stackuser stackuser - 2 months ago 5
C Question

Are the following declarations correct in C?

char array[8] = "Raining";


I think all the comments are true ?

char *p1 = array + 7; --> points to '\0'

char *p2 = array + 8; --> undefined behaviour

char *p3 = array + 9; --> undefined behaviour

char *p4 = array + 10; --> undefined behaviour


Is my understanding correct?

Answer

In your case,

  char *p1 = array + 7;

and

  char *p2 = array + 8;

are valid, because, you are legally allowed to point to

  • any object inside the array length
  • one past the last array element until you dereference it.

OTOH,

  char *p3 = array + 9;

and

  char *p4 = array + 10; 

are undefined.


Quoting C11, chapter §6.5.6, "Additive operators" (emphasis mine)

[same in C99, chapter §6.5.6/p8, for anybody interested]

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. [...]