\n");
}
return 0;
}
It will not make code simpler only harder to read.
One of the loops example:
printf("multiply of the matrix=\n");
i = 0;
while (i < r)
{
j = 0;
while(j < c)
{
mul[i][j] = 0;
k = 0;
while(k < c)
{
mul[i][j] += a[i][k] * b[k][j];
k++;
}
j++;
}
i++;
}
I need to multiply two square matrixes A and B 15x15.
Unfortunately, I'm getting this kind of error.
I know the problem is in pointers while calculating matrix C.
C[i][j] += *(A + k) * *(B + k)
I hope you can explain me what's wrong. I'm a beginner xD.
Thank you in advance.
#include <stdio.h>
#define N 15
#define _CRT_SECURE_NO_WARNINGS
int main() {
int A[N][N];
int B[N][N];
int C[N][N];
printf("Input matrix A.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("Enter your element:\n");
scanf_s("%d", &A[i][j]);
}
printf("\n");
}
printf("Input matrix B.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("Enter your element:\n");
scanf_s("%d", &B[i][j]);
}
printf("\n");
}
printf("Matrix A.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("%d\t", A[i][j]);
}
printf("\n");
}
printf("Matrix B.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("%d\t", B[i][j]);
}
printf("\n");
}
for (int i = 0; i < 15; i++) {
for (int j = 0; j < 15; j++) {
C[i][j] = 0;
for (int k = 0; k < 14; k++) {
C[i][j] += *(A + k) * *(B + k);
k++;
}
}
}
printf("Your result:\n");
printf("Matrix C.\n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("%d\t", C[i][j]);
}
printf("\n");
}
return 0;
}
The problem in the multiplication is that A+k and B+k have type int (*)[15] which means dereferencing it once only makes a pointer out of them; furthermore, you need to take row and column items individually, which means A[i][k] and B[k][j], right? (also, there's no point on using confusing syntax, as the underlying operation is exactly the same).
Here's a fixed and improved version:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#define N 15
/* Improvement 1 (type abstraction) */
typedef int NxN_int_matrix[N][N];
/* Improvement 2 (input function & wrapper) */
#define input_matrix(var) input_matrix_ex((var), #var)
static void input_matrix_ex(NxN_int_matrix dst, char *name)
{
printf("Input matrix %s.\n", name);
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
/* Improvement 3 (nicer prompt) */
printf("%s[%2d][%2d]: ", name, i, j);
fflush(stdout);
scanf_s("%d", &dst[i][j]);
}
}
printf("\n");
}
/* Improvement 4 (print function) */
#define print_matrix(var) print_matrix_ex(#var, (var))
static void print_matrix_ex(char *name, NxN_int_matrix M)
{
printf("Matrix %s.\n", name);
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
printf("%d\t", M[i][j]);
}
printf("\n");
}
}
/* Improvement 5 (move multiplication to a function too, and fix it) */
static void mult_matrix(NxN_int_matrix dst, NxN_int_matrix a, NxN_int_matrix b)
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
/* Improvement 6 (don't write out intermediate values) */
int tmp = 0;
for (int k = 0; k < N; k++)
tmp += a[i][k] * b[k][j];
dst[i][j] = tmp;
}
}
}
int main()
{
NxN_int_matrix A, B, C;
input_matrix(A);
input_matrix(B);
print_matrix(A);
print_matrix(B);
mult_matrix(C, A, B);
printf("Your result:\n");
print_matrix(C);
return 0;
}
/* Possible further improvements:
* - using a transposed B might make multiplication faster
*/
Write a program which will accept 2-dimensional square matrix and find out the transpose of it.
Program should not make use of another matrix
Hi I am trying to transpose a 2*2 matrix without using another matrix.
Is there anything wrong with my transpose logic?
I am a newbie
#include <stdio.h>
int main()
{
int mat[2][2];
int i, j, temp;
for (i = 0; i < 2; i++) {
printf("\nEnter elements of %d row of first matrix: ", i + 1); //i+1 so that it can print 1 row, 2 row, 3 row etc
for (j = 0; j < 2; j++) { //loop inside to loop to get value for a[0][0],a[0][1],a[0][2]
scanf("%d", &mat[i][j]);
}
}
printf("The matrix\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
//transpose logic using same matrix
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
temp = mat[i][j];
mat[i][j] = mat[j][i];
mat[j][i] = temp;
}
}
printf("The transpose of the matrix is\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
}
EDIT: I found an easier way to do it however I still don't understand why my transpose logic by using this
temp = mat[i][j];
mat[i][j] = mat[j][i];
mat[j][i] = temp;
cannot get it to transpose.
Below is my corrected answer
#include <stdio.h>
int main()
{
int mat[2][2];
int i, j, temp;
for (i = 0; i < 2; i++) {
printf("\nEnter elements of %d row of first matrix: ", i + 1);//i+1 so that it can print 1 row, 2 row, 3 row etc
for (j = 0; j < 2; j++) {//loop inside to loop to get value for a[0][0],a[0][1],a[0][2]
scanf("%d", &mat[i][j]);
}
}
printf("The matrix\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
printf("The transpose of the matrix is\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[j][i]);
}
printf("\n");
}
}
Your program is a good attempt, but transposing the matrix is like reversing an array: you must stop half way to avoid swapping the transposed values twice, leading to the original matrix as you observe.
You should stop the inner loop when j == i, hence change the inner loop to:
for (j = 0; j < i; j++) { // j < i instead of j < 2
Here is a modified version:
#include <stdio.h>
int main() {
int mat[2][2];
int i, j, temp;
for (i = 0; i < 2; i++) {
printf("\nEnter elements of %d row of first matrix: ", i + 1);
for (j = 0; j < 2; j++) {
if (scanf("%d", &mat[i][j]) != 1)
return 1;
}
}
printf("The matrix:\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
//transpose logic using same matrix
for (i = 0; i < 2; i++) {
for (j = 0; j < i; j++) {
temp = mat[i][j];
mat[i][j] = mat[j][i];
mat[j][i] = temp;
}
}
printf("The transpose of the matrix is:\n");
for (i = 0; i < 2; i++) {
for (j = 0; j < 2; j++) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
return 0;
}
Your corrected answer does not transpose the matrix at all, it merely outputs the transposed matrix. The matrix mat in memory is unchanged.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(){
int N;
scanf("%d", &N);
int **arr;
arr = (int**)malloc(sizeof(int) * N);
for(int i = 0; i < N; i++){
arr[i] = (int*)malloc(sizeof(int) * N);
}
for(int i = 0; i < N; i++){
for(int j = 0; j < N; j++){
if(i == 0 || j == 0)
arr[i][j] = 1;
else
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
for(int i = 0; i < N; i++){
for(int j = 0; j < N; j++){
printf("%d ", arr[i][j]);
}
printf("\n");
}
for(int i = 0; i < N; i++)
free(arr[i]);
free(arr);
return 0;
}
This code makes a 2-dimensional array with a pointer; I want assign value arr[i][j] = 1. However, an error EXC_BAD_ACCESS occurs in XCode on a Mac. I don't know how to solve the problem. What is my mistake? The assigning for loop is where the trouble occurs.
Good day to everybody. My task is to determine if a rectangular matrix has two rows of positive elements. I write the code below. At end I try to chect statement about positive row, but it's not working at all. Please explain me how to correct get the access to the each row and column in matrix, and meaybe edit my code.
#include <stdio.h>
#include <conio.h>
#include <locale.h>
#define M 3
#define N 4
int main(){
setlocale(LC_ALL, "Rus");
float a[M][N]; //set matrix with 3 row and 4 column
int i, j; // row and column index
int count;
for (i = 0; i < M; i++){
for (j = 0; j < N; j++)
scanf_s("%f", &a[i][j]);
}
for (i = 0; i < M; i++){
printf("%d-я строка:", i + 1);
for (j = 0; j < N; j++)
printf("%f", a[i][j]);
printf("\n");
}
count = 0;
for (i = 0; i < M; i++){
for (j = 0; j < N; j++)
if (a[i][j] > 0){
count++;
printf("%d", count);
}
}
_getch();
return 0;
}
When counting the number of positive elements in a row, you need to set the count back to zero for each row.
So instead of:
count = 0;
for (i = 0; i < M; i++){
for (j = 0; j < N; j++)
if (a[i][j] > 0){
count++;
printf("%d", count);
}
}
you need
for (i = 0; i < M; i++){
count = 0;
for (j = 0; j < N; j++) {
if (a[i][j] > 0){
count++;
}
}
printf("row %d has %d positive elements\n", i, count);
}