何时使用 std::size_t？

Question

I'm just wondering should I use std::size_t for loops and stuff instead of int ? For instance:

#include <cstdint>

int main()
{
    for (std::size_t i = 0; i < 10; ++i) {
        // std::size_t OK here? Or should I use, say, unsigned int instead?
    }
}

In general, what is the best practice regarding when to use std::size_t ?

Answer 1

A good rule of thumb is for anything that you need to compare in the loop condition against something that is naturally a std::size_t itself.

std::size_t is the type of any sizeof expression and as is guaranteed to be able to express the maximum size of any object (including any array) in C++. By extension it is also guaranteed to be big enough for any array index so it is a natural type for a loop by index over an array.

If you are just counting up to a number then it may be more natural to use either the type of the variable that holds that number or an int or unsigned int (if large enough) as these should be a natural size for the machine.

Answer 2

size_t is the result type of the sizeof operator.

Use size_t for variables that model size or index in an array. size_t conveys semantics: you immediately know it represents a size in bytes or an index, rather than just another integer.

Also, using size_t to represent a size in bytes helps making the code portable.

Answer 3

The size_t type is meant to specify the size of something so it's natural to use it, for example, getting the length of a string and then processing each character:

for (size_t i = 0, max = strlen (str); i < max; i++)
    doSomethingWith (str[i]);

You do have to watch out for boundary conditions of course, since it's an unsigned type. The boundary at the top end is not usually that important since the maximum is usually large (though it is possible to get there). Most people just use an int for that sort of thing because they rarely have structures or arrays that get big enough to exceed the capacity of that int .

But watch out for things like:

for (size_t i = strlen (str) - 1; i >= 0; i--)

which will cause an infinite loop due to the wrapping behaviour of unsigned values (although I've seen compilers warn against this). This can also be alleviated by the (slightly harder to understand but at least immune to wrapping problems):

for (size_t i = strlen (str); i-- > 0; )

By shifting the decrement into a post-check side-effect of the continuation condition, this does the check for continuation on the value before decrement, but still uses the decremented value inside the loop (which is why the loop runs from len .. 1 rather than len-1 .. 0 ).

Answer 4

By definition, size_t is the result of the sizeof operator. size_t was created to refer to sizes.

The number of times you do something (10, in your example) is not about sizes, so why use size_t ? int , or unsigned int , should be ok.

Of course it is also relevant what you do with i inside the loop. If you pass it to a function which takes an unsigned int , for example, pick unsigned int .

In any case, I recommend to avoid implicit type conversions. Make all type conversions explicit.

Answer 5

short answer:

almost never

long answer:

Whenever you need to have a vector of char bigger that 2gb on a 32 bit system. In every other use case, using a signed type is much safer than using an unsigned type.

example:

std::vector<A> data;
[...]
// calculate the index that should be used;
size_t i = calc_index(param1, param2);
// doing calculations close to the underflow of an integer is already dangerous

// do some bounds checking
if( i - 1 < 0 ) {
    // always false, because 0-1 on unsigned creates an underflow
    return LEFT_BORDER;
} else if( i >= data.size() - 1 ) {
    // if i already had an underflow, this becomes true
    return RIGHT_BORDER;
}

// now you have a bug that is very hard to track, because you never 
// get an exception or anything anymore, to detect that you actually 
// return the false border case.

return calc_something(data[i-1], data[i], data[i+1]);

The signed equivalent of size_t is ptrdiff_t , not int . But using int is still much better in most cases than size_t. ptrdiff_t is long on 32 and 64 bit systems.

This means that you always have to convert to and from size_t whenever you interact with a std::containers, which not very beautiful. But on a going native conference the authors of c++ mentioned that designing std::vector with an unsigned size_t was a mistake.

If your compiler gives you warnings on implicit conversions from ptrdiff_t to size_t, you can make it explicit with constructor syntax:

calc_something(data[size_t(i-1)], data[size_t(i)], data[size_t(i+1)]);

if just want to iterate a collection, without bounds cheking, use range based for:

for(const auto& d : data) {
    [...]
}

here some words from Bjarne Stroustrup (C++ author) at going native

For some people this signed/unsigned design error in the STL is reason enough, to not use the std::vector, but instead an own implementation.

Answer 6

size_t是一种非常易读的方式来指定项目的大小维度 - 字符串的长度、指针占用的字节数等。它还可以跨平台移植 - 你会发现 64 位和 32 位在系统函数和size_t - unsigned int可能不会做的事情（例如，什么时候应该使用unsigned long

Answer 7

Use std::size_t for indexing/counting C-style arrays.

For STL containers, you'll have (for example) vector<int>::size_type , which should be used for indexing and counting vector elements.

In practice, they are usually both unsigned ints, but it isn't guaranteed, especially when using custom allocators.

Answer 8

Soon most computers will be 64-bit architectures with 64-bit OS:es running programs operating on containers of billions of elements. Then you must use size_t instead of int as loop index, otherwise your index will wrap around at the 2^32:th element, on both 32- and 64-bit systems.

Prepare for the future!

Answer 9

size_t is returned by various libraries to indicate that the size of that container is non-zero. You use it when you get once back :0

However, in the your example above looping on a size_t is a potential bug. Consider the following:

for (size_t i = thing.size(); i >= 0; --i) {
  // this will never terminate because size_t is a typedef for
  // unsigned int which can not be negative by definition
  // therefore i will always be >= 0
  printf("the never ending story. la la la la");
}

the use of unsigned integers has the potential to create these types of subtle issues. Therefore imho I prefer to use size_t only when I interact with containers/types that require it.

Answer 10

When using size_t be careful with the following expression

size_t i = containner.find("mytoken");
size_t x = 99;
if (i-x>-1 && i+x < containner.size()) {
    cout << containner[i-x] << " " << containner[i+x] << endl;
}

You will get false in the if expression regardless of what value you have for x. It took me several days to realize this (the code is so simple that I did not do unit test), although it only take a few minutes to figure the source of the problem. Not sure it is better to do a cast or use zero.

if ((int)(i-x) > -1 or (i-x) >= 0)

Both ways should work. Here is my test run

size_t i = 5;
cerr << "i-7=" << i-7 << " (int)(i-7)=" << (int)(i-7) << endl;

The output: i-7=18446744073709551614 (int)(i-7)=-2

I would like other's comments.

Answer 11

It is often better not to use size_t in a loop. For example,

vector<int> a = {1,2,3,4};
for (size_t i=0; i<a.size(); i++) {
    std::cout << a[i] << std::endl;
}
size_t n = a.size();
for (size_t i=n-1; i>=0; i--) {
    std::cout << a[i] << std::endl;
}

The first loop is ok. But for the second loop:
When i=0, the result of i-- will be ULLONG_MAX (assuming size_t = unsigned long long), which is not what you want in a loop.
Moreover, if a is empty then n=0 and n-1=ULLONG_MAX which is not good either.

Answer 12

size_t is an unsigned type that can hold maximum integer value for your architecture, so it is protected from integer overflows due to sign (signed int 0x7FFFFFFF incremented by 1 will give you -1) or short size (unsigned short int 0xFFFF incremented by 1 will give you 0).

It is mainly used in array indexing/loops/address arithmetic and so on. Functions like memset() and alike accept size_t only, because theoretically you may have a block of memory of size 2^32-1 (on 32bit platform).

For such simple loops don't bother and use just int.

Answer 13

I have been struggling myself with understanding what and when to use it. But size_t is just an unsigned integral data type which is defined in various header files such as <stddef.h>, <stdio.h>, <stdlib.h>, <string.h>, <time.h>, <wchar.h> etc.

It is used to represent the size of objects in bytes hence it's used as the return type by the sizeof operator. The maximum permissible size is dependent on the compiler; if the compiler is 32 bit then it is simply a typedef (alias) for unsigned int but if the compiler is 64 bit then it would be a typedef for unsigned long long. The size_t data type is never negative(excluding ssize_t) Therefore many C library functions like malloc, memcpy and strlen declare their arguments and return type as size_t .

/ Declaration of various standard library functions.
  
// Here argument of 'n' refers to maximum blocks that can be
// allocated which is guaranteed to be non-negative.
void *malloc(size_t n);
  
// While copying 'n' bytes from 's2' to 's1'
// n must be non-negative integer.
void *memcpy(void *s1, void const *s2, size_t n);
  
// the size of any string or `std::vector<char> st;` will always be at least 0.
size_t strlen(char const *s);

size_t or any unsigned type might be seen used as loop variable as loop variables are typically greater than or equal to 0.

Answer 14

size_t is an unsigned integral type, that can represent the largest integer on you system. Only use it if you need very large arrays,matrices etc.

Some functions return an size_t and your compiler will warn you if you try to do comparisons.

Avoid that by using a the appropriate signed/unsigned datatype or simply typecast for a fast hack.

Answer 15

size_t is unsigned int. so whenever you want unsigned int you can use it.

I use it when i want to specify size of the array , counter ect...

void * operator new (size_t size); is a good use of it.