将参数传递给函数

Question

The output of the following ode segment is 9. I think the argument a of the function foo is passed by value. Is my assumption correct. If so, how the output become 9?

 #include <stdio.h>

void foo(int[][3]);

int main(void)
{
   int a[3][3] = { {1, 2, 3}, {4, 5, 6}, {7, 8, 9} };

   foo(a);
   printf("%d\n", a[2][1]);

   return 0;
}

void foo(int b[][3])
{
   ++b;
   b[1][1] = 9;
}

Answer 1

The [] syntax for a function parameter's type is just syntactic sugar. A pointer is passed in this case, just like with any other use of an array in a function call context. That means your foo() function's signature is equivalent to:

void foo(int (*b)[3])

That is, it accepts a pointer to an array of three integers. With that in mind, let's look at what your implementation does. First, ++b advances the pointer to point to the next array of 3 integers in memory. That is, it advances b by 3 * sizeof(int) . In the case of your program, that means it's pointing at the "middle row" of a - the { 4, 5, 6 } row.

Next, you access element [2][1] of the new b . That is, row two, element one of this new offset logical array. That causes your program to write to an address past the end of a , which is undefined behaviour. At this stage, all bets are off. The output of your program could be anything .

You can get back to defined behaviour by changing that line to:

b[1][2] = 157;

for example. Then, print a[2][2] to see the change from the context of your main() function.

Answer 2

Compiler won't copy entire array. if its pass by value how about passing an array of 100000 elements. so its passed as "Pointer" . The address is passed as a copy i guess. so changing the contents of address 'll change original thing too.

seems like pass by reference by pointer.

Answer 3

According to: 6.2.3.1/3 & 6.7.5.3/7

In the call to foo, the array expression arr isn't an operand of either sizeof or &, its type is implicitly converted from "array of int" to "pointer to int" according to . Thus, foo will receive a pointer value, rather than an array value.

So when you do:

 void foo(int b[][3])

it is being implicitly converted to:

 void foo((*b)[3])

Answer 4

The output of your program is 8 as expected.Nothing unexpected here.What you are doing in b[2][1]=9 after incrementing b is writing 9 to the location a[3][1] .It doesn't affect the output of a[2][1] which remains 8 before and after you call foo() .

http://ideone.com/eWFxht

You certainly need to recheck your output.