问题
For example result of this code snippet depends on which machine: the compiler machine or the machine executable file works?
sizeof(short int)
回答1:
sizeof is a compile time operator.
回答2:
It depends on the machine executing your program. But the value evaluates at compile time. Thus the compiler (of course) has to know for which machine it's compiling.
回答3:
sizeof is evaluated at compile time, but if the executable is moved to a machine where the compile time and runtime values would be different, the executable will not be valid.
回答4:
As of C99, sizeof is evaluated at runtime if and only if the operand is a variable-length array, e.g. int a[b], where b is not known at compile time. In this case, sizeof(a) is evaluated at runtime and its result is the size (in bytes) of the entire array, i.e. the size of all elements in the array, combined. To get the number of elements in the array, use sizeof(a) / sizeof(b). From the C99 standard:
The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.
Note that all of this is different from what you'd get if you allocated an array on the heap, e.g. int* a = new int[b]. In that case, sizeof(a) would just give you the size of a pointer to int, i.e. 4 or 8 bytes, regardless of how many elements are in the array.
来源：https://stackoverflow.com/questions/2615203/is-sizeof-in-c-evaluated-at-compilation-time-or-run-time