I tried a lot of things to do but it still show me the same, that there is unhandled exception: access violation writing location in VS. But it doesn't happen when i sorting 1d array. What can I try next?
int main(void) {
static int a[3][4]{}, ab[3][4]{};
int i, j, k, N, M;
int* a1=nullptr;
printf("Matrica mora da ima velicinu 3 sa 4\n");
printf("Enter the order \n\n\t");
scanf_s("%d%d",&N ,&M);
for (i = 0;i < M;++i)
{
for (j = 0;j < N;++j)
{
scanf_s(" %d", &a[i][j]);
ab[i][j] = a[i][j];
}
printf("\n");
}
for (i = 0;i < M;++i) {
for (j = 0;j < N;++j) {
printf(" %d", a[i][j]);
}
printf("\n ");
}
//classic sorting
for (i=0; i < M; ++i)
{
for (j = 0;j < N;++j)
{
for (k = j + 1;j < N;++k)
if (a[i][j] > a[i][k])
{
*a1 = a[i][j]; // there is exception thrown
a[i][j] = a[i][k];
a[i][k] = *a1;
}
}
}
First off, there is a problem with static allocation of arrays, but there is no sanitization of N and M after the user inputs them. That means that you allocate only a matrix of 3x4, but the user can input and write to a matrix of any dimensions (e.g. 10x10), which could lead to access violation.
I'd recommend either having sanitation of the input values, e.g.
// It's always better to have limits as constant.
const int MAX_N = 3;
const int MAX_M = 4;
static int a[MAX_N][MAX_M];
...
scanf_s("%d%d",&N ,&M);
// Check if the dimensions can be fitted into the statically allocated array.
if(N > MAX_N || N <= 0 || M > MAX_M || M < 0)
{
// indicate invalid dimensions, either by returning from main with -1
// or calling exit(-1), or throwing an exception.
return -1;
}
In case the input didn't exceed 3x4, another thing that could be problematic - i goes from 0 to M, not N (what I would expect), which could also be problematic. The way matrix addressing works in C/Cpp is that the matrix is linearized into an array, and accessing it with a[i][j] leads to accessing the array with a[i*MAX_J + j]. In your case, the array has 12 elements (3x4), and MAX_J=4, so accessing it with a reverse set of indexes a[4][3] will access a[4*4+3]=a[19], which will access memory from outside of the array.
On the access violation writing problem, a1 isn't allocated, so when you try do execute *a1= ... you are writing to nullptr, which is a protected address, hence the access violation when writing. The way to solve this is either to:
have a1 be a int variable (not a pointer)
first allocate memory for a1 by executing a1 = malloc(sizeof(int)); and then freeing it after use with free(a1) (but since it's only a single element, I'd recommend converting a1 to int instead)
assign the address of the matrix element like a1=&a[i][j], but that would not be valid logically in your case (after that, you write into the location the pointer is pointing to, so the original value will be lost).
The reason why it's not happening for the 1d array is probably because of the inverted dimensions - the matrix would probably be 1x4, but you will be accessing it as 4x1, and you are sorting all the values with j index from 0 to 1, and since there is only one value you would not enter the k loop.
Related
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;
}
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:
An array
The length of the array
An initial value to compare against
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);
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.
So basically according to definition of array we cannot change array size. But if I am adding element to a same array by shifting other elements to the right of array, so the array size is going to increase.
How this is possible?
#include<stdio.h>
int main() {
int n, j, k, item;
printf("Enter size of array:\n");
scanf("%d", &n);
printf("Enter element to insert and position of element:\n");
scanf("%d,%d", &item, &k);
int a[n];
for (j = 0; j < n; j++) {
printf("Enter a[%d] element:\n", j);
scanf("%d", &a[j]);
}
j = n - 1;
while (j >= k - 1) {
a[j + 1] = a[j];
j = j - 1;
}
a[k - 1] = item;
for (j = 0; j <= n; j++) {
printf("%d\n", a[j]);
}
}
Shifting the contents of the array to the right will not resize the array. If the array was not already large enough to hold the result of the shift, then you have overrun the array object, and have induced undefined behavior.
There is no way to dynamically increase the size of a variable with static or auto duration (e.g., global or local variables), and this includes arrays. If your compiler supports variable length arrays (VLAs), changing the value of the expression controlling the dimension of the array does not affect the array's size.
int main (void) {
int n = 3;
int v[n];
printf("%zu\n", sizeof(v));
++n;
printf("%zu\n", sizeof(v));
}
The program above will print the same value twice.
I am not entirely sure what you're asking, but for any readers interested in knowing how to dynamically change the size of an array in C: if an array is declared in stack memory, its size cannot change. However, a block of memory intended to be used as an array is declared on the heap (i.e. with malloc or calloc), can be reallocated with a different size if necessary:
int *data = malloc(10 * sizeof(int)), *data2 = NULL;
int i;
if(data == NULL)
{
perror("malloc");
exit(EXIT_FAILURE);
}
for (i = 0; i < 10; i++)
{
data[i] = i;
}
data2 = realloc(data, 11 * sizeof(int));
if(data2 == NULL)
{
free(data);
perror("realloc");
exit(EXIT_FAILURE);
}
else
{
data = data2;
}
data[10] = 10;
for (i = 0; i < 11; i++)
printf("%d ", data[i]);
free(data);
data = NULL;
Shifting elements in an array down one element will not change its size.
If you declare an array as
T a[N]; // assume N is a constant expression
then a can only ever hold N elements of type T - no more, no less. You cannot add extra elements to the array, nor can you remove elements from the array.
However...
C does not force any bounds checking on array subscripting, so it's possible that you can read or write past the end of the array such as
a[N + 2] = x;
The behavior on doing so is undefined - your program may work as expected, or it may crash immediately, or you may corrupt other objects in the program. The runtime environment will (most likely) not throw an IndexOutOfBounds-type exception.
There is a thing called a variable-length array that was added in C99, where the array size is not a constant expression:
size_t size = some_value();
T a[size];
Variable length arrays are only variable length in the sense that their size isn't determined until runtime - however, once defined, their size is fixed throughout their lifetime, and like regular arrays, they cannot grow as new items are added.
If you dynamically allocate a chunk of memory using
T *a = malloc( sizeof *a * some_size );
then you can grow or shrink that chunk of memory using realloc:
T *tmp = realloc( a, sizeof *a * (some_size * 2) );
if ( tmp )
{
a = tmp;
some_size *= 2;
}
.... array we cannot change .. But if I (do something special) ... the array size is going to increase.
How this is possible?
Undefined behavior
Arrays cannot change size once defined.
Code attempts to assign a[j + 1] with j = n-1 and that is a[n]. This is outside array a[] and so undefined behavior. Rest of code is irrelevant for at that point anything is possible, code crash, error report, even apparent successful array expansion, etc.
int a[n];
...
j = n - 1;
while (j >= k - 1) {
a[j + 1] = a[j]; // To attempt access to `a[n]` is UB
Hi i need to check if the array is symmetry or not. i have a function that takes in a two-dimensional array of integer numbers M and the array sizes for rows and columns as parameters, and returns 1 if M is symmetric or 0 otherwise. I tried many times but the output will be either yes to non-symmetric array or no to symmetric array
Here is my code:
#include <stdio.h>
#define SIZE 10
#define INIT_VALUE -1
int symmetry2D(int M[][SIZE], int rowSize, int colSize);
int main()
{
int M[SIZE][SIZE], i, j, result = INIT_VALUE;
int rowSize, colSize;
printf("Enter the array size (rowSize, colSize): \n");
scanf("%d %d", &rowSize, &colSize);
printf("Enter the matrix (%dx%d): \n", rowSize, colSize);
for (i = 0; i < rowSize; i++)
for (j = 0; j < colSize; j++)
scanf("%d", &M[i][j]);
result = symmetry2D(M, rowSize, colSize);
if (result == 1)
printf("symmetry2D(): No\n");
else if (result == 0)
printf("symmetry2D(): Yes\n");
else
printf("Error\n");
return 0;
}
int symmetry2D(int M[][SIZE], int rowSize, int colSize)
{
int h, k, temp;
int result;
for (h = 0; h < rowSize; h++)
{
for (k = 0; k < colSize; k++)
{
M[h][k] = M[k][h];
}
}
result = 0;
for (h = 0; h < rowSize && result; h++)
{
for (k = 0; k < colSize; k++)
{
//if it is not equal to its transpose
if (M[h][k] != M[h][k])
{
result = 1;
break;
}
}
}
if (result == 0)
{
for (h = 0; h < rowSize; h++)
{
for (k = 0; k < colSize; k++)
{
return result = 0;
}
}
}
else
return result = 1;
}
Several issues:
By your definition, a matrix is symmetric if and only if it is equal to its transpose. That can be the case only for square matrices, yet you accommodate non-square matrices as well, for no apparent reason.
Your symmetry2D() function contains serious logical flaws:
It makes the input symmetric via the loop that performs M[h][k] = M[k][h]
Even if it did not do so, it would never find the input non-symmetric, because its test for that is if (M[h][k] != M[h][k]), which must always fail.
It's unclear what you think the if/else and loop nest at the end of symmetry2D() are achieving for you, but provided that rowSize and colSize are both greater than zero, the actual effect of the whole construct is the same as a simple return result;.
It looks like the idea might have been to create an array containing the transpose of the input, and then compare the input to that. That would have worked, despite being rather grotesquely inefficient, but you never in fact create that separate array for the transpose. If you're going to test without creating the transpose -- which you should -- then
Do not modify the input array (so remove the first loop nest altogether).
Get your indexing right for the symmetry comparisons: M[h][k] != M[k][h]
For best efficiency, avoid redundant and needless comparisons. For example, if you have already tested the M[1][2] == M[2][1] then you do not need to test whether M[2][1] == M[1][2]. And you never need to test elements on the main diagonal. You could achieve this efficiency pretty easily with a better choice of loop bounds.
Also, if indeed the symmetry2D() function is supposed to avoid modifying the input array, consider declaring the element type for its first argument to be const int instead of plain int (but do not modify the type of the corresponding variable in main()). If you had written it that way in the first place then the compiler would have noticed the function's logically erroneous attempt to modify the array elements, and rejected the code.
I'm writing a program that converts an array of integer vectors to an adjacency matrix , an (n+1)x(n+1) array. When in the function i have written to do this the dynamic memory allocation from setting up the int** seems to overwrite the n+1 and n+n elements of the integer vector.
int** makeAdjMatrix(IntVec* Vec, int length) { //allocates a new nxn array
printf(" madm test %d \n" , Vec[1]->data[6]);
//confirming that the intvec entered okay
int** new;
new = (int**)malloc(length+1*length+1*sizeof(int*))
printf(" madm test %d \n" , Vec[1]->data[6]);
// confirming that something happenend to the intvec
for (int i = 0; i <= length + 1; i++) {
new[i] = (int*)malloc(length + 1*sizeof(int));
}
for (int i = 1; i <= length+1; i++) {
for (int j = 1; j <= length+1; j++) {
new[i][j] = 0;
}
}
outputs normally for all elements in the vector data structure except those n+1 and above. n in this case being 5.
the above code prints
test 1
test 33
segfault
(because there is no 33rd indices in the array which the code referrences later)
is this memory exhaustion? how am I overwriting a previously allocated array on the heap? this might be a bit vague, this is my first question go easy on me. by the way this only happens when the vector array has repeated identical input.
how bad did I mess up?
I am able to declare in a good way two matrices A and B.
But, when using the memcpy (to copy B from A), B gives me arrays of 0s.
How can I do? Is my code correct for using memcpy?
int r = 10, c = 10, i, j;
int (*MatrixA)[r];
MatrixA=malloc(c * sizeof(*MatrixA));
int (*MatrixB)[r];
MatrixB=malloc(c * sizeof(*MatrixB));
memcpy(MatrixB,MatrixA,c * sizeof(MatrixA));
for(i=1;i<r+1;i++)
{
for (j = 1; j < c+1; j++)
{
MatrixA[i][j]=j;
printf("A[%d][%d]= %d\t",i,j,MatrixA[i][j]);
}
printf("\n");
}
printf("\n");printf("\n");printf("\n");printf("\n");printf("\n");
for(i=1;i<r+1;i++)
{
for (j = 1; j < c+1; j++)
{
printf("B[%d][%d]= %d\t",i,j,MatrixB[i][j]);
}
printf("\n");
}
You copied contents before initializing MatrixA .And also you access index out of bound (r+1 evaluates 11 which is out of bound) causing UB. Do this instead -
for(i=0;i<r;i++) // i starts from 0
{
for (j =0; j < c; j++) // j from 0
{
MatrixA[i][j]=j;
printf("A[%d][%d]= %d\t",i,j,MatrixA[i][j]);
}
printf("\n");
}
memcpy(MatrixB,MatrixA,c * sizeof(*MatrixA)); // copy after setting MatrixA
for(i=0;i<r;i++) // similarly indexing starts with 0
{
for (j =0; j < c; j++)
{
printf("B[%d][%d]= %d\t",i,j,MatrixB[i][j]);
}
printf("\n");
}
Is my code correct for using memcpy?
No, your code is wrong, but that's less of a memcpy problem. You're simply doing C arrays wrong.
int r = 10, c = 10, i, j;
int (*MatrixA)[r];
MatrixA=malloc(c * sizeof(*MatrixA));
Ok, MatrixA is now a pointer to a 10-element array of integers right? So the compiler reserves memory for ten ints; however, in the malloc line, you overwrite that with a pointer to a memory region of ten times the size of a single integer. A code analysis tool will tell you that you've built a memory leak.
These mistakes continue throughout your code; you will have to understand the difference between statically allocated C arrays and dynamic allocation using malloc.