why sizeof(a pointer) returns incorrect value? [duplicate]

问题

Possible Duplicate:
Sizeof string literal

On my windows7 mingw environment, I tried this:

char str[] = "hello";

The value of sizeof(str) is 6, not 4 or 8.

How can this happen?

回答1:

sizeof(str) returns the number of bytes in the string which includes the terminating null character. I.e., "hello\0" makes for 6 bytes in your array.

If instead you had had

char* str = "hello";

then sizeof(str) would have returned the number of bytes in the pointer str (4 for 32-bit systems).

You may find this SO question Why does sizeof return different values for same string in C? of interest.

回答2:

The "sizeof()" operator returns the number of bytes. In the case of your string, it happens to be 6 (5 characters plus the null byte). For a 6 element integer array, it's 24 (at least on my 32-bit Linux system). And for a pointer, it will be 4 (again, for a 32-bit system).

#include <stdio.h>

int
main (int argc, char *argv[])
{
  char str[] = "Hello";
  char *s = "Hello";
  int a[] = {1, 2, 3, 4, 5, 6};
  printf ("sizeof (str)=%d, sizeof(s)=%d, sizseof(a)=%d\n",
    sizeof (str), sizeof(s), sizeof(a));
  return 0;
}

RESULT:

sizeof (str)=6, sizeof(s)=4, sizeof(a)=24

回答3:

It's because str is an array of chars. sizeof(array) will tell you the number of bytes that the array occupies, and since str is an array of chars, each element is only 1 byte, including the null terminating character, for 6 bytes.

You can get the number of elements in an array using: sizeof(array)/sizeof(array[0]), and so this could be more explicitly written: sizeof(str)/sizeof(str[0]), but if str is an array of char then this is equivalent to just sizeof(str).

Note that this isn't generally applicable to C strings, for the length of a string, use strlen. For instance:

char *str = "hello";
sizeof(str);  // returns 4, since str is declared as char*, not char[]

来源：https://stackoverflow.com/questions/11919369/why-sizeofa-pointer-returns-incorrect-value

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