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I want to pass the B int array pointer into func function and be able to change it from there and then view the changes in main function

#include <stdio.h> int func(int *B[10]){ } int main(void){ int *B[10]; func(&B); return 0; }

the above code gives me some errors:

In function 'main':| warning: passing argument 1 of 'func' from incompatible pointer type [enabled by default]| note: expected 'int **' but argument is of type 'int * (*)[10]'|

EDIT: new code:

#include <stdio.h> int func(int *B){ *B[0] = 5; } int main(void){ int B[10] = {NULL}; printf("b[0] = %d\n\n", B[0]); func(B); printf("b[0] = %d\n\n", B[0]); return 0; }

now i get these errors:

||In function 'func':| |4|error: invalid type argument of unary '*' (have 'int')| ||In function 'main':| |9|warning: initialization makes integer from pointer without a cast [enabled by default]| |9|warning: (near initialization for 'B[0]') [enabled by default]| ||=== Build finished: 1 errors, 2 warnings ===|
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5
  • Well, the error message explains it. You pass a pointer to an array of 10 int *, but func expects an int** (which is expected to be a pointer to the first element of an array of (10, presumably) int*s). How to fix it depends on what func does. Commented Dec 5, 2012 at 19:22
  • func will simply edit B values like B[0], B[1] etc.. Commented Dec 5, 2012 at 19:24
  • Then you probably want to pass just B. Since B is actually an array, passing &B is typically not useful, since B can't be changed (but its contents can be changed, and that's what you want to do). Commented Dec 5, 2012 at 19:26
  • updated question with new code with errors when i try to edit an value of B in func function Commented Dec 5, 2012 at 19:29
  • The answer to your original question is that func should be declared like this: int func(int (*B)[10]) Commented Dec 5, 2012 at 21:04

9 Answers 9

Reset to default
45

In your new code,

int func(int *B){ *B[0] = 5; }

B is a pointer to int, thus B[0] is an int, and you can't dereference an int. Just remove the *,

int func(int *B){ B[0] = 5; }

and it works.

In the initialisation

int B[10] = {NULL};

you are initialising anint with a void* (NULL). Since there is a valid conversion from void* to int, that works, but it is not quite kosher, because the conversion is implementation defined, and usually indicates a mistake by the programmer, hence the compiler warns about it.

int B[10] = {0};

is the proper way to 0-initialise an int[10].

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12

Maybe you were trying to do this?

#include <stdio.h> int func(int * B){ /* B + OFFSET = 5 () You are pointing to the same region as B[OFFSET] */ *(B + 2) = 5; } int main(void) { int B[10]; func(B); /* Let's say you edited only 2 and you want to show it. */ printf("b[0] = %d\n\n", B[2]); return 0; }
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3 Comments

if i try to edit B[2] in func i get some errors... code: *B[2] = 5;
@fxuser: "I get some errors" is not a useful problem description. Ask for help as if you were the person being asked. You are getting an error because B is a pointer to int and B[2] returns an int, not a pointer. So, you just want B[2] = 5;
This is the answer I was looking for. Why would you do array + 2 instead of array[2]?
8

If you actually want to pass an array pointer, it's

#include <stdio.h> void func(int (*B)[10]){ // ptr to array of 10 ints. (*B)[0] = 5; // note, *B[0] means *(B[0]) //B[0][0] = 5; // same, but could be misleading here; see below. } int main(void){ int B[10] = {0}; // not NULL, which is for pointers. printf("b[0] = %d\n\n", B[0]); func(&B); // &B is ptr to arry of 10 ints. printf("b[0] = %d\n\n", B[0]); return 0; }

But as mentioned in other answers, it's not that common to do this. Usually a pointer-to-array is passed only when you want to pass a 2d array, where it suddenly looks a lot clearer, as below. A 2D array is actually passed as a pointer to its first row.

void func( int B[5][10] ) // this func is actually the same as the one above! { B[0][0] = 5; } int main(void){ int Ar2D[5][10]; func(Ar2D); // same as func( &Ar2D[0] ) }

The parameter of func may be declared as int B[5][10], int B[][10], int (*B)[10], all are equivalent as parameter types.

Addendum: you can return a pointer-to-array from a function, but the syntax to declare the function is very awkward, the [10] part of the type has to go after the parameter list:

int MyArr[5][10]; int MyRow[10]; int (*select_myarr_row( int i ))[10] { // yes, really return (i>=0 && i<5)? &MyArr[i] : &MyRow; }

This is usually done as below, to avoid eyestrain:

typedef int (*pa10int)[10]; pa10int select_myarr_row( int i ) { return (i>=0 && i<5)? &MyArr[i] : &MyRow; }
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1 Comment

And I note that C++ is getting templateable typedefs, so it should be possible to write ptr_to_arr_of<int,10> select_myarr_row ... .
2

Make use of *(B) instead of *B[0]. Here, *(B+i) implies B[i] and *(B) implies B[0], that is
*(B+0)=*(B)=B[0].

#include <stdio.h> int func(int *B){ *B = 5; // if you want to modify ith index element in the array just do *(B+i)=<value> } int main(void){ int B[10] = {}; printf("b[0] = %d\n\n", B[0]); func(B); printf("b[0] = %d\n\n", B[0]); return 0; }
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1

In new code assignment should be,

B[0] = 5

In func(B), you are just passing address of the pointer which is pointing to array B. You can do change in func() as B[i] or *(B + i). Where i is the index of the array.

In the first code the declaration says, 

int *B[10]

says that B is an array of 10 elements, each element of which is a pointer to a int. That is, B[i] is a int pointer and *B[i] is the integer it points to the first integer of the i-th saved text line.

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0 
main() { int *arr[5]; int i=31, j=5, k=19, l=71, m; arr[0]=&i; arr[1]=&j; arr[2]=&k; arr[3]=&l; arr[4]=&m; for(m=0; m<=4; m++) { printf("%d",*(arr[m])); } return 0; }
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0

Using the really excellent example from Greggo, I got this to work as a bubble sort with passing an array as a pointer and doing a simple -1 manipulation.

#include<stdio.h> void sub_one(int (*arr)[7]) { int i; for(i=0;i<7;i++) { (*arr)[i] -= 1 ; // subtract 1 from each point printf("%i\n", (*arr)[i]); } } int main() { int a[]= { 180, 185, 190, 175, 200, 180, 181}; int pos, j, i; int n=7; int temp; for (pos =0; pos < 7; pos ++){ printf("\nPosition=%i Value=%i", pos, a[pos]); } for(i=1;i<=n-1;i++){ temp=a[i]; j=i-1; while((temp<a[j])&&(j>=0)) // while selected # less than a[j] and not j isn't 0 { a[j+1]=a[j]; //moves element forward j=j-1; } a[j+1]=temp; //insert element in proper place } printf("\nSorted list is as follows:\n"); for(i=0;i<n;i++) { printf("%d\n",a[i]); } printf("\nmedian = %d\n", a[3]); sub_one(&a); return 0; }

I need to read up on how to encapsulate pointers because that threw me off.

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0

The argument of func is accepting double-pointer variable. Hope this helps...

#include <stdio.h> int func(int **B){ } int main(void){ int *B[10]; func(B); return 0; }
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-2

In the function declaration you have to type as

VOID FUN(INT *a[]); /*HERE YOU CAN TAKE ANY FUNCTION RETURN TYPE HERE I CAN TAKE VOID AS THE FUNCTION RETURN TYPE FOR THE FUNCTION FUN*/ //IN THE FUNCTION HEADER WE CAN WRITE AS FOLLOWS void fun(int *a[]) //in the function body we can use as a[i]=var
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