How does the following code give the answer as 2036, 2036, 2036.
What is the output of the following C code? Assume that the address of x is 2000 (in decimal) and an integer requires four bytes of memory.
int main()
{
unsigned int x[4][3] = {{1, 2, 3}, {4, 5, 6},
{7, 8, 9}, {10, 11, 12}};
printf("%u, %u, %u", x+3, *(x+3), *(x+2)+3);
}
As you've just found out, people can be a bit harsh on SO if you don't post carefully crafted code. To print addresses, use either %p or macros from <inttypes.h> such as PRIXPTR and casts to uintptr_t.
Using just %p, your code should probably read:
#include <stdio.h>
int main(void)
{
unsigned int x[4][3] =
{
{ 1, 2, 3 }, { 4, 5, 6 },
{ 7, 8, 9 }, { 10, 11, 12 }
};
printf("%p\n", (void *)x);
printf("%p, %p, %p\n", (void *)(x+3), (void *)(*(x+3)), (void *)(*(x+2)+3));
return 0;
}
Now you have defined behaviour, though the values won't be convenient like 2000 and 2036.
However, x+3 is the address of the fourth array of 3 int after the start of the array, so assuming sizeof(int) == 4 as stated, that is 36 bytes after the start of x — 2036 if x is at 2000.
*(x+3) is the address of the start of the fourth array; that is at 2036 under the same assumptions.
*(x+2)+3 adds 3 to the address of the third array of 3 int and then adds 3 more to it. That too ends up 36 bytes after the start of x, or at 2036 if x is at 2000.
On my 64-bit machine, the output reads:
0x7fff5800a440
0x7fff5800a464, 0x7fff5800a464, 0x7fff5800a464
Hex 0x24 is 36 decimal, of course.