CODEWITHSUNDEEP
c
Alex Zhang
Alex Zhang

Reputation: 19

C program why the value of "a" also changed

#include <stdio.h>

int main( int argc, char *argv[] )
{
    int a[3][3] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
    int *b  = a[1];
    b[2] = 99;
    for ( int i = 0; i < 3; i++ )
    {
        printf( "%d,", a[1][i] );
    }
}

ouput is ：4,5,99,

#include <stdio.h>

int main( int argc, char *argv[] )
{
        int     a[3]    = { 1, 2, 3 };
        int     b       = a[1];
        b = 99;
        for ( int i = 0; i < 3; i++ )
        {
                printf( "%d,", a[i] );
        }
}

output is ：1,2,3

why I got such different results? i thought in the first part a should also stay unchanged

Upvotes: 1

Views: 142

Answers (3)

Manjunath N
Manjunath N

Reputation: 1375

NOTE: Here the int is assumed to be 2 bytes.

case 1: int a[3][3] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };

Will be allocated as follows:

a[0]:

a[0][0] a[0][1] a[0][2]
+----+----+----+
| 1  | 2  | 3  |
+----+----+----+
 1000 1002 1004

a[1]:

a[1][0] a[1][1] a[1][2]
+----+----+----+
| 4  | 5  |  6 | 
+----+----+----+
 1006 1008 1010

a[2]:

a[2][0] a[2][1] a[2][2]   
+----+----+----+
|  7 |  8 |  9 |
+----+----+----+
 1012 1014 1016

int *b  = a[1];

  b
+----+
|1006| // 1006 is the address of the a[1] 
+----+
 11000

b[2] = 99;

b[2] is the third element in the second array which has address location 1010 .

since b is a pointer to an second(i.e, a[1]) array and hence b[2] is nothing but *(b+2) its as offset from the first element and therefore a[1][2] will get afftected.

  b
+----+
|1006| // 1006+2 will be the location that is 1010 (as int takes 2 bytes). Here 
+----+ // address location are taken for just reference purpose.
 11000

a[1]:

a[1][0] a[1][1] a[1][2]
+----+----+----+
| 4  | 5  | 22 | 
+----+----+----+
 1006 1008 1010

printf( "%d,", a[1][i] );

Outputs: 4,5,99

case 2 :

int a[3] = { 1, 2, 3 };

a[3]:

 a[0] a[1] a[2]
+----+----+----+
| 1  | 2  | 3  |
+----+----+----+
 1000 1002 1004

int b = a[1];

  b
+----+
| 2  |
+----+
 5000

b = 99;

The value of b gets overwritten. 

  b
+----+
| 99 | // Since b is a normal variable and hence will not affect array memory location.
+----+
 5000

 printf( "%d,", a[i] );

As you can see the array does not got affected.

 Output: 1,2,3

Upvotes: 1

Gopi
Gopi

Reputation: 19874

int *b

b is a pointer which points to the second row of your 2D array and it has the memory location of the row a[1] and it can change the contents of the memory location by dereferencing it.

b[2] = *(b+2) = 99;

Whereas

int b;

b is not a pointer. You can't reference a variable like a pointer to store value in particular memory location. Hence you see that the value is not changed when you do

b = 99;

in the second code snippet

Upvotes: 1

TartanLlama
TartanLlama

Reputation: 65770

When int *b = a[1] executes, b is pointing to the same memory location as a[1], i.e. an int[3]. Thus, when you update b[2], a[1][2] is also updated.

In the second case, int b = a[1] sets b to be the same value as a[1], but at a different memory location, so when b is updated, a is not.

I would read up more about how pointers work in C.

Upvotes: 4

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