CODEWITHSUNDEEP
carrays
Abhishek Choubey
Abhishek Choubey

Reputation: 883

Unable to understand behaviour of arrays

I have the following code : 

 #include<stdio.h>
   void func(int [][3]);

   int main(){
           int a[][3] = {{1,2,3}, {4,5,6}, {7,8,9}};
           func(a);
           printf("%d", a[2][1]);
   }       

  void func(int b[][3]){
          ++b;
          b[1][1] = 17;
  }

Problem :
I expected the printf statement to print 8 but its printing 17.
I dont understand why ?

Thanks

Upvotes: 3

Views: 98

Answers (3)

dmitry_romanov
dmitry_romanov

Reputation: 5425

You can take a look on memory layout of your code with extended code:

     #include <stdio.h>

     void print_addr(int b[][3])
     {
        for (int i = 0 ; i < 3 ; i++) {
           for (int j = 0 ; j < 3 ; j++)
              printf("%p ", &b[i][j]);
           printf("\n");
        }

     }

     void func(int b[][3]){
        print_addr(b);
        printf("sizeof(b): %d   sizeof(b[0][0]) %d\n", sizeof(b), sizeof(b[0][0]));
        ++b;
        print_addr(b);
        b[1][1] = 17;
     }

     int main()
     {
        int a[][3] = {{1,2,3}, {4,5,6}, {7,8,9}};
        func(a);
        printf("%d\n", a[2][1]);
     }

It will show that after the shift you make, the b[1][1] address will point into a[2][1] cell because of b is a pointer to lines of array and b++ shifts down by one line.

Also you pass value by pointer, so function func accesses the same memory where a sits.

If I may suggest, just print everything while you learn, and also take a look on this nice presentation: http://www.slideshare.net/olvemaudal/deep-c?qid=1fcb7c99-e916-4bcd-baa2-1a1f974d1c68

Upvotes: 5

Stanley Xu
Stanley Xu

Reputation: 61

In C, array parameters like int b[][3] are just pointers to the memory where the array resides. So an update in func() to the array persists, since you're accessing the same memory in main and func().

b currently points to first array {1, 2, 3}. ++b does pointer arithmetic, so b increments to point to the next array {4, 5, 6}. So b[1][1] will update main's a[2][1], which is what you printed. Try the following code to see that you are indeed updating the same memory addresses.

#include<stdio.h>
void func(int [][3]);

int main(){
    int a[][3] = {{1,2,3}, {4,5,6}, {7,8,9}};
    printf("%p\n", a+1);
    func(a);
    printf("%d\n", a[2][1]);
}

void func(int b[][3]){
    ++b;
    printf("%p\n", b);
    b[1][1] = 17;
}

Upvotes: 4

Yu Hao
Yu Hao

Reputation: 122493

Note ++b in func(), after this, b (originally pointing to a[0][0]), now points to a[1][0], so the following

b[1][1] = 17;

modifies a[2][1] outside.

Upvotes: 7

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