c中的sizeof运算符的大小是多少？

Question

In recent I was attend one Interview on "SureSoft Technology"...... In that interview, they ask one question lik "What is size for the sizeof operator in c? "

If any one Know answer Share with me?

Answer 1

The sizeof operator itself has no size. Its result will generally* turn into a constant value at compile time.

As for the value it returns, that would be the size, in bytes, of the argument. The type of the result is size_t (defined in <stdlib.h> ) (§6.5.3.4.4)

* - with the notable exception of dynamically-sized automatic arrays.

Answer 2

You'll find the answers here and here : the result of sizeof is of type size_t, and:

"The actual type of size_t is platform-dependent; a common mistake is to assume size_t is the same as unsigned int, which can lead to programming errors when moving from 32 to 64-bit architecture, for example. According to the 1999 ISO C standard (C99), size_t is an unsigned integer type of at least 16 bits."

Answer 3

在stddef.h中定义的size_t类型。

Answer 4

The answer is simple - it's the size in bytes of the given data structure that you pass to sizeof.

For example:

sizeof(char) // one bytes
sizeof(int) // four bytes

Hope that helps.