试图理解：对二进制表达式无效的操作数，C

Question

Last line yields "invalid operands to binary expression". Trying to understand why. Does it mean that "p2-p1" is an invalid operand to the binary expression "-" that lies to the right of p3? Any rule I can follow here? Confusing to me because "3-2-1" integers are valid.

  int array[3] = {1,2,3};
  int* p1 = &array[0];
  int* p2 = &array[1];
  int* p3 = &array[2];

  p3-p2-p1;

Answer 1

You are doing address arithmetic. Given operator precedence, it is evaluating p1-p2-p3 as (p1-p2)-p3 . p1-p2 yields not an address but an integer. Then you are attempting to subtract an address from an integer, which isn't valid. You could do p1-(p2-p3) , then it's taking p2-p3 , yielding an integer, and subtracting that as an integer offset from an address ( p1 ), which will compile. However, [Thanks to @EOF for this reference in his comment] such subtraction (of integer from a pointer) would only be valid if it points somewhere within the allocation for p1 . It's subject to the C11 standard described specifically in section 6.5.6, excerpted below:

When an expression that has integer type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the pointer operand points to an element of an array object, and the array is large enough, the result points to an element offset from the original element such that the difference of the subscripts of the resulting and original array elements equals the integer expression. In other words, if the expression P points to the i-th element of an array object, the expressions (P)+N (equivalently, N+(P)) and (P)-N (where N has the value n) point to, respectively, the i+n-th and i−n-th elements of the array object, provided they exist. Moreover, if the expression P points to the last element of an array object, the expression (P)+1 points one past the last element of the array object, and if the expression Q points one past the last element of an array object, the expression (Q)-1 points to the last element of the array object. If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined. If the result points one past the last element of the array object, it shall not be used as the operand of a unary * operator that is evaluated.

Answer 2

In your code, p1 , p2 and p3 are all pointers to integers , not integers.

To get what you want, you probably want:

*p3 - *p2 - *p1;

where the * operator is the dereference operator . It dereferences pointers, so in this case *p3 etc are of type int . You can think of it as the inverse of the & address-of operator.