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1

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

3 Answers

up vote 4 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].

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Finally a correct answer... +1 –  Ed S. Dec 5 '12 at 19:41
up vote 1 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.

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