I am getting confused with size_t
in C. I know that it is returned by the sizeof
operator. But what exactly is it? Is it a data type?
Let's say I have a for
loop:
for(i = 0; i < some_size; i++)
Should I use int i;
or size_t i;
?
According to the 1999 ISO C standard (C99),
size_t
is an unsigned integer type of at least 16 bit (see sections 7.17 and 7.18.3).
size_t
is an unsigned data type defined by several C/C++ standards, e.g. the C99 ISO/IEC 9899 standard, that is defined instddef.h
.1 It can be further imported by inclusion ofstdlib.h
as this file internally sub includesstddef.h
.This type is used to represent the size of an object. Library functions that take or return sizes expect them to be of type or have the return type of
size_t
. Further, the most frequently used compiler-based operator sizeof should evaluate to a constant value that is compatible withsize_t
.
As an implication, size_t
is a type guaranteed to hold any array index.
size_t
is an unsigned type. So, it cannot represent any negative values(<0). You use it when you are counting something, and are sure that it cannot be negative. For example, strlen()
returns a size_t
because the length of a string has to be at least 0.
In your example, if your loop index is going to be always greater than 0, it might make sense to use size_t
, or any other unsigned data type.
When you use a size_t
object, you have to make sure that in all the contexts it is used, including arithmetic, you want non-negative values. For example, let's say you have:
size_t s1 = strlen(str1);
size_t s2 = strlen(str2);
and you want to find the difference of the lengths of str2
and str1
. You cannot do:
int diff = s2 - s1; /* bad */
This is because the value assigned to diff
is always going to be a positive number, even when s2 < s1
, because the calculation is done with unsigned types. In this case, depending upon what your use case is, you might be better off using int
(or long long
) for s1
and s2
.
There are some functions in C/POSIX that could/should use size_t
, but don't because of historical reasons. For example, the second parameter to fgets
should ideally be size_t
, but is int
.
If you are the empirical type,
echo | gcc -E -xc -include 'stddef.h' - | grep size_t
Output for Ubuntu 14.04 64-bit GCC 4.8:
typedef long unsigned int size_t;
Note that stddef.h
is provided by GCC and not glibc under src/gcc/ginclude/stddef.h
in GCC 4.2.
Interesting C99 appearances
malloc
takes size_t
as an argument, so it determines the maximum size that may be allocated.
And since it is also returned by sizeof
, I think it limits the maximum size of any array.
Since nobody has yet mentioned it, the primary linguistic significance of size_t
is that the sizeof
operator returns a value of that type. Likewise, the primary significance of ptrdiff_t
is that subtracting one pointer from another will yield a value of that type. Library functions that accept it do so because it will allow such functions to work with objects whose size exceeds UINT_MAX on systems where such objects could exist, without forcing callers to waste code passing a value larger than "unsigned int" on systems where the larger type would suffice for all possible objects.
To go into why size_t
needed to exist and how we got here:
In pragmatic terms, size_t
and ptrdiff_t
are guaranteed to be 64 bits wide on a 64-bit implementation, 32 bits wide on a 32-bit implementation, and so on. They could not force any existing type to mean that, on every compiler, without breaking legacy code.
A size_t
or ptrdiff_t
is not necessarily the same as an intptr_t
or uintptr_t
. They were different on certain architectures that were still in use when size_t
and ptrdiff_t
were added to the Standard in the late 1980s, and becoming obsolete when C99 added many new types but not gone yet (such as 16-bit Windows). The x86 in 16-bit protected mode had a segmented memory where the largest possible array or structure could be only 65,536 bytes in size, but a far
pointer needed to be 32 bits wide, wider than the registers. On those, intptr_t
would have been 32 bits wide but size_t
and ptrdiff_t
could be 16 bits wide and fit in a register. And who knew what kind of operating system might be written in the future? In theory, the i386 architecture offers a 32-bit segmentation model with 48-bit pointers that no operating system has ever actually used.
The type of a memory offset could not be long
because far too much legacy code assumes that long
is exactly 32 bits wide. This assumption was even built into the UNIX and Windows APIs. Unfortunately, a lot of other legacy code also assumed that a long
is wide enough to hold a pointer, a file offset, the number of seconds that have elapsed since 1970, and so on. POSIX now provides a standardized way to force the latter assumption to be true instead of the former, but neither is a portable assumption to make.
It couldn’t be int
because only a tiny handful of compilers in the ’90s made int
64 bits wide. Then they really got weird by keeping long
32 bits wide. The next revision of the Standard declared it illegal for int
to be wider than long
, but int
is still 32 bits wide on most 64-bit systems.
It couldn’t be long long int
, which anyway was added later, since that was created to be at least 64 bits wide even on 32-bit systems.
So, a new type was needed. Even if it weren’t, all those other types meant something other than an offset within an array or object. And if there was one lesson from the fiasco of 32-to-64-bit migration, it was to be specific about what properties a type needed to have, and not use one that meant different things in different programs.
size_t
and int
are not interchangeable. For instance on 64-bit Linux size_t
is 64-bit in size (i.e. sizeof(void*)
) but int
is 32-bit.
Also note that size_t
is unsigned. If you need signed version then there is ssize_t
on some platforms and it would be more relevant to your example.
As a general rule I would suggest using int
for most general cases and only use size_t
/ssize_t
when there is a specific need for it (with mmap()
for example).
size_t
is an unsigned integer data type which can assign only 0 and greater than 0 integer values. It measure bytes of any object's size and is returned by sizeof
operator.
const
is the syntax representation of size_t
, but without const
you can run the program.
const size_t number;
size_t
regularly used for array indexing and loop counting. If the compiler is 32-bit
it would work on unsigned int
. If the compiler is 64-bit
it would work on unsigned long long int
also. There for maximum size of size_t
depending on the compiler type.
size_t
already defined in the <stdio.h>
header file, but it can also be defined by the
<stddef.h>
, <stdlib.h>
, <string.h>
, <time.h>
, and <wchar.h>
headers.
const
)#include <stdio.h>
int main()
{
const size_t value = 200;
size_t i;
int arr[value];
for (i = 0 ; i < value ; ++i)
{
arr[i] = i;
}
size_t size = sizeof(arr);
printf("size = %zu\n", size);
}
Output: size = 800
const
)#include <stdio.h>
int main()
{
size_t value = 200;
size_t i;
int arr[value];
for (i = 0; i < value; ++i)
{
arr[i] = i;
}
size_t size = sizeof(arr);
printf("size = %zu\n", size);
}
Output: size = 800
In general, if you are starting at 0 and going upward, always use an unsigned type to avoid an overflow taking you into a negative value situation. This is critically important, because if your array bounds happens to be less than the max of your loop, but your loop max happens to be greater than the max of your type, you will wrap around negative and you may experience a segmentation fault (SIGSEGV). So, in general, never use int for a loop starting at 0 and going upwards. Use an unsigned.
size_t or any unsigned type might be seen used as loop variable as loop variables are typically greater than or equal to 0.
When we use a size_t object, we have to make sure that in all the contexts it is used, including arithmetic, we want only non-negative values. For instance, following program would definitely give the unexpected result:
// C program to demonstrate that size_t or
// any unsigned int type should be used
// carefully when used in a loop
#include<stdio.h>
int main()
{
const size_t N = 10;
int a[N];
// This is fine
for (size_t n = 0; n < N; ++n)
a[n] = n;
// But reverse cycles are tricky for unsigned
// types as can lead to infinite loop
for (size_t n = N-1; n >= 0; --n)
printf("%d ", a[n]);
}
Output
Infinite loop and then segmentation fault
int
ifsome_size
is signed,size_t
if it is unsigned. – Nateint i
may not be enough to address a huge array. So by usingsize_t i
you can address more indices, so even if you have a huge array that should not be a problem.size_t
is a data type: usually aunsigned long int
but this depends on your system. – bruno