This function f is to find common elements in an array and return result array and i am using 4 four loops to accomplish this task which i feel is no the best use of the loops,
Another problem is, how can i determine the size of the returned array so that my loop is within bounds
here is the code
#include <stdio.h>
#include <stdlib.h>
int *f(int first[], int second[], int size_first, int size_second);
int main(void) {
int arr1[]={1, 8, 3, 2, 6};
int arr2[]= {2, 6, 1};
int size1 = sizeof(arr1)/sizeof(arr1[0]);
int size2 = sizeof(arr2)/sizeof(arr2[0]);
int *intersection = f(arr1, arr2, size1, size2);
for(int i=0;i<3; i++){
printf("%d ", intersection[i]);
}
return 0;
}
// function to find common elements in 2 arrays
int *f(int first[], int second[], int size_first, int size_second){
int k=0, count=0;
//loop through the array to find the number common elements and store in count for dynamic memory allocation in future
for(int i=0;i<size_first;i++){
for(int j=0;j<size_second;j++){
if(first[i]==second[j]){
count ++;
}
}
}
// allocate memory for the common elements by making use of count
int * common_elements = (int*)malloc(count*sizeof(int));
// store the common elements in the new memory location
for(int i=0;i<size_first;i++){
for(int j=0;j<size_second;j++){
if(first[i]==second[j]){
common_elements[k]=first[i];
k++;
}
}
}
return common_elements;
free(common_elements);
}
If you are allowed to waste some memory, note that the intersection cannot have cardinality larger than the number of elements in the smaller set. Therefore, you can allocate more memory than you might need and avoid having to count first and allocate later.
Or, you can realloc as you go.
In general, you need a good data structure for checking set membership more quickly than scanning an entire array although for small sizes which fit in various caches, the linear scan will not perform too shabbily either.
For larger sets, however, you'll want to load the larger of the sets into an AVL tree or Scapegoat tree.
For really large data sets, you'll need to look into Bloom filters and related data structures depending on the use case.
I am including below the most naive improvement in your code which still has the nested loop and wastes memory up to the size of the smaller set to avoid counting common elements first.
#include <stdlib.h>
#include <stdint.h>
#include <stdio.h>
// TODO: What about duplicates in smaller set?
int *
int_set_intersection(
const int *first,
const int *second,
const size_t size_first,
const size_t size_second,
size_t *n
)
{
size_t K = 0; // number of common elements
const int is_first_smaller = (size_first < size_second);
// Done this way so I can declare variables as consts
const int *set_smaller = is_first_smaller ? first : second;
const int *set_larger = is_first_smaller ? second : first;
const size_t size_smaller = is_first_smaller ? size_first : size_second;
const size_t size_larger = is_first_smaller ? size_second : size_first;
int *common = malloc(size_smaller * sizeof(*common));
if (!common) {
fprintf(stderr, "Failed to allocate memory for %z ints\n", size_smaller);
perror("Cannot allocate memory for common elements");
exit(EXIT_FAILURE);
}
for (size_t i = 0; i < size_smaller; ++i) {
for (size_t j = 0; j < size_larger; ++j) {
if (set_smaller[i] == set_larger[j]) {
common[K] = set_smaller[i];
++K;
break;
}
}
}
*n = K;
return common;
}
void
int_set_print(const int *set, size_t n, FILE *f)
{
FILE *out = f ? f : stdout;
size_t i = 0;
fputs("{ ", out);
for (i = 0; i < n - 1; ++i) {
fprintf(out, "%d, ", set[i]);
}
fprintf(out, "%d }\n", set[i]);
}
int
main(void) {
int arr1[] = {1, 8, 3, 2, 6};
int arr2[] = {2, 5, 1};
size_t n = 0;
const int *intersection = int_set_intersection(
arr1,
arr2,
sizeof(arr1)/sizeof(arr1[0]),
sizeof(arr2)/sizeof(arr2[0]),
&n
);
int_set_print(intersection, n, NULL);
free(intersection); // not really needed, but good hygiene
return 0;
}
For larger arrays, one option is to sort the contents first to make it easier to check for common elements, as shown in the code below. If the original array contents cannot be changed, first copy them into dynamically allocated memory. Dynamically allocated memory is also needed to hold the list of common elements, but that can use the same storage as one of the copies.
OP's original function returns a pointer to dynamically allocated memory containing the array of common elements, but does not indicate the length of the array. I added a parameter to return the length.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int *f(int first[], int second[], int size_first, int size_second, int *size_common);
int main(void) {
int arr1[]={1, 8, 3, 2, 6};
int arr2[]= {2, 6, 1};
int size1 = sizeof(arr1)/sizeof(arr1[0]);
int size2 = sizeof(arr2)/sizeof(arr2[0]);
int size_common;
int *intersection = f(arr1, arr2, size1, size2, &size_common);
for(int i=0;i<size_common; i++){
printf("%d ", intersection[i]);
}
free(intersection);
return 0;
}
static int cmp_int(const void *ap, const void *bp) {
int a = *(const int *)ap;
int b = *(const int *)bp;
return (a > b) - (a < b);
}
// function to find common elements in 2 arrays
int *f(int first[], int second[], int size_first, int size_second,
int *size_common) {
int *copy1;
int *copy2;
int idx1;
int idx2;
int count;
// allocate memory local copies of the arrays
copy1 = malloc(size_first * sizeof (int));
copy2 = malloc(size_second * sizeof (int));
if (!copy1 || !copy2) {
// allocation error
free(copy1);
free(copy2);
*size_common = -1; // use -1 to report error
return NULL;
}
// copy the arrays
memcpy(copy1, first, size_first * sizeof (int));
memcpy(copy2, second, size_second * sizeof (int));
// sort the copies in ascending order
qsort(copy1, size_first, sizeof (int), cmp_int);
qsort(copy2, size_second, sizeof (int), cmp_int);
// find common elements
idx1 = 0;
idx2 = 0;
count = 0;
while (idx1 < size_first && idx2 < size_second) {
if (copy1[idx1] < copy2[idx2]) {
idx1++;
} else if (copy1[idx1] > copy2[idx2]) {
idx2++;
} else {
// common element found!
// use copy1[] to store common elements
copy1[count] = copy1[idx1];
count++;
idx1++;
idx2++;
}
}
// common elements are in copy1[].
// finished with copy2, so free it.
free(copy2);
if (count == 0) {
// no common elements
free(copy1); // free the memory
copy1 = NULL; // and make the function return NULL
} else {
// try to reduce memory for common elements
copy2 = realloc(copy1, count * sizeof (int));
if (copy2) {
// reallocation successful
copy1 = copy2;
} // else, never mind, copy1 is still valid
}
// return the common elements
*size_common = count;
return copy1;
}
If your arrays are of comparable elements (you use integers, which are comparable), the best way in my opinion is to sort both arrays and traverse both in parallel, looking at both sides and comparing the elements at both sides. If there's a lowest element, advance its pointer, leaving the other waiting.... if there's a match (they are equal), you can mark it (more on this later) and advance both pointers, until you reach the end in one array (the sortest). You will get the marks on the matching positions, but if you reorder the array, exchanging the found element with the first of the yet, unmatched elements, you will have all matching elements in the first positions of both arrays, letting you to return only the number of matches from the function and the matches themselves in the first positions of both arrays.
The complexity of this algorithm should be O(n*log(n)) (because of the quicksorts) if you use quicksort, plus O(n) (which doesn't affect the final O) for the matching, so O(n*log(n)) should be the big O complexity, as a general case. Below is a sample code, with a run:
comp.c
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#define N(arr) (sizeof(arr)/sizeof((arr)[0]))
void swap(int *ref_a, int *ref_b)
{
if (ref_a == ref_b)
return; /* nothing to do. */
int temp = *ref_a;
*ref_a = *ref_b;
*ref_b = temp;
}
int int_cmp(const void *_a, const void *_b)
{
const int *a = _a, *b = _b;
return *a - *b;
}
void print(int v[], int v_sz, const char *fmt, ...)
{
va_list p;
va_start(p, fmt);
vprintf(fmt, p);
va_end(p);
char *sep = "[";
for (int i = 0; i < v_sz; i++) {
printf("%s%d", sep, v[i]);
sep = ", ";
}
printf("]\n");
}
int find_matches(int a[], int b[], int a_sz, int b_sz)
{
print(a, a_sz, "a(unsorted)");
print(b, b_sz, "b(unsorted)");
qsort(a, a_sz, sizeof a[0], int_cmp);
qsort(b, b_sz, sizeof b[0], int_cmp);
print(a, a_sz, "a(sorted)");
print(b, b_sz, "b(sorted)");
int i = 0;
for (int i_a = 0, i_b = 0; i_a < a_sz && i_b < b_sz;) {
if (a[i_a] < b[i_b]) {
i_a++;
continue;
} else if (a[i_a] > b[i_b]) {
i_b++;
continue;
}
/* a[i_a] == b[i_b] */
swap(&a[i_a], &a[i]);
swap(&b[i_b], &b[i]);
print(a, a_sz, "after #%d, a:", i);
print(b, b_sz, "after #%d, b:", i);
i_a++; i_b++; i++;
}
return i;
}
int main()
{
int arr1[] = {1, 8, 3, 2, 6, 7};
int arr2[] = {2, 6, 1, 7, 4, 1, 9, 6};
int size1 = N(arr1);
int size2 = N(arr2);
int match = find_matches(arr1, arr2, size1, size2);
for (int i = 0; i < match; i++) {
printf("Match #%d: %d\n", i, arr1[i]);
}
}
It will produce:
$ comp
a(unsorted)[1, 8, 3, 2, 6, 7]
b(unsorted)[2, 6, 1, 7, 4, 1, 9, 6]
a(sorted)[1, 2, 3, 6, 7, 8]
b(sorted)[1, 1, 2, 4, 6, 6, 7, 9]
after #0, a:[1, 2, 3, 6, 7, 8]
after #0, b:[1, 1, 2, 4, 6, 6, 7, 9]
after #1, a:[1, 2, 3, 6, 7, 8]
after #1, b:[1, 2, 1, 4, 6, 6, 7, 9]
after #2, a:[1, 2, 6, 3, 7, 8]
after #2, b:[1, 2, 6, 4, 1, 6, 7, 9]
after #3, a:[1, 2, 6, 7, 3, 8]
after #3, b:[1, 2, 6, 7, 1, 6, 4, 9]
Match #0: 1
Match #1: 2
Match #2: 6
Match #3: 7
$ _
A good interface is to switch in both algorithms the matched elements with the first of the non-yet-matched elements in both arrays, so in this way you can return an integer (the one you use to know the start of the non-yet-matched elements) that tells you the number of matched elements, and you will get them from any of the two arrays.
If the elements are not comparable, and they can be just be compared for equity, then you have to compare each element with any other for a match, take them off from the arrays (this can be done swapping them with the first of the not yet matched elemnts, and advance the pointers), and start again with the reduced versions of them. Some way of doing this is, when you find a match, to exchange them with the first, second, third elements of each array, and use a variation of the above algorithm (you reorder as you match) In this case you compare at first time n*m (but not all), when you get a match, (n-1)*(m-1), ... and so until the last comparition in which you fail all comparitions to (m-k)*(n-k). This is, in the average, m*n/2 + (m-1)*(n-1)/2 +...+ (m-k)*(n-k). something in the range of m(m-1)*n(n-1)/k^2, which is O(m^2*n^2):
comp2.c
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#define N(arr) (sizeof(arr)/sizeof((arr)[0]))
void swap(int *ref_a, int *ref_b)
{
if (ref_a == ref_b)
return; /* nothing to do. */
int temp = *ref_a;
*ref_a = *ref_b;
*ref_b = temp;
}
int int_cmp(const void *_a, const void *_b)
{
const int *a = _a, *b = _b;
return *a - *b;
}
void print(int v[], int v_sz, const char *fmt, ...)
{
va_list p;
va_start(p, fmt);
vprintf(fmt, p);
va_end(p);
char *sep = "[";
for (int i = 0; i < v_sz; i++) {
printf("%s%d", sep, v[i]);
sep = ", ";
}
printf("]\n");
}
int find_matches(int a[], int b[], int a_sz, int b_sz)
{
print(a, a_sz, "a(unsorted)");
print(b, b_sz, "b(unsorted)");
int i = 0;
loop:
for (int i_a = 0; i_a + i < a_sz; i_a++) {
for (int i_b = 0; i_b + i < b_sz; i_b++) {
/* we can only compare for equality */
if (a[i + i_a] == b[i + i_b]) {
swap(&a[i + i_a], &a[i]);
swap(&b[i + i_b], &b[i]);
i++;
goto loop;
}
}
}
print(a, a_sz, "a(final)");
print(b, b_sz, "b(final)");
return i;
}
int main()
{
int arr1[] = {1, 8, 3, 2, 6, 7};
int arr2[] = {2, 6, 1, 7, 4, 1, 9, 6};
int size1 = N(arr1);
int size2 = N(arr2);
int match = find_matches(arr1, arr2, size1, size2);
for (int i = 0; i < match; i++) {
printf("Match #%d: %d\n", i, arr1[i]);
}
}
which produces, when running, the following output:
$ comp2
a(unsorted)[1, 8, 3, 2, 6, 7]
b(unsorted)[2, 6, 1, 7, 4, 1, 9, 6]
a(final)[1, 2, 6, 7, 3, 8]
b(final)[1, 2, 6, 7, 4, 1, 9, 6]
Match #0: 1
Match #1: 2
Match #2: 6
Match #3: 7
$ _
You can reorder the values, there's no difference, in this case you had a two level for loop, mixed with a third level go back to the beginning and start again loop. The loop is warranted to finish, as when you go back to the top, you have increased i, which means the nested for loops will be shorter each time. We can rewrite the find_matches routine in this case by adjusting the array start points, in this manner:
comp3.c
/* ... as before */
int find_matches(int a[], int b[], int a_sz, int b_sz)
{
print(a, a_sz, "a(unsorted)");
print(b, b_sz, "b(unsorted)");
int i = 0;
loop:
for (int i_a = 0; i_a < a_sz; i_a++) {
for (int i_b = 0; i_b < b_sz; i_b++) {
/* we can only compare for equality */
if (a[i_a] == b[i_b]) {
swap(&a[i_a], &a[0]);
swap(&b[i_b], &b[0]);
i++;
print(a++, a_sz--, "a(after match)");
print(b++, b_sz--, "b(after match)");
goto loop;
}
}
}
print(a, a_sz, "a(final)");
print(b, b_sz, "b(final)");
return i;
}
/* ... as before */
that will produce this result (I changed the initial sort order to see how it affects the final result):
$ comp3
a(unsorted): [7, 8, 2, 3, 6, 1]
b(unsorted): [2, 6, 1, 7, 4, 1, 9, 6]
a(after match): [7, 8, 2, 3, 6, 1]
b(after match): [7, 6, 1, 2, 4, 1, 9, 6]
a(after match): [2, 8, 3, 6, 1]
b(after match): [2, 1, 6, 4, 1, 9, 6]
a(after match): [6, 3, 8, 1]
b(after match): [6, 1, 4, 1, 9, 6]
a(after match): [1, 8, 3]
b(after match): [1, 4, 1, 9, 6]
a(final): [8, 3]
b(final): [4, 1, 9, 6]
Match #0: 7
Match #1: 2
Match #2: 6
Match #3: 1
$ _
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Need to make a function that can replace first n elements from array A with last n elements of array B.
arguments in function should be like (arrayA, arrayB, n)
Works like:
int A[ ] = {1, 2, 3, 4, 5};
int B[ ] = {6, 7, 8, 9, 10};
function name(A[ ], B[ ], 2)
Output:
int A[ ] = {9, 10, 3, 4, 5};
int B[ ] = {6, 7, 8, 1, 2};
void *mymemcpy(void *a, const void *b, size_t size)
{
unsigned char *ac = a;
const unsigned char *bc = b;
while(size--) *ac++ = *bc++;
return a;
}
void replace(void *a, const void *b, size_t nelem, size_t elemsize, size_t size_b)
{
const char *bc = b;
mymemcpy(a, b + (size_b - nelem) * elemsize, nelem * elemsize);
}
int main(void)
{
int a[] = {1,2,3,4,5,6};
int b[] = {1,2,3,4,5,6,7,8,9,0};
replace(a,b,3,sizeof(a[0]), sizeof(b)/sizeof(b[0]));
for(size_t index = 0; index < sizeof(a)/sizeof(a[0]); index++)
{
printf("%2d ", a[index]);
}
printf("\n");
}
I would not use it directly as your teacher probably will not believe you.
Here you are.
#include <stdio.h>
void swap_n( int *a, int *b, size_t n )
{
while ( n-- )
{
int tmp = *a;
*a++ = *b;
*b++ = tmp;
}
}
int main(void)
{
int a[] = { 1, 2, 3, 4, 5 };
int b[] = { 6, 7, 8, 9, 10 };
const size_t N1 = sizeof( a ) / sizeof( *a );
const size_t N2 = sizeof( b ) / sizeof( *b );
for ( size_t i = 0; i < N1; i++ )
{
printf( "%2d ", a[i] );
}
putchar( '\n' );
for ( size_t i = 0; i < N2; i++ )
{
printf( "%2d ", b[i] );
}
putchar( '\n' );
size_t n = 2;
swap_n( a, b + N2 - n, n );
for ( size_t i = 0; i < N1; i++ )
{
printf( "%2d ", a[i] );
}
putchar( '\n' );
for ( size_t i = 0; i < N2; i++ )
{
printf( "%2d ", b[i] );
}
putchar( '\n' );
return 0;
}
The program output is
1 2 3 4 5
6 7 8 9 10
9 10 3 4 5
6 7 8 1 2
Pay attention to that the argument n of the function swap_n shall be not greater than the number of elements in the array a and in the array b.
Using the function swap_n you can exchange any parts of the size n of two arrays. For example to exchange two elements third and fourth of the array a with second and third elements of the array b you could write
size_t n = 2;
swap_n( a + 2, b + 1, n );
In this case the program output will be
1 2 3 4 5
6 7 8 9 10
1 2 7 8 5
6 3 4 9 10
I have the following:
struct matrix {
int h;
int w;
int** data;
};
int m1[2][2] = {
{1, 2},
{3, 4}
};
int m2[3][3] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
struct matrix matrix1 = {2, 2, m1};
struct matrix matrix2 = {3, 3, m2};
This gives the error 'initialisation from incompatible pointer type'. What pointer type should I be using?
You may be interested in the variable length arrays of C99. This solution does not directly answer your question about how to initialize the structure with a properly typed data (one can't); but you can use a simple pointer to store the array's address and then cast to the variable length array pointer when using struct matrix.
The user side would just call functions like printFroMat() which receive a single argument of type struct matrix; the code inside these functions (so to speak, the library implementation) would perform the somewhat unsightly casts, as demonstrated. The typedef makes the cast perhaps a little more understandable because it demonstrates where the variable name in a declaration would go.
Note that the funny sizeof(m2)/sizeof(*m2) etc. are not strictly necessary, you can just say 3. But the sizeof expression automatically stays in sync with the actual matrix size, which quickly becomes a real asset.
You can pass "arrays" (in fact: still just addresses, but of a known array type) together with their dimensions as parameters to functions, and index them the normal way (below in printMatrix). Example:
#include<stdio.h>
#include<string.h>
struct matrix {
int h;
int w;
int *data; // first element of matrix
};
int m2[4][3] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9},
{10, 11, 12}
};
void printMatrix(int dim1, int dim2, int mat[][dim2] )
{
printf("Address of matrix: %p\n", (void *)mat);
for(int i1=0; i1<dim1; i1++)
{
for(int i2=0; i2<dim2; i2++)
{
printf("%d ", mat[i1][i2]);
}
putchar('\n');
}
}
void printFromMat(struct matrix mat)
{
printMatrix(mat.h, mat.w, (int (*)[mat.w])mat.data);
// or:
typedef int (*mT)[mat.w];
printMatrix(mat.h, mat.w, (mT)mat.data);
}
int main()
{
printMatrix( sizeof(m2) /sizeof(*m2), // number of highest-order elements
sizeof(*m2)/sizeof(**m2), // number of second-order elements per highest-order
m2 ); // address of the first sub-array
struct matrix mat = { sizeof(m2) /sizeof(*m2), sizeof(*m2)/sizeof(**m2), *m2 };
printFromMat(mat);
return 0;
}
Sample session:
$ gcc -std=c99 -Wall -o 2d-matrix 2d-matrix.c && ./2d-matrix
Address of matrix: 0x100402020
1 2 3
4 5 6
7 8 9
10 11 12
Address of matrix: 0x100402020
1 2 3
4 5 6
7 8 9
10 11 12
Address of matrix: 0x100402020
1 2 3
4 5 6
7 8 9
10 11 12
2d array is not a pointer to pointer.
How to use the void *.
#include <stdio.h>
struct matrix {
int h;
int w;
void *data;
};
int m1[2][2] = {
{1, 2},
{3, 4}
};
int m2[3][3] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
struct matrix matrix1 = {2, 2, m1};
struct matrix matrix2 = {3, 3, m2};
int main(void){
int i, j;
int (*matp)[matrix2.w] = matrix2.data;//Back from the void* to pointer to array
for (i=0; i<matrix2.h; i++){
for(j=0; j<matrix2.w; j++)
printf("%d ", matp[i][j]);
puts("");
}
printf("\n");
return 0;
}
A matrix, as you declared, is not a pointer to pointer. Use a simple pointer to point its element.
#include <stdio.h>
struct matrix {
int h;
int w;
int* data;
};
int m1[2][2] = {
{1, 2},
{3, 4}
};
int m2[3][3] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
struct matrix matrix1 = {2, 2, &m1[0][0]};
struct matrix matrix2 = {3, 3, &m2[0][0]};
int main(int argc, char **argv)
{
int i, j;
for (i=0; i<matrix2.h; i++)
for(j=0; j<matrix2.w; j++)
printf("%d ", matrix2.data[(i*matrix2.h)+j]);
printf("\n");
}
EDIT
To answer to the comment, you can use compound literal as below. In this way you could access it with [i][j]
#include <stdio.h>
struct matrix {
int h;
int w;
int** data;
};
int *m2[3] = {
(int[]){1, 2, 3},
(int[]){4, 5, 6},
(int[]){7, 8, 9}
};
struct matrix matrix2 = {3, 3, m2};
int main(int argc, char **argv)
{
int i, j;
for (i=0; i<matrix2.h; i++)
for(j=0; j<matrix2.w; j++)
printf("%d ", matrix2.data[i][j]);
printf("\n");
}
You may not initialize the structure such a way like
struct matrix matrix1 = {2, 2, m1};
struct matrix matrix2 = {3, 3, m2};
because there is no conversion from types int ( * )[2] and int ( * )[3] to type int **
I suggest to allocate memory for copies of the arrays dynamically.
The approach can look the following way as it is shown in the demonstrative program
#include <stdlib.h>
#include <stdio.h>
struct matrix
{
size_t h;
size_t w;
int **data;
};
int m1[2][2] =
{
{ 1, 2 },
{ 3, 4 }
};
int m2[3][3] =
{
{ 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 }
};
struct matrix init( size_t h, size_t w, int a[h][w] )
{
struct matrix matrix = { 0 };
matrix.data = malloc( h * sizeof( int * ) );
if ( matrix.data )
{
matrix.h = h;
matrix.w = w;
for ( size_t i = 0; i < h; i++ )
{
matrix.data[i] = malloc( w * sizeof( int ) );
if ( matrix.data[i] )
{
for ( size_t j = 0; j < w; j++ ) matrix.data[i][j] = a[i][j];
}
}
}
return matrix;
}
int main( void )
{
struct matrix matrix1 = init( 2, 2, m1 );
struct matrix matrix2 = init( 3, 3, m2 );
for ( size_t i = 0; i < matrix1.h; i++ )
{
for ( size_t j = 0; j < matrix1.w; j++ ) printf( "%d ", matrix1.data[i][j] );
printf( "\n" );
}
printf( "\n" );
for ( size_t i = 0; i < matrix2.h; i++ )
{
for ( size_t j = 0; j < matrix2.w; j++ ) printf( "%d ", matrix2.data[i][j] );
printf( "\n" );
}
printf( "\n" );
// free allocated arrays of matrix1 and matrix2
}
The program output is
1 2
3 4
1 2 3
4 5 6
7 8 9
You need also to write a function that will free the allocated memory for the structure.
The compiler must support variable length arrays.
I want to decrement a pointer to the last element of an array and check for the condition that if the value of the pointer was smaller than 5, do nothing and go to the next round of loop and if not, that is if the value is equal or bigger than 5, print the value. So, for the array in the example, I want only 6 to be printed, that is the first encounter of a value equal or bigger than 5. I tried the code below but while being compiled with no error, it doesn't print any value. I'd appreciate your comments on this.
#include<stdio.h>
//The is a C program to decrement an array from the last element to the first.
int x[11] = {5, 6, -4, -3, -2, -1, 4, 3, 2, 1, -2};
int *lastElem, count;
int main (void) {
lastElem = &x[10];
for (count = 0; count < 11; count++)
if (abs(*lastElem) < 5)
continue;
else
printf("%d\n", *lastElem--);
return 0;
}
There is a problem in your decrement logic. If the value is less than 5, you're missing the decremet.
check the below code.
#include<stdio.h>
//The is a C programme to decrement an array from the last element to the first.
int x[11] = {5, 6, -4, -3, -2, -1, 4, 3, 2, 1, -2};
int *lastElem, count;
int main (void) {
lastElem = &x[10];
for (count = 0; count < 11; count++)
{
if (*lastElem >= 5)
{
printf("%d\n", *lastElem);
break ;
}
lastElem--;
}
return 0;
}
EDIT:
As you modified your question to include the absolute value comparions, the changes are like below.
Note : When making some major changes [which will change the behaviour and output] to the original question asked, do make a note of that, and if possible, mention the edit properly.
#include<stdio.h>
#include <stdlib.h>
//The is a C programme to decrement an array from the last element to the first.
int x[11] = {5, -6, -4, -3, -2, -1, 4, 3, 2, 1, -2};
int *lastElem, count;
int main (void) {
lastElem = &x[10];
for (count = 0; count < 11; count++)
{
if ( abs(*lastElem) >= 5)
{
printf("%d\n", *lastElem);
break;
}
lastElem--;
}
return 0;
}
You never assign another address to lastElem than the inital value (that is the 10th element). Also, if you want to go backwards, set count to 10 and let it count to 0. In each loop you have to assign &x[count] to lastElem (or decrement it, as it is a pointer and the address will be decremented according to the object it points to).
The line printf("%d\n", *lastElem--); is not always executed it is only executed when abs(*lastElem) >= 5. This is how you should do it
#include<stdio.h>
//The is a C programme to decrement an array from the last element to the first.
int x[11] = {5, 6, -4, -3, -2, -1, 4, 3, 2, 1, -2};
int *lastElem, count, value;
int main (void)
{
lastElem = &x[10];
for (count = 0; count < 11; count++)
{
value = *(lastElem--);
if (value < 5)
continue;
printf("%d\n", value);
break;
}
return 0;
}
This will do it with continue.
Here is the only correct code among presented here in other answers that does the task.:)
In fact you need a program that searches backward an alement of an array that satisfies a given condition. And if there is such an element then to output it.
#include <stdio.h>
#include <stdlib.h>
int main( void )
{
int a[] = { 5, 6, -4, -3, -2, -1, 4, 3, 2, 1, -2 };
const size_t N = sizeof( a ) / sizeof( *a );
int *first = a, *last = a + N;
while ( first != last && ( unsigned int )abs( *( last - 1 ) ) < 5 ) --last;
if ( first != last ) printf( "%d\n", *--last );
return 0;
}
The output is
6
Below there is demonstrated a general approach for such tasks
#include <stdio.h>
#include <stdlib.h>
int * find_backward( const int a[], size_t n, int ( *predicate )( int ) )
{
const int *first = a, *last = a + n;
while ( first != last && !predicate( *( last -1 ) ) ) --last;
return ( int * )last;
}
int is_greater_or_equal_to_5( int x )
{
return 5 <= ( unsigned int )abs( x );
}
int main( void )
{
int a[] = { 5, 6, -4, -3, -2, -1, 4, 3, 2, 1, -2 };
const size_t N = sizeof( a ) / sizeof( *a );
int *target = find_backward( a, N, is_greater_or_equal_to_5 );
if ( target != a ) printf( "%d\n", *--target );
return 0;
}
Using this approach you can use any integer arrays (even with zero size) and any conditions that are set by means of predicates.
For example if you want to find the last element of an array that is divisible by 3 you can write
#include <stdio.h>
#include <stdlib.h>
int * find_backward( const int a[], size_t n, int ( *predicate )( int ) )
{
const int *first = a, *last = a + n;
while ( first != last && !predicate( *( last -1 ) ) ) --last;
return ( int * )last;
}
int divisible_by_3( int x )
{
return x % 3 == 0;
}
int main( void )
{
int a[] = { 5, 6, -4, -3, -2, -1, 4, 3, 2, 1, -2 };
const size_t N = sizeof( a ) / sizeof( *a );
int *target = find_backward( a, N, divisible_by_3 );
if ( target != a ) printf( "%d\n", *--target );
return 0;
}
The output is
3
Take into account that this function allows also to deal with constant arrays.:)