无效的算术指针*

Question

So I need to create a call to strange_step that results in it printing out the if statement:

void strange_step(void* value)
{
    if(*(int*)(value+5) == 15)
    printf("4: Illinois\n");
else
    printf("4: ERROR\n");
}

I've tried the following (since to my knowledge, with gcc, it treats void* like a char* so increments 1 byte):

int arr[10];
arr[4] = 15;
arr[5] = 15;
arr[3] = 15;
arr[2] = 15;
arr[6] = 15;
arr[7] = 15;
arr[8] = 15;
arr[9] = 15;
//arr[10] = 15;
strange_step(arr);

But that just seems to print out random numbers. What am I missing?

Answer 1

You've got two issues there:

• Doing pointer arithmetics with void* is not allowed
• Reading an int from a non-aligned pointer may result in an error

However, if you would like to shoehorn 15 into the proper offset, you could do this:

char data[32];
*((int*)&data[5]) = 15;
strange_step(data);

This prints what you expect ( demo ). However, unless you modify strange_step to stop reading unaligned data, your program would rely on undefined behavior.

Answer 2

When you type value+5 , it's moving the pointer 5 bytes before casting it as an int* . Depending on your platform, an int is 4 or 8 bytes. On a little-endian machine

0F0000000F0000000F0000000F0000000F0000000F0000000F0000000F000000
^         ^
value     value+5

You're moving off the int boundary, so you're probably finding values that are off by some multiple of 256. I'm guessing you're seeing 251658240 a lot? That's why.

Answer 3

You need to change the function to:

void strange_step(void* value)
{
  // Treat the input pointer as an `int*`.
  int* ip = (int*)value

  // Use the 6-the value of the array.
  // You could also use if ( ip[5] == 15 )
  if(*(ip+5) == 15)
    printf("4: Illinois\n");
  else
    printf("4: ERROR\n");
}