i'm a bit confused about using pointers.
I want to hand over the array to the pointer and then get the result of the function from the pointer.
The function just returns the minimum of an array.
int *minimum(int (*A)[5]){
int min = 0;
int A_result[1];
for (int i = 0; i < 5; i++) {
if ((*A)[i]< (*A)[min])
min = i;
}
A_result[0] = (*A)[min];
return A_result; //line 39
}
int main(void) {
int A[5] = {5, 7, 3, 6, 4};
int (*array_ptr)[5] = minimum(&A); //line 45
printf("%d ", *array_ptr);
return EXIT_SUCCESS;
}
by now i get the following errors:
line 45: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=]
initialization from incompatible pointer type [-Wincompatible-pointer-types]
line 39:
function returns address of local variable [-Wreturn-local-addr]
There is no need to pass pointer to array to the function second_lowest. Also, no point in returning a pointer from the function second_lowest. I modified your code :
int second_lowest(int *A) {
int min = 0;
int A_result;
for (int i = 0; i < 5; i++) {
if (A[i] < A[min])
min = i;
}
A_result = A[min];
return A_result; //line 39
}
int main(void) {
int A[5] = { 5, 7, 3, 6, 4 };
int res = second_lowest(A); //line 45
printf("%d ", res);
return EXIT_SUCCESS;
}
Hope it helps.
Related
I dont understand some small actions that has taken in this code, for example
i) why do we need to do &rs, why cant we just write
int returnSize;
p = twoSum(a, sizeof(a)/sizeof(a[0]), target, int*returnSize);
if(returnSize == 0)
instead of &rs... what is wrong with doing this?
i ran the code here:
#include <stdio.h>
#include <stdlib.h>
int* twoSum(int* nums, int numsSize, int target, int* returnSize){
int i, j;
int* ret_arr = (int*)malloc(2*sizeof(int));
if(ret_arr == NULL){
// *returnSize = 0;
returnSize = 0;
return NULL;
}
for(i = 0; i< numsSize; i++){
for(j = i+1; j< numsSize; j++){
if(nums[i] + nums[j] == target){
//*returnSize = 2;
returnSize = 2;
ret_arr[0] = I;
ret_arr[1] = j;
return ret_arr;
}
}
}
//*returnSize = 0;
returnSize = 0;
free(ret_arr);
return NULL;
}
int main()
{
int a[] = {2,7,11,15};
int returnSize, target = 18;
int *p = NULL;
p = twoSum(a, sizeof(a)/sizeof(a[0]), target, int*returnSize);
//if(*returnSize == 0){
if(returnSize == 0){
printf("target not found");
}else{
printf("target found at indices %d and %d/n", p[0], p[1]);
free(p);
}
return 0;
}
error result:
main.c: In function ‘twoSum’:
main.c:34:28: warning: assignment to ‘int *’ from ‘int’ makes
pointer from integer without a cast [-Wint-conversion]
34 | returnSize = 2;
| ^
main.c: In function ‘main’:
main.c:57:48: error: expected expression before ‘int’
57 | p = twoSum(a, sizeof(a)/sizeof(a[0]), target,
int*returnSize);
| ^~~
full code: https://onlinegdb.com/7OpTpwxZy3
i dont know how this doesn't make sense to the compiler.
ii) Pointer array conceptual Q
we need to declare pointer p in line 60 because to access returned array elements (of two sum function) and print them (like they are used in line ), on screen from main. QUESTION: WHY DO WE NEED POINTER FOR TWO SUM FUNCTION WHEN WE ALREADY HAVE pointer p DECLARED IN MAIN, ITS LIKE POINTER POINTING TO A POINTER(LIKE int*a, int*p, p = a), why does the code work?
error i get if i remove "int*" in line 4(2nd pic), in front of twosum function.
main.c:28:1: warning: return type defaults to ‘int’ [-Wimplicit-int]
28 | twoSum(int* nums, int numsSize, int target, int* returnSize){
| ^~~~~~
main.c: In function ‘twoSum’:
main.c:34:16: warning: returning ‘void *’ from a function with return type ‘int’ makes integer from pointer without a cast [-Wint-conversion]
34 | return NULL;
| ^~~~
main.c:43:24: warning: returning ‘int *’ from a function with return type ‘int’ makes integer from pointer without a cast [-Wint-conversion]
43 | return ret_arr;
| ^~~~~~~
main.c:49:12: warning: returning ‘void *’ from a function with
return type ‘int’ makes integer from pointer without a cast [-
Wint-conversion]
49 | return NULL;
| ^~~~
main.c: In function ‘main’:
main.c:60:4: warning: assignment to ‘int *’ from ‘int’ makes
pointer from integer without a cast [-Wint-conversion]
60 | p = twoSum(a, sizeof(a)/sizeof(a[0]), target, &rs);
| ^
here is the full code for reference:
#include <stdio.h>
#include <stdlib.h>
twoSum(int* nums, int numsSize, int target, int* returnSize){
int i, j;
int* ret_arr = (int*)malloc(2*sizeof(int));
if(ret_arr == NULL){
*returnSize = 0;
return NULL;
}
for(i = 0; i< numsSize; i++){
for(j = i+1; j< numsSize; j++){
if(nums[i] + nums[j] == target){
*returnSize = 2;
ret_arr[0] = I;
ret_arr[1] = j;
return ret_arr;
}
}
}
*returnSize = 0;
free(ret_arr);
return NULL;
}
int main()
{
int a[] = {2,7,11,15};
int rs, target = 18;
int *p = NULL;
p = twoSum(a, sizeof(a)/sizeof(a[0]), target, &rs);
if(rs == 0){
printf("target not found");
}else{
printf("target found at indices %d and %d/n", p[0], p[1]);
free(p);
}
return 0;
}
i hope i provided enough information to understand the question, Question might be fairly simple for experienced programmers but i am just a beginner. Also please explain in easy simple words if you can thinking you are explaining to a beginner.
writing image as texts: i am not sure what to do because in this case image seems easier that text.
i) i tried changing the code as shown below in main function as well as in twosum function.
int returnSize;
p = twoSum(a, sizeof(a)/sizeof(a[0]), target, int*returnSize);
if(returnSize == 0)
ii) i have tried removing int* in line 4(2nd pic), in front of twosum function but i got an error shown here:
There seems to be a basic misunderstanding of what pointers are, how they works and why are used in the posted program.
The function twoSum is trying to find the two elements in an array whose sum equals a given target. The author of this snippet made the following design choices:
// The function returns a POINTER to some memory, allocated in the free store,
// where the two indeces will be stored, or NULL if no elements satisfy the constraint.
int*
twoSum( int* nums, int numsSize // The array is passed as pointer to its first
// element and size (number of elements)
, int target
, int* returnSize ) // This POINTER is used to "return" the size
// of the memory allocated in this function
// to the caller. This may be referred to as
// an "out parameter", it allows the function
// to change the value of a variable stored elsewhere.
{
int i, j;
int* ret_arr = (int*)malloc(2*sizeof(int));
if(ret_arr == NULL){
*returnSize = 0;
// ^ The pointer must be DEREFERENCED to access the value of the pointee.
return NULL;
}
/* [...] */
}
int main()
{
int a[] = {2,7,11,15};
int returnSize, target = 18;
//^^^^^^^^^^^^ This variable is declared as an int.
int *p = NULL;
p = twoSum(a, sizeof(a)/sizeof(a[0]), target, &returnSize);
// We need to pass its ADDRESS here ^^^^^^^^^^^
// because this function is expecting a POINTER to int.
if( returnSize == 0){
// ^^^^^^^^^^ Beeing an int, it must be used as an int value.
printf("target not found");
} else {
printf("target found at indices %d and %d/n", p[0], p[1]);
free(p); // <- Correct. The caller has this responsability.
}
return 0;
}
To answer the first point, it needs the & because it needs to take the address of the variable returnSize declared in main as an int. That address is passed to the function where it is used to initialize the local variable returnSize, which is declared in the signature of the function as a parameter of type int *, a pointer. Please note that those are two different objects, in different scopes, with different lifetimes and different type.
The call can't be written as
p = twoSum(a, sizeof(a)/sizeof(a[0]), target, int*returnSize);
// ^^^^ Syntax error, this isn't a declaration
why do we need pointer for twoSum function when we already have poiner p declared in main?
I'm not sure what the OP's doubts are here. The variable declared in main needs to be assigned a meaningful value, so it uses the value returned by the function. Inside twoSum, a variable of type pointer is needed to allocate the required memory and that is returned to main.
This isn't the only way to accoplish this task, in fact we might prefer to not allocate any memeory inside the function and pass an address instead:
#include <stdio.h>
#include <stdlib.h>
// If there exist two elements whose sum equals the 'target', the function returns 1
// and the indices of these elements are stored in the array 'indices'.
// Otherwise, the function returns 0 and doesn't modify 'indices'.
int
find_indices_of_sum( size_t n, int const* arr
, int target
, int *indices )
{
for(size_t i = 0; i < n; ++i)
{
for(size_t j = i + 1; j < n; ++j)
{
if( arr[i] + arr[j] == target )
{
indices[0] = i;
indices[1] = j;
return 1;
}
}
}
return 0;
}
int main(void)
{
int a[] = {2, 7, 11, 15};
size_t a_size = sizeof a / sizeof *a;
int target = 18;
// Declares the array in main.
int indices[2];
if ( find_indices_of_sum(a_size, a, target, indices) )
{ // Pass the address of its first element ^^^^^^^, the size is known (2).
printf("Target found at indices %d and %d\n", indices[0], indices[1]);
// ^^
// No need to free anything. 'indices' has automatic storage duration.
}
else
{
puts("Target not found");
}
return 0;
}
We can also directly return a struct with all the needed informations from the function, without using any "out parameter".
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
struct idx_pair
{
size_t first, second;
};
static inline bool is_valid(struct idx_pair p)
{
return p.first != p.second;
}
// The function returns both the indices wrapped in a single struct.
struct idx_pair
find_inidces_of_sum( size_t n, int const *arr
, int target )
{
for(size_t i = 0; i < n; ++i)
{
for( size_t j = i + 1; j < n; ++j)
{
if( arr[i] + arr[j] == target )
{
return (struct idx_pair){ .first = i, .second = j };
// ^^^^^^^^^^^^^^^^^ This is not a cast, it's a
// compound literal with designated initializers.
}
}
}
return (struct idx_pair){ .first = 0, .second = 0 };
}
int main(void)
{
int a[] = {2, 7, 11, 15};
size_t a_size = sizeof a / sizeof *a;
int target = 18;
struct idx_pair solution = find_indices_of_sum(a_size, a, target);
if ( is_valid(solution) )
{
printf("Target found at indices %d and %d\n", solution.first, solution.second);
}
else
{
puts("Target not found");
}
return 0;
}
Note though, that your compiler might not be able to optimize out all the copies and, even if the involved structure isn't that big, the generated assembly might be not optimal on some hardware.
With my programm I try to change the order of numbers in the int array. To the first function, I just passed both arrays and printed the array called arraytemp with the changed order. After that I printed in the main function the same array, just to see if the array was filled too. I havented used any pointers in the first function - how did the array got filled? Does the arrays adress get passed to functions anyway?
Then I wanted to pass arrays with the same content to the second function, but this time I used pointers. I have no clue, how to get the same result printed, because I get a stack smashing error. I am kinda comfused with '*' and '&'. So, how should I pass these arrays when using pointers?
#include <stdio.h>
void switchnum (int arraytemp[6], int array[], int laenge) {
printf("\n\nAfter (in function 1):\n");
for(int i = 0 ; i<laenge ; i++) {
arraytemp[i] = array[laenge-1-i];
printf("%d ", arraytemp[i]);
}
return 0;
}
void switchnum2 (int *arraytemp2[6], int array2[], int laenge2) {
printf("\nAfter (in function2):\n");
for(int j = 0 ; j<laenge2 ; j++) {
arraytemp2[j] = array2[laenge2-1-j];
printf("%d ", arraytemp2[j]);
}
return 0;
}
int main() {
int array[] = {4,8,1,3,0,9};
int arraytemp[6];
printf("Before (main):\n");
for(int i = 0 ; i<6 ; i++) {
printf("%d ", array[i]);
}
switchnum(arraytemp, array, 6);
printf("\nAfter (in main):\n");
for(int i = 0 ; i<6 ; i++) {
printf("%d ", arraytemp[i]);
}
int array2[] = {4,8,1,3,0,9};
int arraytemp2[6];
switchnum2(arraytemp2, array2, 6);
return 0;
}
The compiler adjusts a parameter having an array type to pointer to the array element type.
So this function declaration
void switchnum (int arraytemp[6], int array[], int laenge);
is equivalent to the following declaration
void switchnum (int arraytemp[], int array[], int laenge);
and the same way is equivalent to the following declaration
void switchnum (int *arraytemp, int *array, int laenge);
As for this function declaration
void switchnum2 (int *arraytemp2[6], int array2[], int laenge2);
then it is adjusted by the compiler to the declaration
void switchnum2 (int **arraytemp2, int *array2, int laenge2);
So the used argument expression and the function parameter have incompatible pointer types.
Pay attention to that in this call
switchnum(arraytemp, array, 6);
the both arrays are converted to pointers to their first elements of the type int *.
In fact this call is equivalent to
switchnum( &arraytemp[0], &array[0], 6);
How to pass pointer of arrays to functions
In this case you are trying to pass an array of pointers to a function and not a pointer of arrays:
void switchnum2(
int* arraytemp2[6], int array2[], int laenge2)
Maybe it helps to see the output of this example program, as it shows the output of many of your cases
Example
#include <stdio.h>
void test_arr(int*[6]);
int main(void)
{
int array[] = {4, 8, 1, 3, 0, 9};
int* pArr[6] = {0}; // 6 pointers to int
printf("original vector in main(): ");
for (int i = 0; i < 6; i += 1) printf("%d ", array[i]);
printf("\n");
for (int i = 0; i < 6; i++) pArr[i] = &array[i];
test_arr(pArr);
printf("\nIn main() &array[0] = %p\n", &array[0]);
return 0;
}
void test_arr(int* pInt[6])
{
printf("In test_array(): ");
for (int i = 0; i < 6; i += 1)
printf("%d ", *pInt[i]);
printf("\n");
int* myP = *pInt;
printf("*pInt\tpoints to value %d\n", *myP);
myP = pInt[0];
printf("pInt[0]\tpoints to value %d\n", *myP);
int x = *pInt[0];
printf("*pInt[0] = %d\n", x);
printf("\ntest_array() &pInt[0] = %X\n", pInt[0]);
return;
}
output
original vector in main(): 4 8 1 3 0 9
In test_array(): 4 8 1 3 0 9
*pInt points to value 4
pInt[0] points to value 4
*pInt[0] = 4
test_array() &pInt[0] = 197BFB00
In main() &array[0] = 000000CF197BFB00
Your program with some changes in the functions
I changed some lines in your code to get the expected result
#include <stdio.h>
void switch1(const int[],int[],const int);
void switch2(const int[],int*[],const int);
void show_array(const int[6],const char*);
int main(void)
{
int arr_out[] = {0,0,0,0,0,0};
show_array(arr_out, "arr_out in main()");
// call 1st function
printf("switch1() uses int[] as output\n");
switch1((int[6]){6, 5, 4, 3, 2, 1}, arr_out, 6);
show_array(arr_out, "arr_out using 6..1 array as input and 1st function");
// for 2nd function we need an array of pointers
int* pArr[6] = {0}; // 6 pointers to int
for (int i = 0; i < 6; i++) pArr[i] = &arr_out[i];
printf("switch2() uses int*[] as output\n");
switch2((int[6]){1, 2, 3, 4, 5, 6}, pArr, 6);
show_array(arr_out, "arr_out using 1..6 array as input and 2nd function");
return 0;
}
void switch1(const int in[], int out[], const int laenge)
{
for (int i = 0; i < laenge; i++)
out[i] = in[laenge - 1 - i];
}
void switch2(const int in[], int* out[], const int laenge)
{
for (int i = 0; i < laenge; i++)
*out[i] = in[laenge - 1 - i];
}
void show_array(const int array[6], const char* msg)
{
printf("%s:\t", msg);
for (int i = 0; i < 6; i++) printf("%d ", array[i]);
printf("\n");
}
output of the modified code
arr_out in main(): 0 0 0 0 0 0
switch1() uses int[] as output
arr_out using 6..1 array as input and 1st function: 1 2 3 4 5 6
switch2() uses int*[] as output
arr_out using 1..6 array as input and 2nd function: 6 5 4 3 2 1
about the changes
void show_array(const int array[6], const char* msg)
{
printf("%s:\t", msg);
for (int i = 0; i < 6; i++) printf("%d ", array[i]);
printf("\n");
}
This function is a helper to show the array contents and accepts a title. Very convenient here
the 2 functions has no output (printf() calls)
parameters are declared const so we can build the vector at the function call
I am using shorter names and changed the order of arguments to input and then output
void switch1(const int in[], int out[], const int laenge)
{
for (int i = 0; i < laenge; i++)
out[i] = in[laenge - 1 - i];
}
void switch2(const int in[], int* out[], const int laenge)
{
for (int i = 0; i < laenge; i++)
*out[i] = in[laenge - 1 - i];
}
Here you see the difference between the 2 functions: a single asterisk.
But in order of using the second function you need to build the vector of pointers as here
// call 1st function
printf("switch1() uses int[] as output\n");
switch1((int[6]){6, 5, 4, 3, 2, 1}, arr_out, 6);
show_array(arr_out, "arr_out using 6..1 array as input and 1st function");
// for 2nd function we need an array of pointers
int* pArr[6] = {0}; // 6 pointers to int
for (int i = 0; i < 6; i++) pArr[i] = &arr_out[i];
printf("switch2() uses int*[] as output\n");
switch2((int[6]){1, 2, 3, 4, 5, 6}, pArr, 6);
show_array(arr_out, "arr_out using 1..6 array as input and 2nd function");
i have the following program in c lanquage:
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
bool is_sorted ( int *array , int num , int prev , int *index);
int main ()
{
int N;
int i;
int prev;
int data[100];
bool flag;
printf("Enter length:\n");
scanf("%d",&N);
printf("Enter %d integers:\n" ,N);
for (i =0; i<N; i++)
{
scanf("%d",&data[i]);
}
printf("Enter previous number:\n");
scanf("%d",&prev);
int *index= NULL;
flag = is_sorted(data,N,prev,index);
if ( !flag )
{
printf("%d ", *index);
}
}
bool is_sorted ( int *array , int num , int prev , int *index)
{
if ( prev > array[0] )
{
index=prev;
return false;
}
for ( int i=0; i<num; i++)
{
if ( array[i] > array[i+1] )
{
index = array[i];
return false;
}
}
return true;
}
The function is_sorted takes as input an array of integers and another one random integer and returns true if prev < array[0] < array[1] < ... < array[n].
I am using a pointer in order to find which is the first element to spoil the serie's order but i am a little bit confused with pointer's syntax.
Running it i am getting the following results:
pointers.c:43:14: warning: assignment to ‘int *’ from ‘int’ makes pointer from integer without a cast [-Wint-conversion]
43 | index=prev;
| ^
pointers.c:51:19: warning: assignment to ‘int *’ from ‘int’ makes pointer from integer without a cast [-Wint-conversion]
51 | index = array[i];
index is a pointer to an integer. You need to dereference the pointer to assign to the variable that it points to in the caller.
*index = prev;
...
*index = array[i];
I need help please;
this is the sample code:
#include <stdio.h>
const int n = 4;
const int m = 4;
void display(int arr[n][m]){
for(int i=0; i<n; i++) {
for(int j=0;j<m;j++) {
printf("%d ", arr[i][j]);
}
printf("\n");
}
}
int *constrast(int arr[n][m])
{
int temp[n][m];
int max = 25;
int min = 10;
int uBound = 255;
int lBound = 0;
for(int i=0; i<n; i++) {
for(int j=0;j<m;j++) {
temp[i][j] = ((arr[i][j] - max) *((int)(uBound-lBound)/(max-min))+uBound;
}
}
return temp;
}
int main()
{
int disp[4][4] = {
{10, 11, 12, 13},
{14, 15, 16, 17},
{18, 19, 20, 21},
{22, 23, 24, 25}
};
printf("Image Before Stretching:\n");
display(disp);
printf("Image After Stretching:\n");
display(constrast(disp));
return 0;
}
This is the error message I get after attempting to compile:
contrast.c: In function ‘display’:
contrast.c:6:4: error: ‘for’ loop initial declarations are only allowed in C99 mode
for(int i=0; i<n; i++) {
^
contrast.c:6:4: note: use option -std=c99 or -std=gnu99 to compile your code
contrast.c:7:7: error: ‘for’ loop initial declarations are only allowed in C99 mode
for(int j=0;j<m;j++) {
^
contrast.c: In function ‘constrast’:
contrast.c:21:4: error: ‘for’ loop initial declarations are only allowed in C99 mode
for(int i=0; i<n; i++) {
^
contrast.c:22:7: error: ‘for’ loop initial declarations are only allowed in C99 mode
for(int j=0;j<m;j++) {
^
contrast.c:23:82: error: expected ‘)’ before ‘;’ token
temp[i][j] = ((arr[i][j] - max) *((int)(uBound-lBound)/(max-min))+uBound;
^
contrast.c:24:7: error: expected ‘;’ before ‘}’ token
}
^
contrast.c:28:4: warning: return from incompatible pointer type [enabled by default]
return temp;
^
contrast.c:28:4: warning: function returns address of local variable [-Wreturn-local-addr]
contrast.c: In function ‘main’:
contrast.c:44:1: warning: passing argument 1 of ‘display’ from incompatible pointer type [enabled by default]
display(constrast(disp));
^
contrast.c:4:6: note: expected ‘int (*)[(sizetype)m]’ but argument is of type ‘int *’
void display(int arr[n][m]){
^
int *constrast(int arr[n][m])
{
int temp[n][m];
int max = 25;
int min = 10;
int uBound = 255;
int lBound = 0;
for(int i=0; i<n; i++) {
for(int j=0;j<m;j++) {
temp[i][j] = ((arr[i][j] - max) *((int)(uBound-lBound)/(max-min))+uBound;
}
}
return temp;
}
Besides the type of temp which is int[4][4] doesn't match the function type which is int * you shall never return the address of a local variable since such variables are gone when the function ends and a pointer to it won't be valid any longer.
You cannot return an array from a function in C. A workaround would be to either wrap the array in a struct or to allocate the memory dynamically. Even better: Pass the output array as a parameter.
When you are trying to execute row return temp;, it returns only the pointer to the array and then goes out of the function. But that array was allocated in system stack when your code went in int *constrast(int arr[n][m]) function. On return, it deletes that variable and you are trying to work with a bad pointer.
As a solution, you can pass result array pointer as a parameter too, or use global variables.
#include<stdio.h>
void selsort(int a[],int s)
{
int min,temp,i,j;
for(i = 0; i < s ; i++)
{
min = 0;
for(j = i+1 ; j < s; j++)
{
if(min > a[j])
{
min = j;
}
}
temp = a[i];
a[i] = a[min];
a[min] = temp;
for (j=0;j<s;j++)
{
printf("%d",a[i]);
}
printf("\n");
}
}
main()
{
int i,a[5];
int size = 5;
printf("Enter elements of the array");
for(i=0;i<size;i++)
{
scanf("%d",&a[i]);
}
selsort(a[5],size);
}
The error is as follows:
selsort.c:35:2: warning: passing argument 1 of ‘selsort’ makes pointer from integer without a cast [enabled by default]
selsort.c:2:1: note: expected ‘int *’ but argument is of type ‘int’
any tips on how to avoid this problem in the future will be really helpful
You should be calling your function like this:
selsort(a, size);
a[5] means "element at index 5" (past the end of your array, by the way: the largest legal element in int a[5] is at index 4).
You should also replace
min = 0;
with
min = i;
and use it like this:
if(a[min] > a[j]) ...
In the function call selsort(a[5], size); you are only passing the 5th [nonexistent] member of a. You want selsort(a, size);