passing the array inside a function

int f(int b[][3]);

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

int f(int b[][3])
{
    ++b;
    b[1][1] = 1;
}

3x3 => 9 elements contained in the 2-D array a. When it's passed, then b will contain the the base address of the a. If suppose base address is 1000 then ++b how does it 开发者_开发技巧to 3 locations and not 9 locations ahead? Are we doing typecasting when the variable a is passed to b[][3] as only the three elements?

How does b[1][1] correspond to the address of 8 and not 5?

We can't do incrementing or decrementing in an array as array is a const pointer, but how is that they are incrementing ++b as its an array?

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The function heading

 int f(int b[][3])

is a nothing more than a confusing way to write (and is exactly equivalent to)

 int f(int (*b)[3])

The type of b is "pointer to three-element array of int". When you increment the b parameter you adjust it to point to the next three-element array of int -- now it points to {4,5,6}. Then b[1] indexes once more and gives you the array {7,8,9} and finally b[1][1] gives you the oneth element of that array, namely 8.

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C multidimensional arrays are really linear, except that there is syntactic sugar to do the arithmetic correctly.

so with b[][3], it excepts a 1-D array and implicitly translates b[i][j] --> b[3*i+j]

++b works as follows: (++b)[i][j] = ORIGINAL_b[i+1][j]. So in your case, you are accessing ORIGINAL_b[1+1][1] = ORIGINAL_b[2*3+1] = ORIGINAL_b[7] (the 8th element)

Note: this is in stark contrast to the dynamic malloc version (in **b, b is a array of pointers)

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How is b[1][1] corresponds to the address of 8 and not address of 5?

This is expected behavior:

int f(int b[][3])
{
    //at this point b[0][0] is 1, b[1][1] is 5

    ++b;
    //now b[0][0] is 4, b[1][1] is 8

    b[1][1]=1;
}

The pointer has incremented to point to the next memory slot, which is the second slot of array a. Basically:

b -> a[0]
++b -> a[1]