In my code the following function exists:
int Count_border(int loc[], int search[], int search_c){
int count = 0, i, j;
for(j = -1; j < 2; j += 2){
if(In_array(BOARD[loc[0] + j][loc[1]], search, search_c) == 1) count++;
}
for(j = -1; j < 2; j += 2){
if(In_array(BOARD[loc[0]][loc[1] + j], search, search_c) == 1) count++;
}
return count;
}
In this function I am searching for values in the array search. How it is done doesn't matter for this question. My question is however, how can I input a "manual" array, like this: Count_border(con_input, {-1, 0, 1}, 3);
This syntaxis isn't allowed by the compiler. And I don't want to create an array before the function, I really want to hardcode it.
Thank you for your help!
EDIT:
Now I am getting this error:
In function 'main':
file.c:40:1: error: expected ';' before '}' token
}
^
file.c:85:1: error: expected declaration or statement at end of input
}
Where this is my whole code, PLEASE help me out.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
void Put(int letter, int number);
void Set(int letter, int number, int who);
void Init_board();
int Count_border(int loc[], int search[], int search_c);
int In_array(int val, int arr[], int size);
int BOARD[9][9]; // 0 = empty, 1 = me, 2 = enemy
int STONES[2][81][9][9], STONE_COUNTER[2];
int main(void){
char input[5];
int con_input[2], t;
Init_board();
memset(STONES, 0, sizeof(STONES));
memset(STONE_COUNTER, 0, sizeof(STONE_COUNTER));
scanf("%s", input);
if(strcmp(input,"Start") == 0){
Put(4, 4);
}
scanf("%s", input); //get the first input after start
do{
con_input[0] = input[0]-'a'; /* Convert the input */
con_input[1] = input[1];
Set(con_input[0], con_input[1], 2);
t = Count_border(con_input, (int[]){-1, 0, 1}, 3);
printf("%i\n", t);
scanf("%s", input); /* Get the next input */
} while(strcmp(input, "Quit") != 0)
}
void Init_board(){
int i,j;
memset(BOARD, -1, sizeof(BOARD));
for(i = 0; i < 9; i++){
for(j = 0; j < 9; j++){
BOARD[i][j] = 0;
}
}
}
void Put(int letter, int number){
char t = letter + 'a';
printf("%c%i\n", t, number);
//fflush(stdout);
Set(letter, number, 1);
}
void Set(int letter, int number, int who){
BOARD[letter][number] = who;
}
int Count_border(int loc[], int search[], int search_c){
int count = 0, i, j;
for(j = -1; j < 2; j += 2){
if(In_array(BOARD[loc[0] + j][loc[1]], search, search_c) == 1) count++;
}
for(j = -1; j < 2; j += 2){
if(In_array(BOARD[loc[0]][loc[1] + j], search, search_c) == 1) count++;
}
return count;
}
int In_array(int val, int arr[], int size){
int i;
for (i=0; i < size; i++) {
if (arr[i] == val)
return 1;
}
return 0;
}
/* notes:
fflush(stdout);
*/
If you have a C99 (or newer) compiler just do
Count_border(con_input, (int[]){-1, 0, 1}, 3);
this (int[]){ something } is called a compound literal in C jargon and defines a temporary object (here an int array with 3 elements) that you can pass to your function.
Something like this?
#include <stdio.h>
void f(char arr[]);
int main(int argc, char *argv[])
{
f((char [4]){'1', '2', '3', '5'});
return 0;
}
void f(char arr[4])
{
int i;
for (i = 0; i < sizeof(arr)/sizeof(*arr); i++)
printf("%c ", arr[i]);
putchar('\n');
}
Related
When I simply run it in Clion, it works all right. While I am trying to debug it, it only printf after all the scanf is completed.
Below is a simple program to verify whether the array containing three values can add up to a target with either two of the three values.
This is the effect after debugging.
#include "stdio.h"
#include "stdlib.h"
int twoNums(int *nums, int numsSize, int target, int *returnSize);
int main() {
system("chcp 65001");
int string[3], target;
int numsSize = 3;
int isTrue;
int *returnSize = malloc(sizeof(int) * 2);
printf("Please input the array to be queried：\n");
scanf("[%d,%d,%d]", &string[0], &string[1], &string[2]);
printf("Please input the target value：\n");
scanf("%d", &target);
isTrue = twoNums(string, numsSize, target, returnSize);
if (isTrue)
printf("indexs of these is (%d, %d)", returnSize[0], returnSize[1]);
else
printf("something went wrong!");
free(returnSize);
}
int twoNums(int *nums, int numsSize, int target, int *returnSize) {
int i, j, cnt;
int flag = 0;
for (i = 0; i < numsSize; i++) {
cnt = target - nums[i];
for (j = i + 1; j < numsSize; j++) {
if (nums[j] == cnt) {
flag = 1;
break;
}
}
if (flag == 1)
break;
}
if (flag == 1) {
returnSize[0] = i;
returnSize[1] = j;
return 1;
} else {
return 0;
}
}
Why would this happen? How can I solve this problem?
#include <stdio.h>
#include <stdlib.h>
int *squares(int max_val) {
int *result = malloc(max_val * sizeof(int));
int i;
for(i = 1; i <= max_val; i++) {
result[i-1] = i*i;
}
return(result);
}
int main() {
int *sq = squares(10);
int i;
for(i = 0; i < 10; i++) {
printf("%d\t", sq[i]);
}
printf("\n");
return(0);
}
Basically take's an integer and returns a integer array of its squares. (Above works)
How would I do this without malloc or pointers?
#include <stdio.h>
#include <stdlib.h>
int[] squares(int max_val) {
int result[max_val];
int i;
for(i = 1; i <= max_val; i++) {
result[i-1] = i*i;
}
return(result);
}
int main() {
int sq[] = squares(10);
int i;
for(i = 0; i < 10; i++) {
printf("%d\t", sq[i]);
}
printf("\n");
return(0);
}
Above errors because of the function call. Is this possible? Or do we have to do it with pointers?
How about something like this.. ?
#include <stdio.h>
#include <stdlib.h>
void squares(int values[], int max_val) {
for(int i = 1; i <= max_val; i++) {
values[i - 1] = i * i;
}
}
int main() {
int sq[10];
squares(sq, 10);
int i;
for(i = 0; i < 10; i++) {
printf("%d\t", sq[i]);
}
printf("\n");
return(0);
}
This
int[] squares(int max_val) {
int result[max_val];
int i;
for(i = 1; i <= max_val; i++) {
result[i-1] = i*i;
}
return(result);
}
is not valid C:
a.c:3:4: error: expected identifier or ‘(’ before ‘[’ token
int[] squares(int max_val)
It should be
int *squares(int max_val) {
...
}
Putting that aside, your second squares returns a pointer to a local array.
This array ceases to exist once squares ends it's execution, so you are
returning an pointer that become invalid the moment it returns.
Also
int sq[] = squares(10);
is invalid C as well
a.c:9:13: error: invalid initializer
int sq[] = squares(10);
^~~~~~~
you cannot assign a function value to an array. The correct
version is
int *sq = squares(10);
So, if you don't want squares to allocate memory with malloc for the result, then you can
either allocate the memory in main or create an array in main and pass it to
squares:
int squares(int *result, int max_val) {
if(result == NULL || max_val <= 0)
return 0;
int i;
for(i = 1; i <= max_val; i++) {
result[i-1] = i*i;
}
return 1;
}
// version 1
int main() {
int *sq = malloc(10 * sizeof *sq);
// NEVER forget to check the return value of malloc
if(sq == NULL)
{
fprintf(stderr, "not enough memory\n");
return 1;
}
squares(sq, 10);
int i;
for(i = 0; i < 10; i++) {
printf("%d\t", sq[i]);
}
printf("\n");
free(sq); // NEVER forget to free
return 0;
}
// version 2
int main() {
int sq[10];
squares(sq, sizeof sq / sizeof sq[0]);
int i;
for(i = 0; i < 10; i++) {
printf("%d\t", sq[i]);
}
printf("\n");
return 0;
}
I have no idea what's wrong with the function goes by the name "int Count_largest_even". It's supposed to take the largest digit found in the given array (by the function "int find") and find how many times the digit appears in the array.
#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int Count_largest_even(int size, int *array, int large);
void ArrayPrint(int a[], int size);
int find(int array[], int size);
int arr1[16] = { 2, 22, 1, 3, 24, 94, 93, 12, 12, 66666, 21, 24, 8888, 21, 2, 33 };
int main() {
int mount;
int even;
ArrayPrint(arr1,16);
even = find(arr1, 16);
mount = Count_largest_even(16, arr1, even);
printf("\n The biggest even digit is : %d\n %d", even,mount);
system("pause");
return 0;
}
int find(int array[], int size){
int i = 0, digit, edigit = 0;
for (i = 0; i<size; i++){
while (array[i]!=0)
{
digit = abs(array[i] % 10);
if (digit > edigit)//checking condition for large
{
if (digit % 2 == 0)
{
edigit = digit;
}
}
array[i] = array[i] / 10;
}
}
return edigit;
}
void ArrayPrint(int a[], int size)
{
int i;
for (i = 0; i<size; i++){
printf("%d\n", a[i]);
}
}
int Count_largest_even(int size, int *array, int large)
{
int i;
int count = 0, digit;
for (i = 0; i < size; i++){
while ((array[i]!=0))
{
digit = abs(array[i] % 10);
if (digit == large)
{
count++;
}
array[i] = array[i] / 10;
}
}
return count;
}
As Ian Abbott said, you should not modify the array inside your loop.
But you can also do this in a single pass - some pseudo code:
int count = 0;
int largest_digit = 0;
for each digit:
if(digit > largest_digit) {
largest_digit = digit;
count = 1;
}
else if(digit == largest_digit)
count++;
I'm trying to find the union of two sets of 10 digit numbers, I'm passing along three int arrays: first, second, and comp (this will hold the union set).
So far, I've added first and second into one array. I was thinking about finding the matching indices in comp[] then filter through deleting them. I figure there's a much easier way. Can anyone give me a hint?
Basically I'm taking
first[] = [1,2,3,4,5,6,7,8,9,10];
second[] = [1,2,3,4,11,12,13,14,15,16];
and I want to return
comp[] = [5,6,7,8,9,10,11,12,13,14,15,16];
The numbers won't necessarily be in order.
int compound(int first[],int second[],int comp[]){
int i=0;
int indicies[20];
for(int j = 0; j<SIZE; j++){
comp[i]=first[j];
i++;
}
for(int k = 0; k<SIZE; k++){
comp[i]=second[k];
i++;
}
int z=0;
for(int l = 0; l<SIZE*2; l++){
for(int m = 0; m<SIZE*2; m++){
if(comp[l]==comp[m]){
indicies[z]=m;
z++;
}}}
return 0;
}
A first good step is (nearly) always sorting.
Sort both input-sets (unless you know they are already sorted).
Then iterate over both at once (two indices) and only add those elements to the output which fulfill your criteria (seems to be union minus intersection, thus only in one).
Bonus: The output-set will be sorted.
I suggest you start by writing a contains(int[], int) method like
#include <stdio.h>
#include <stdbool.h>
bool contains(int arr[], int val) {
int offset;
for (offset = 0; arr[offset] != '\0'; offset++) {
if (arr[offset] == val) return true;
}
return false;
}
Then your compound method could be implemented using it with something like
int compound(int first[],int second[],int comp[]){
int i=0;
int j;
for(j = 0; first[j] != '\0'; j++){
int val = first[j];
if (contains(second, val) && !contains(comp, val))
comp[i++] = val;
}
return i;
}
Finally, you can test it like
int main(int argc, char *args[]) {
int a[] = {1,2,3,'\0'};
int b[] = {2,3,4,'\0'};
int c[3];
int count = compound(a,b,c);
int i;
for (i = 0; i < count; i++) {
printf("%i\n", c[i]);
}
}
Output is
2
3
If the numeric range is small you could do this:
#include <stdio.h>
#define MAX 20 // small numeric range
#define sz(a) (sizeof(a)/sizeof(*(a)))
int xunion(int *a, int sa, int *b, int sb, int *c) {
int n[MAX] = {0};
for (int i=0; i<sa; i++) n[a[i]] = 1;
for (int i=0; i<sb; i++) n[b[i]] = 1;
int j=0;
for (int i=0; i<MAX; i++) if (n[i]) c[j++] = i;
return j;
}
void prn(int *a, int s) {
while (s-- > 0) printf("%d ", *a++);
putchar('\n');
}
int main() {
int a[] = {6, 3, 4, 7, 5};
int b[] = {2, 4, 5, 7, 6, 3};
int c[MAX];
prn(a, sz(a));
prn(b, sz(b));
int n = xunion(a, sz(a), b, sz(b), c);
prn(c, n);
return 0;
}
This prog is to accept an array of chars n compress them....(aaaabbbcc-->a4b3c2)....my prog is showing error at the point where im equating the addr of the 2d array to 1d array. This is my code:
/* size1 defined as 5 and size2 as 10.... (consts)*/
void compress(char data[SIZE1][SIZE2]);
int main()
{
char data[SIZE1][SIZE2];
printf("Enter a 5x10 matrix of characters:\n");
scanf("%c", &data);
compress(data[SIZE1][SIZE2]);
_getch();
return 0;
}
void compress(char data[SIZE1][SIZE2])
{
int hold[SIZE1*SIZE2];
int cnt = 0;
hold[SIZE1*SIZE2] = data[SIZE1][SIZE2];
for (int i = 0; i < (SIZE1*SIZE2); i++)
{
if (hold[i] == hold[i + 1])
{
cnt++;
continue;
}
else
{
printf("%c%d", hold[i], cnt);
}
}
}
This didn't work so I tried to use pointers:
void compress(char data[SIZE1][SIZE2])
{
int *hold[SIZE1*SIZE2];
int cnt = 0;
hold = data[SIZE1][SIZE2];
for (int i = 0; i < (SIZE1*SIZE2); i++)
{
if (*(hold+i) == *(hold+i+1))
{
cnt++;
}
else
{
printf("%c%d", *(hold+i), cnt);
}
}
}
I thought that the addrs of 2d arrays are stored linearly, hence they can be directly =to that of 1d.But the error says "'=':left operand must be an l-value".Im very new to pointers.Any help or corrections ....pls?
#include <stdio.h>
#define SIZE1 3
#define SIZE2 3
void compress(char data[SIZE1][SIZE2]);
int main(){
char data[SIZE1][SIZE2];
printf("Enter a %dx%d matrix of characters:\n", SIZE1, SIZE2);
for(int i=0;i<SIZE1;++i)
for(int j=0;j<SIZE2;++j)
scanf("%c", &data[i][j]);//aaaabbbcc
compress(data);
(void)getchar();
return 0;
}
void compress(char data[SIZE1][SIZE2]){
char *hold = &data[0][0];
int cnt = 1, size = SIZE1*SIZE2;
for (int i = 0; i < size; i++){
if (i < size -1 && hold[i] == hold[i + 1]){
cnt++;
//continue;
} else {
printf("%c%d", hold[i], cnt);//a4b3c2
cnt = 1;
}
}
}