I have this in my code:
int x = 4;
char* array = malloc(x*sizeof(char));
size_t arraysize = sizeof (array);
printf("arraysize: %zu\n", arraysize);
This code prints out,
arraysize: 8
Why is it 8 and not 4? (Since 4*sizeof(char) = 4 * 1)
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Chris
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1@modifiablelvalue: Yes I've meant zu.I've changed it. Well,it's the first time I am pointed to that site, thank you for the info! You can calm down by the way, I am a student learning to program, and yes it was the first time I got this problem! So once again, do calm down! – Chris Mar 26 '13 at 09:27
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It returns ptr size. Even if you make x value as 1000 it will return only 8. – Jeyamaran Jul 30 '14 at 06:39
5 Answers
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array is a pointer in your code. sizeof(array) therefore returns the size of that pointer in C bytes (reminder: C's bytes can have more than 8 bits in them).
So, 8 is your pointer size.
Also, the correct type specifier for size_t in printf()'s format strings is %zu.
Alexey Frunze
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1*C's bytes can have more than 8 bits in them?* Is it ?? I need a link to read – Grijesh Chauhan Mar 26 '13 at 08:13
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1@GrijeshChauhan [Here's one](http://www.gnu.org/software/libc/manual/html_node/Width-of-Type.html). – Alexey Frunze Mar 26 '13 at 08:14
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array is a pointer, hence it will be size of a pointer(sizeof(void *) or sizeof(char *)), and not sizeof(char) as you might expect. It seems that you are using 64bit computer.
Aniket Inge
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@AlexeyFrunze, is it not sizeof(int) almost everytime? The exceptions are always around but it is usually sizeof(int) or `intptr_t`? And I have mentioned both in the answer. – Aniket Inge Mar 26 '13 at 02:31
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2@Aniket Almost, but not everytime/everywhere. In 64-bit programs on x86 systems you typically have 32-bit ints and 64-bit pointers. – Alexey Frunze Mar 26 '13 at 02:33
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@AlexeyFrunze ok edited my answer and removed the reference to `sizeof(int)` – Aniket Inge Mar 26 '13 at 02:34
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`void *` and `char *` are required to have the same representation, so their sizes are the same. – R.. GitHub STOP HELPING ICE Mar 26 '13 at 02:36
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[**sizeof (int) == sizeof (void*)?**](http://stackoverflow.com/questions/502811/sizeof-int-sizeof-void) – Grijesh Chauhan Mar 26 '13 at 08:10
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@Aniket Are all pointers the same size? *Hint: Refer to [this page](http://www.c-faq.com/null/machexamp.html) in your answer.* – autistic Mar 28 '13 at 01:08
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sizeof doesn't have a return value because it isn't a function, it's a C language construct -- consider the fact that you can write sizeof array without the parentheses. As a C language construct, its value is based entirely on compile-time information. It has no idea how big your array is, only how big the array pointer variable is. See http://en.wikipedia.org/wiki/Sizeof for complete coverage of the subject.
Jim Balter
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Are you trying to say `sizeof` expressions are always evaluated during translation? – autistic Mar 28 '13 at 01:15
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sizeof(array) returns the size of a pointer.
Eric Urban
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2Only if `array` *is* a pointer. Given `char array[999];`, `sizeof(array)` is 999. – Jim Balter Mar 26 '13 at 04:43
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You are calculating the size of the full array (4*sizeof(char)) But you spect to know the number of items on the array (4).
You can do
size_t arraysize = sizeof (array)/sizeof(char);
It will return 8/2 = 4
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2sizeof(char) is *always* 1 -- the C Standard requires that. See other answers for the correct response. – Jim Balter Mar 26 '13 at 04:36