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up vote 5 down vote favorite
2

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 ===|
share|improve this question
    
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. –  Daniel Fischer Dec 5 '12 at 19:22
    
func will simply edit B values like B[0], B[1] etc.. –  fxuser Dec 5 '12 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). –  Daniel Fischer Dec 5 '12 at 19:26
    
updated question with new code with errors when i try to edit an value of B in func function –  fxuser Dec 5 '12 at 19:29
    
The answer to your original question is that func should be declared like this: int func(int (*B)[10]) –  newacct Dec 5 '12 at 21:04

5 Answers 5

up vote 10 down vote accepted

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].

share|improve this answer
    
Finally a correct answer... +1 –  Ed S. Dec 5 '12 at 19:41
up vote 3 down vote

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;
}
share|improve this answer
    
if i try to edit B[2] in func i get some errors... code: *B[2] = 5; –  fxuser Dec 5 '12 at 19:27
    
@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; –  Ed S. Dec 5 '12 at 19:35
    
@fxuser I solved your problem take a look –  Alberto Bonsanto Dec 5 '12 at 19:38
up vote 0 down vote

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.

share|improve this answer
up vote 1 down vote

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

#include <stdio.h>

int 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.

int 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;
}
share|improve this answer
    
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 ... . –  greggo Dec 13 '13 at 15:37
up vote 0 down vote 
main()
{
int *arr[4];
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]));
   }
getch();
return 0;
}
share|improve this answer
2  
Explanation will be more useful along with code. –  CODE FISH May 13 at 6:46

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