Hi im trying to use a function to initialize an array with random number. somehow i get this error that i couldnt be able to solve
void arrayInit(int *A, int n){
int i;
for ( i = 0; i < n; ++i)
{
A[i] = rand();
}
}
call arrayInit() in main
int main(void){
int array1[1000];
arrayInit(&array1, 1000);
return 0;
}
I get error saying:
csort.c:62:13: warning: passing argument 1 of ‘arrayInit’ from incompatible pointer type
arrayInit(&array1, 1000);
^
csort.c:8:6: note: expected ‘int *’ but argument is of type ‘int (*)[1000]’
void arrayInit(int *A, int n){
^
You need the call for arrayInit to be like this:
arrayInit(array1, 1000);
Instead of this:
arrayInit(&array1, 1000);
The name of an array decays to be a pointer to the first element, which is what your function needs.
When you have
int array1[1000];
The type of &array1 is int (*)[1000], which is not what arrayInit expects. By using
arrayInit(arra1, 1000);
you are letting compiler decay the array to a pointer, which of of type int* in your case.
An array does not decay to a pointer in couple of cases:
When you use the & operator.
When you use the sizeof operator.
The identifier of any array is a type * itself.
So you need to call your function like this:
arrayInit(array1, 1000);.
Calling your function with an address of a pointer(&array1) would require a int** A(a pointer to a pointer type) in your function definition, which is certainly not the thing to do in this case, since the indexing in this case would also give an error.
you can call arrayInit() using the '&' as:
arrayInit(&array1[0], 1000);
Related
I implemented a queue and want a function that prints its elements, regardless of the type. So I made a function printQueue as follows:
void printQueue(queue_t *queue, void (*f)(void *)) {
node_t *node = (*queue).front;
while(node != NULL) {
(*f)(node->item);
puts("");
node = (*node).next;
}
}
The idea is that it will receive the queue, and a function to print an element from the queue (the queue was implemented to be able to hold any kind of element). So this function would iterate the queue and call that function to print each of it's elements.
Then I made a function to print arrays of ints which contain their own size in it's first element as follows:
void printSpecialIntArray(int *array) {
int i;
for(i=0; i<array[0]; i++) {
printf("%d ", array[i]);
}
}
However if I initialize a queue say Q with that kind of arrays and call printQueue(Q, printSpecialIntArray) it seems to work as intended but I get a warning:
warning: passing argument 2 of ‘printQueue’ from incompatible pointer type [-Wincompatible-pointer-types]
printQueue(Q, printSpecialIntArray);
^~~~~~~~~~~~
In file included from main.c:3:0:
queue.c:32:6: note: expected ‘void (*)(void *)’ but argument is of type ‘void (*)(int *)’
void printQueue(queue_t *queue, void (*f)(void *)) {
Why does this warning happen and how can it be fixed? Is there a cleaner way to do this?
Also if you're wondering, I'm using an array which has it's own size as an element because I'm implementing a process simulator, and those "Special int arrays" are actually supposed to represent a process. If you know a cleaner way to do that I would also appreciate if you include in your answer/comment.
The error message is very straight forward. The function void printQueue(queue_t *queue, void (*f)(void *)) expects f to be a pointer to a function returning void and taking exactly one argument and its type should be void*. The function printSpecialIntArray does not fulfill this, since it's argument is int*.
One solution is to change f so that it takes int* as argument instead of void*. Another solution is to rewrite printSpecialIntArray:
void printSpecialIntArray(void *a) {
int *array = (int*) a;
int i;
for(i=0; i<array[0]; i++) {
printf("%d ", array[i]);
}
}
Note:
Even though a void pointer can be safely promoted to an int pointer, the function f still needs to be a function that takes a void pointer as argument. It's not the argument that's the problem, it's the function declaration itself.
Your function signature does not match.
You need to provide the expected signature and cast the pointer according to your needs:
void printSpecialIntArray(void *arr) {
int array = (int *)arr;
int i;
for(i=0; i<array[0]; i++) {
printf("%d ", array[i]);
}
}
I have an array:
char gameBoard[6][6];
// Initilize the array
for(int i = 0; i < 6; i++) {
for(int o = 0;o < 6;o++) {
gameBoard[i][o] = ' ';
}
}
Later in my code I have a function I am trying to use that will modify this array:
void placePiece(char piece, char *gameBoard) {
int x = 0;
int y = 0;
gameBoard[posXInArray][posYInArray] = piece;
}
I am calling the function from the same scope as where the gameBoard array is created and initialized.
placePiece('X', gameBoard);
The warning I get doing this:
warning: passing argument 2 of ‘placePiece’ from incompatible pointer type [enabled by default]
placePiece('X', gameBoard);
^
note: expected ‘char *’ but argument is of type ‘char (*)[6]’
void placePiece(char piece, char *gameBoard);
I am confused as to what I am doing wrong? I assumed to modify the array from another function I would pass the pointer so that I could directly modify the array. Is this incorrect? I have also tried &gameBoard in the call with the same warning message. I'm guessing I am doing something silly that is incorrect?
You can't pass the 2-dimensional array to this function
void placePiece(char piece, char *gameBoard)
I think this would work fine for you
void placePiece(char piece, unsigned int size, char gameBoard[size][size])
you can use it this way
placePiesce('X', 6, gameBoard);
or better
placePiesce('X', sizeof(gameBoard) / sizeof(gameBoard[0]), gameBoard);
and since the type of gameBoard[0] is char this will also work
placePiesce('X', sizeof(gameBoard), gameBoard);
because sizeof(gameBoard[0]) == 1 in that case.
hi I'm attempting to create a program that accepts a 7 element array as an argument and returns the third through fifth element of that array to a smaller array however i'm currently getting this error
assign8p7.c: In function 'main':
assign8p7.c:18:2: warning: passing argument 1 of 'copysect' makes pointer from
integer without a cast [enabled by default]
assign8p7.c:3:6: note: expected 'int *' but argument is of type 'int'
from what i can tell the warning has a problem with me passing it an array in the arguments does anyone know how i might fix this? also any other advice for my code is welcome.
#include <stdio.h>
int *copysect(int ar[],int start,int end)
{
int i;
static int retar[3];
for(i = 0; i<3;i++)
{
retar[i+start]=ar[i+start];
}
return retar;
}
int main(int argc, char const *argv[])
{
int arry[7] = {1,2,3,4,5,6,7};
int miniarry[3];
miniarry[0] = *copysect(arry[0],3,5);
return 0;
}
int *copysect(int ar[],int start,int end)
Okay, copysect takes as its first parameter an array of integers.
miniarry[0] = *copysect(arry[0],3,5);
Oops, you passed it a single integer instead of an array.
You are calling the function copysect with the first element in the array, not the pointer to the array. The correct call is:
copysect(arry,3,5);
You could calculate the difference of the array dynamically. Now the caller of copysect function has to know that the difference between start and end is 2.
int retar[end - start + 1]
The assignment in the for loop is wrong. You are dereferencing a value that is out of scope of retar array
retar[i]=ar[i+start];
When calling the copysect function, you are assigning only the first element in the miniarry by dereferencing the array that the function returns, instead of the whole array.
It's not the best idea to have a static array in a function (that would be problematic if you called the function more than once, etc). Instead, you could declare the smaller array elswhere and pass it as a parameter to the function.
void copysect(int ar[], int retar[], int start,int end, )
Straight to the code:
#define PRO_SIGNAL( func, param ) (*func)(param)
void PRO_SIGNAL( paint[0], Pro_Window* );
signal->paint = realloc( signal->paint, sizeof( void (*)(Pro_Window*) ) * signal->paint_count );
The Error:
error: incompatible types when assigning to type 'void (*[])(struct Pro_Window *)' from type 'void *'|
It appears that you are assigning to an array not a pointer.
From your output error message:
'void (*[])(struct Pro_Window *)' from type 'void *'|
Note the [] in there (and it certainly isn't a lambda!) rather than a *
If this is an "extendable" struct you need to realloc the entire struct not just the array member.
By the way, a tip: if realloc fails it returns a NULL pointer and if you assign it to the variable that was being realloc'ed, the original memory it was pointing to will be lost forever. So always realloc into a temp first, check the value, and then assign back to the original pointer if it worked.
You don't show us the definition of singal->paint, but I infer from the error message that it's declared as an array of function pointers, meaning signal is a struct with a flex array (paint[]). You can't assign to an array, you need to realloc the whole struct.
Not sure what you're trying to do, but this works perfectly here:
#include <stdlib.h>
int main(int argc, char ** argv)
{
void (**foobar) (int a, int b);
void (**tmp) (int a, int b);
foobar = NULL;
if (!(foobar = malloc(sizeof(*foobar)*4))
return 1;
if (!(tmp = realloc(foobar, sizeof(*foobar)*5)) {
free(foobar);
return 1;
} else {
foobar = tmp;
}
free(foobar);
return 0;
}
So, either you're trying to realloc an array like Kevin says, or perhaps you're compiling in C++ mode, where I believe that the cast is not implicit.
Edit: I've added some error handling.
Hi
I'm new to c language i hava a problem :
i want to send a 2-d array to a function via pointer.
The function should return pointer to 2-d array.
I wrote the following code for this :
#include<stdio.h>
int* put(int *b);
int main()
{
int a[2][3],i,j;
system("clear");
put(a);
for(i=0;i<2;i++)
{
for(j=0;j<3;j++)
{
printf("\na[%d][%d]= %d",i,j,a[i][j]);
}
}
return 0;
}
int* put(int *b)
{
for(i=0;i<2;i++)
{
for(j=0;j<3;j++)
{
b[i][j]=i;
}
}
return b;
}
when i compile it with gcc2de.c it shows following errors :
2de.c: In function ‘main’:
2de.c:9: warning: passing argument 1 of ‘put’ from incompatible pointer type
2de.c:3: note: expected ‘int *’ but argument is of type ‘int (*)[3]’
2de.c: In function ‘put’:
2de.c:28: error: subscripted value is neither array nor pointer
2de.c: In function ‘main’:
2de.c:32: error: expected declaration or statement at end of input
Than i just change the code of function which is following :
#include<stdio.h>
int* put(int **b);
int main()
{
int a[2][3],i,j;
system("clear");
put(a);
for(i=0;i<2;i++)
{
for(j=0;j<3;j++)
{
printf("\na[%d][%d]= %d",i,j,a[i][j]);
}
}
return 0;
}
int* put(int **b)
{
for(i=0;i<2;i++)
{
for(j=0;j<3;j++)
{
b[i][j]=i;
}
}
return b;
}
when i complie it i got following errors:
2de.c: In function ‘main’:
2de.c:9: warning: passing argument 1 of ‘put’ from incompatible pointer type
2de.c:3: note: expected ‘int **’ but argument is of type ‘int (*)[3]’
2de.c: In function ‘put’:
2de.c:31: warning: return from incompatible pointer type
2de.c: In function ‘main’:
2de.c:32: error: expected declaration or statement at end of input
2de.c: In function ‘main’:
2de.c:9: warning: passing argument 1 of ‘put’ from incompatible pointer type
2de.c:3: note: expected ‘int **’ but argument is of type ‘int (*)[3]’
2de.c: In function ‘put’:
2de.c:31: warning: return from incompatible pointer type
2de.c: In function ‘main’:
2de.c:32: error: expected declaration or statement at end of input
what I'm doing wrong ?
can anybody tell me what is the way to pass 2d-array via pointers to a function ?
can anybody tell me how to return two d array via pointer in a function
The first error that you have is that you are not passing a correct type as declared by your function. So to clean up your code with the least amount of corrections, it would probably look something like this:
#include<stdio.h>
void put(int *b);
int main()
{
int a[2][3],i,j;
put(&a[0][0]);
for(i=0;i<2;i++)
{
for(j=0;j<3;j++)
{
printf("\na[%d][%d]= %d", i, j, a[i][j]);
}
}
printf("\n\n");
system("PAUSE"); // Not recommended, but works for now
return 0;
}
void put(int *b)
{
int count = 1;
int i, j;
for(i=0;i<2;i++)
{
for(j=0;j<3;j++)
{
//b[i][j]=i;
*(b + ((i*3) + j)) = count++;
}
}
}
The two major corrections are:
You pass in the start address of your 2-D array explicitly by addressing it as &a[0][0].
Also, note the pointer arithmetic that you'll have to use when you use an int *b as well.
Note also that since you're passing in a pointer, you're modifying the value at that address location. Thus there is no need to return a pointer back at all.
Hope it helps. Cheers!
where are you storing the return value from put ?
the declaration should be int** put( int **) according yo your code.
The first error you have is that you are trying to define a function inside another function. The simplest thing to do is to just define put where you declare it:
int put()
{
/* definition of put */
}
int main()
{
/* body calls put */
}
The second problem is that in neither code snippet are you passing a compatible parameter to put.
If you want to pass a to a function then you should note that arrays as arguments always decay to a pointer to their first element.
a has type int [2][3], i.e. an array of 2 arrays of 3 ints. This will decay to a pointer to an array of 3 ints or int (*)[3]. This should explain the compile error that you are getting. You should declare put either as:
void put(int (*b)[3]);
or as the completely equivalent:
void put(int b[][3]);
Because you cannot pass arrays by value the compiler will automatically convert a function declaration which takes an array parameter to one which takes the equivalent pointer parameter.
I've changed the return type to void as you don't use or need the return value as you are passing the parameter by pointer. You should remove return b; from your definition of put.
Hint: Don't think of int[2][3] as a 2-d array but as an array of arrays.
You can not return a 2d-array from a function in C, you can only return a 1d-array of pointers to the first elements of a 2d-array.
Maybe you could find useful this:
Pass 2d array to function in C?
or this:
C++ Returning multidimension array from function
1.you should declare or define the function before use it,it's different with other popular luanguage.
2.you do not need to return the pointer in the function put ,the data in array has be changed
3.you needed to notice the type ,the type of int array[][] is int **