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I'm trying to do pointer arithmetic with a pointer to array, but I get a wrong value since I can't dereference the pointer properly. Here is the code:

  #include "stdlib.h"
  #include "stdio.h"

  int main()
  {
  int a[] = {10, 12, 34};

  for (int i = 0; i < 3; ++i)
  {
      printf("%d", a[i]);
  }
  printf("\n");

  int (*b)[3] = &a;
  for (int i = 0; i < 3; ++i)
  {
      printf("%d", *(b++));
  }
 printf("\n");

  return 0;
  }

In the second for I can't get to print the correct value.

It doesn't work even if I write

printf("%d", *b[i]);

I'd like to see how to print correctly using the b++ and the b[i] syntax.

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2
  • 1
    printf("%d", *(b++)); --> printf("%d", (*b)[i]); Commented May 12, 2015 at 21:55
  • Replacing ++i with i++ should help. Commented Feb 29, 2016 at 2:31

3 Answers 3

Reset to default
1

The following should work:

printf("%d\n", *( *b+i ));

// * b + i will give you each consecutive address starting at address of the first element a[0].
// The outer '*' will give you the value at that location.

instead of:

printf("%d", *(b++));
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3 Comments

I'll select this has the correct answer but I want more explanations, why this doesn't work, and what it's doing? printf("%d\n", *( *b++ )); And why this one works: printf("%d\n", *( *b)); b = *b + 1; with a warning: incompatible pointer types assigning to 'int (*)[3]' from 'int *
*b de-references 'b' i.e. it gives you what is at the address pointed to by 'b', say address location 1000. When you do a '+ 1' to it, you are pointing to the next element from location 1000. A *b++ will cause 'b' to be incremented i.e. the pointer's address itself will be incremented, not what the pointer is pointing to. Also ... for 1-Dimensional arrays, try using just int *b = a; and use int (*b)[3] = a; for 2-Dim arrays e.g. a[2][3].
when I write *b++ I'm incrementing b why? It is the same as b++? When I do b = *b + 1 I'm incrementing what b points to. To use the postfix ++ I should write: (*b)++, correct?
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You have declared b to be a pointer to arrays of 3 integers and you have initialized it with address of a.

int (*b)[3] = &a;

In the first loop you will print the first element of a array but then you will move 3*sizeof(int) and trigger undefined behavior trying to print whatever there is.

To print it correctly:

int *b = a;
// int *b = &a[0];    // same thing
// int *b = (int*)&a; // same thing, &a[0] and &a both points to same address,
                      // though they are of different types: int* and int(*)[3]
                      // ...so incrementing they directly would be incorrect,
                      // but we take addresses as int* 
for (int i = 0; i < 3; ++i)
{
    printf("%d", (*b++));
}
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3 Comments

Wouldn't you need int *b = a;? Or does C allow the conversion int *b = &a;?
I don't think they are both correct. Your compiler just happens not to emit an error. I get array_ptr2.c:9:10: error: incompatible pointer types initializing 'int *' with an expression of type 'int (*)[3]' [-Werror,-Wincompatible-pointer-types] with clang, compiled in C99 mode.
@juanchopanza True. int* b = a; and int *b = (int*)&a; are both correct, but not int *b = &a;
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gcc will complain about the formatting in the second for loop: it will tell you format specifies type 'int' but the argument has type 'int *

your assignment of a to b should look like this:

int *b = a
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