【转载】 Why size_t matters?

Why size_t matters?

July 3, 2007 Dan Saks

Using size_t appropriately can improve the portability, efficiency, or readability of your code. Maybe even all three.

Numerous functions in the Standard C library accept arguments or return values that represent object sizes in bytes. For example, the lone argument in malloc(n) specifies the size of the object to be allocated, and the last argument in memcpy(s1, s2, n) specifies the size of the object to be copied. The return value of strlen(s) yields the length of (the number of characters in) null-terminated character array s excluding the null character, which isn't exactly the size of s , but it's in the ballpark.

You might reasonably expect these parameters and return types that represent sizes to be declared with type int (possibly long and/or unsigned ), but they aren't. Rather, the C standard declares them as type size_t . According to the standard, the declaration for malloc should appear in as something equivalent to:

void *malloc(size_t n);   

and the declarations for memcpy and strlen should appear in looking much like:

void *memcpy(void *s1, void const *s2, size_t n);
size_t strlen(char const *s);   

The type size_t also appears throughout the C++ standard library. In addition, the C++ library uses a related symbol size_type , possibly even more than it uses size_t .

In my experience, most C and C++ programmers are aware that the standard libraries use size_t , but they really don't know what size_t represents or why the libraries use size_t as they do. Moreover, they don't know if and when they should use size_t themselves.

In this column, I'll explain what size_t is, why it exists, and how you should use it in your code.

A portability problem

Classic C (the early dialect of C described by Brian Kernighan and Dennis Ritchie in The C Programming Language , Prentice-Hall, 1978) didn't provide size_t . The C standards committee introduced size_t to eliminate a portability problem, illustrated by the following example.

Let's examine the problem of writing a portable declaration for the standard memcpy function. We'll look at a few different declarations and see how well they work when compiled for different architectures with different-sized address spaces and data paths.

Recall that calling memcpy(s1, s2, n) copies the first n bytes from the object pointed to by s2 to the object pointed to by s1 , and returns s1 . The function can copy objects of any type, so the pointer parameters and return type should be declared as "pointer to void". Moreover, memcpy doesn't modify the source object, so the second parameter should really be “pointer to const void.” None of this poses a problem.

The real concern is how to declare the function's third parameter, which represents the size of the source object. I suspect many programmers would choose plain int , as in:

void *memcpy(void *s1, void const *s2, int  n);   

which works fine most of the time, but it's not as general as it could be. Plain int is signed–it can represent negative values. However, sizes are never negative. Using unsigned int instead of int as the type of the third parameter lets memcpy copy larger objects, at no additional cost.

On most machines, the largest unsigned int value is roughly twice the largest positive int value. For example, on a 16-bit twos-complement machine, the largest unsigned int value is 65,535 and the largest positive int value is 32,767. Using an unsigned int as memcpy 's third parameter lets you copy objects roughly twice as big as when using int .

Although the size of an int varies among C implementations, on any given implementation int objects are always the same size as unsigned int objects. Thus, passing an unsigned int argument is always the same cost as passing an int .

Using unsigned int as the parameter type, as in:

void *memcpy(void *s1, void const *s2, unsigned int  n);   

works just dandy on any platform in which an s unsigned int can represent the size of the largest data object. This is generally the case on any platform in which integers and pointers have the same size, such as IP16, in which both integers and pointers occupy 16 bits, or IP32, in which both occupy 32 bits. (See the sidebar on C data model notation .)

C data model notation

Of late, I've run across several articles that employ a compact notation for describing the C language data representation on different target platforms. I have yet to find the origins of this notation, a formal syntax, or even a name for it, but it appears to be simple enough to be usable without a formal definition. The general form of the notation appears to be:

I *nI* L *nL* LL *nLL* P *nP*

where each capital letter (or pair thereof) represents a C data type, and each corresponding n is the number of bits that the type occupies. I stands for int , L stands for long , LL stands for long long , and P stands for pointer (to data, not pointer to function). Each letter and number is optional.

For example, an I16P32 architecture supports 16-bit int and 32-bit pointers, without describing whether it supports long or long long . If two consecutive types have the same size, you typically omit the first number. For example, you typically write I16L32P32 as I16LP32, which is an architecture that supports 16-bit int , 32-bit long , and 32-bit pointers.

The notation typically arranges the letters so their corresponding numbers appear in ascending order. For example, IL32LL64P32 denotes an architecture with 32-bit int , 32-bit long , 64-bit long long , and 32-bit pointers; however, it appears more commonly as ILP32LL64.

Unfortunately, this declaration for memcpy comes up short on an I16LP32 processor (16-bits for int and 32-bits for long and pointers), such as the first generation Motorola 68000. In this case, the processor can copy objects larger than 65,536 bytes, but this memcpy can't because parameter n can't handle values that large.

Easy to fix, you say? Just change the type of memcpy 's third parameter:

void *memcpy(void *s1, void const *s2, unsigned long   n);   

You can use this declaration to write a memcpy for an I16LP32 target, and it will be able to copy large objects. It will also work on IP16 and IP32 platforms, so it does provide a portable declaration for memcpy . Unfortunately, on an IP16 platform, the machine code you get from using unsigned long here is almost certainly a little less efficient (the code is both bigger and slower) than what you get from using an unsigned int .

In Standard C, a long (whether signed or unsigned) must occupy at least 32 bits. Thus, an IP16 platform that supports Standard C really must be an IP16L32 platform. Such platforms typically implement each 32-bit long as a pair of 16-bit words. In that case, moving a 32-bit long usually requires two machine instructions, one to move each 16-bit chunk. In fact, almost all 32-bit operations on these platforms require at least two instructions, if not more.

Thus, declaring memcpy 's third parameter as an unsigned long in the name of portability exacts a performance toll on some platforms, something we'd like to avoid. Using size_t avoids that toll.

Type size_t is a s typedef that's an alias for some unsigned integer type, typically unsigned int or unsigned long , but possibly even unsigned long long . Each Standard C implementation is supposed to choose the unsigned integer that's big enough–but no bigger than needed–to represent the size of the largest possible object on the target platform.

Using size_t

The definition for size_t appears in several Standard C headers, namely, <stddef.h>, <stdio.h>, <stdlib.h>, <string.h>, <time.h>, and <wchar.h>. It also appears in the corresponding C++ headers, <cstddef>, <cstdio>, and so on. You should include at least one of these headers in your code before referring to size_t.

Including any of the C headers (in a program compiled as either C or C++) declares size_t as a global name. Including any of the C++ headers (something you can do only in C++) defines size_t as a member of namespace std .

By definition, size_t is the result type of the sizeof operator (sizeof is both an operator and a keyword). Thus, the appropriate way to declare n to make the assignment:

n = sizeof(thing);   

both portable and efficient is to declare n with type size_t. Similarly, the appropriate way to declare a function foo to make the call:

foo(sizeof(thing));   

both portable and efficient is to declare foo 's parameter with type size_t. Functions with parameters of type size_t often have local variables that count up to or down from that size and index into arrays, and size_t is often a good type for those variables.

Using size_t appropriately makes your source code a little more self-documenting. When you see an object declared as a size_t , you immediately know it represents a size in bytes or an index, rather than an error code or a general arithmetic value.

Expect to see me using size_t in other examples in upcoming columns.

Dan Saks is president of Saks & Associates, a C/C++ training and consulting company. For more information about Dan Saks, visit his website at www.dansaks.com. Dan also welcomes your feedback: e-mail him at . For more information about Dan .

本文转载自Why size_t matters?...pdf.

两篇译文（像是机翻，不如直接看原文）：

译文1 译文2

至于为什么要在后面加个_t，可以参考Reference中的链接，也放一下截的图方便一点：

References:

size_t 这个类型的意义是什么？