C/C ++ test interview (1) -- sizeof

Source: Internet
Author: User

Sizeof, a handsome guy, leads countless cainiao to fold his waist. I didn't make too many mistakes at the beginning, and adhered to the great idea of "I am one, happy ten million people, I decided to give a detailed summary as much as possible. However, when I sum up, I find that this problem can be both simple and complex. Therefore, this article is not suitable for beginners, and it is not even necessary to write an article. However, this article may be helpful to you if you want to "know the truth, know more about it. I have not learned much about C ++ in cainiao, but I have many errors. please correct me.

1. Definition:
Sizeof which is the Holy sizeof is an operator in C/C ++. Simply put, it is used to return the memory bytes occupied by an object or type.
The description on msdn is as follows:
The sizeof keyword gives the amount of storage, in bytes, associated with a variable or a type (including aggregate types ).
This keyword returns a value of Type size_t.
The return value type is size_t, which is defined in the header file stddef. h. This is a value dependent on the compilation system, which is generally defined
Typedef unsigned int size_t;
There are many compilers in the world, but as a standard, they will ensure that the sizeof values of char, signed Char, and unsignedchar are 1. After all, char is the minimum data type that we can use for programming.

2. Syntax:
Sizeof has three syntax forms:
1) sizeof (object); // sizeof (object );
2) sizeof (type_name); // sizeof (type );
3) sizeof object; // sizeof object;
So,

int i;sizeof( i ); // oksizeof i; // oksizeof( int ); // oksizeof int; // error

Since writing 3 can be replaced by writing 1, in order to unify the form and reduce the burden on our brains, there are 3rd writing methods. Forget it!
In fact, the size of the sizeof calculation object is also converted to the calculation of the object type, that is, the sizeof values of different objects of the same type are consistent. Here, the object can be further extended to the expression, that is, sizeof can evaluate an expression. The Compiler determines the size based on the final result type of the expression, and generally does not calculate the expression. For example:
Sizeof (2); // The type of 2 is int, so it is equivalent to sizeof (INT );
Sizeof (2 + 3.14); // The type of 3.14 is double, and 2 is also upgraded to double, so it is equivalent to sizeof (double );
Sizeof can also be used to evaluate a function call. The result is the size of the function return type, and the function is not called. Let's take a complete example:

Char Foo () {printf ("Foo () has been called. /n "); Return 'a';} int main () {size_t SZ = sizeof (FOO (); // the return value type of Foo () is Char, so SZ = sizeof (char), Foo () will not be called printf ("sizeof (FOO () = % d/N", SZ );}


According to the c99 standard, functions, expressions of undetermined types, and bit-field members cannot be computed with sizeof values. That is, the following statements are incorrect:

sizeof( foo );// errorvoid foo2() { }sizeof( foo2() );// errorstruct S{unsigned int f1 : 1;unsigned int f2 : 5;unsigned int f3 : 12;};sizeof( S.f1 );// error

3. constants of sizeof
The sizeof calculation occurs at the Compilation Time, so it can be used as a constant expression, such:
Char ary [sizeof (INT) * 10]; // OK
The latest c99 standard stipulates that sizeof can also be calculated at runtime. The following programs can be used in Dev-C ++.
Correct execution:

Int N; n = 10; // n dynamic value assignment char ary [N]; // c99 also supports the dynamic definition of the array printf ("% d/N ", sizeof (ary); // OK. output 10

But it won't work in the compiler that does not fully implement the c99 standard, and the above Code won't be able to compile in vc6. Therefore, we 'd better think that sizeof is executed during the compilation period. This will not cause errors and make the program more portable.


4. sizeof of basic data type
The basic data types here refer to simple built-in data types such as short, Int, long, float, and double. Because they are all related to the system, the values may be different in different systems, this must attract our attention, and try not to cause trouble for porting our programs in this respect. Generally, in a 32-bit compiling environment, the value of sizeof (INT) is 4.

5. sizeof pointer variable
If you have learned the data structure, you should know that the pointer is a very important concept, which records the address of another object. Since the address is stored, it is equal to the width of the computer's internal address bus. Therefore, in a 32-bit computer, the return value of a pointer variable must be 4 (note that the result is in bytes), which can be predicted, in the future 64-bit system, the sizeof result of the pointer variable is 8.

Char * Pc = "ABC"; int * PI; string * pS; char ** PPC = & PC; void (* PF )(); // function pointer sizeof (PC); // The result is 4 sizeof (PI); // The result is 4 sizeof (PS); // The result is 4 sizeof (PPC ); // The result is 4 sizeof (PF); // The result is 4


The sizeof value of the pointer variable has nothing to do with the object indicated by the pointer. It is precisely because all pointer variables occupy the same memory size, therefore, the MFC message processing function uses two parameters wparam and lparam to transmit various complex message structures (using pointers to struct ).

6. sizeof of Array
The sizeof value of the array is equal to the number of memory bytes occupied by the array, for example:

Char A1 [] = "ABC"; int A2 [3]; sizeof (A1); // The result is 4, and a null Terminator sizeof (A2) exists at the end of the character ); // The result is 3*4 = 12 (dependent on INT)


Some friends regard sizeof as the number of array elements at the beginning. Now, you should know that this is not correct. How should we calculate the number of array elements easily, there are usually two ways to write:

Int C1 = sizeof (A1)/sizeof (char); // total length/length of a single element int C2 = sizeof (A1)/sizeof (A1 [0]); // total length/length of the first element


Write it here and ask, what is the value of C3 and C4 below?

void foo3(char a3[3]){int c3 = sizeof( a3 ); // c3 == }void foo4(char a4[]){int c4 = sizeof( a4 ); // c4 == }


Maybe when you try to answer the C4 value, you realize that C3 is wrong. Yes, C3! = 3. Here, the function parameter A3 is no longer an array type, but a pointer, which is equivalent to char * A3. It is hard to understand why, when we call the function foo1, will the program allocate a 3 array on the stack? No! The array is "Address Transfer". The caller only needs to pass the address of the real parameter, so A3 is naturally a pointer type (char *), and the C3 value is 4.

7. sizeof of struct
This is the most frequently asked question for beginners, so it is necessary to pay more attention here. Let's first look at a struct:

struct S1{char c;int i;};

Ask how smart the sizeof (S1) is. You start to think about it. Char occupies 1 byte and INT occupies 4 byte. Then, the total value should be 5. Is that true? Have you tried it on your machine? Maybe you are right, but you are probably wrong! In vc6, the default value is 8.

Extension:

Struct BS {char C1; char C2; int A ;}; int main () {BS t; cout <"sizeof (BS):" <sizeof (BS) <Endl; // output sizeof (BS): 8 cout <"sizeof (t):" <sizeof (t) <Endl; // output sizeof (t): 8}

Struct BS {char C1; int A; char C2 ;}; int main () {BS t; cout <"sizeof (BS):" <sizeof (BS) <Endl; // output sizeof (BS): 12 cout <"sizeof (t):" <sizeof (t) <Endl; // output sizeof (t): 12}

Why is it that why is always hurt? Please don't be frustrated. Let's take a good look at the definition of sizeof -- the result of sizeof is equal to the number of memory bytes occupied by the object or type. Okay, let's take a look at the memory allocation of S1:
S1 S1 = {'A', 0 xffffffff };
After defining the above variable, add a breakpoint and run the program. Observe the memory where S1 is located. what do you find? Take my vc6.0 as an example. The S1 address is 0x0012ff78, and its data content is as follows:
0012ff78: 61 CC FF
What are the three-byte CC in the middle? Let's take a look at the description on msdn:
When applied to a structure type or variable, sizeof returns the actual size, which may include Padding Bytes inserted for alignment.
Originally, this is the legendary byte alignment!

An important topic has emerged. Why do we need byte alignment principles to teach us how to speed up the computer's data acquisition speed? Otherwise, it will take a longer command cycle. Therefore, the compiler will process the struct by default (in fact, the number
According to the variable), so that the basic data type (short, etc.) with the width of 2 is located on the address that can be divisible by 2, so that the basic data type with the width of 4 (INT, etc) are located on the address that can be divisible by four, and so on. In this way, the two numbers may need to be added to the padding byte, so the sizeof value of the entire struct increases.
Let's exchange the positions of char and INT in S1:

struct S2{int i;char c;};


Let's look at the number of sizeof (S2) results. How can we still look at the memory at 8? The original member C still has three Padding Bytes. Why? Don't worry. The following summarizes the rules.

The details of byte alignment are related to compiler implementation, but generally three criteria are met:
1) The first address of the struct variable can be divisible by the size of its widest basic type member;
2) The offset (offset) of each member of the struct to the first address of the struct is an integer multiple of the member size. If necessary, the compiler will add the internal adding between the members );
3) the total size of the struct is an integer multiple of the size of the widest basic type of the struct. If necessary, the compiler will add the trailing padding after the last member ).

There are several notes for the above principles:
1) I didn't mean that the address of a struct member is an integer multiple of its size. How can I talk about the offset? Because a 1st point exists, we can only consider the member offset, this is easy to think about. Think about why.
The offset of a member of the struct to the first address of the struct can be obtained through the macro offsetof (). This macro is also defined in stddef. H, as follows:
# Define offsetof (S, m) (size_t) & (S *) 0)-> m)
For example, to get the offset of C in S2, the method is
Size_t Pos = offsetof (S2, c); // POS equals 4


2) The basic type refers to the built-in data types such as char, short, Int, float, and double mentioned above. The "data width" here refers to the size of its sizeof. Because the struct member can be a composite type, for example, another struct, when looking for the widest basic type member, it should include the Child Member of the composite type member, instead of seeing a composite member as a whole. However, when determining the offset of a composite member, the composite type is regarded as a whole.
This is a bit difficult to describe here, and it seems a bit confusing to think about it. Let's take a look at the example (the specific value is still taken as an example of vc6 and will not be explained later ):

struct S3{char c1;S1 s;char c2};


The type of the widest and simplest member of S1 is int. When considering the widest and simplest type members, S3 splits S1, so the widest and simplest type of S3 is int. In this way, for variables defined by S3, the first address of the bucket needs to be divisible by four, and the entire sizeof (S3) value should also be divisible by four. The offset of C1 is 0, and the offset of S is a whole. As a struct variable, S satisfies the preceding three criteria, so its size is 8 and its offset is 4, three bytes are required between C1 and S, and the offset between C2 and S is 12, the size of C2 is 13, and 13 cannot be fully divided by 4. Therefore, three fill bytes must be added at the end. Finally, the value of sizeof (S3) is 16.
Through the above description, we can get a formula:
The size of the struct is equal to the offset of the last member plus the size of the struct plus the number of filled bytes at the end, that is:

Sizeof (struct) = offsetof (last item) + sizeof (trailing padding)

Here, friends should have a new understanding of the sizeof struct, but don't be too happy. An important parameter that affects sizeof has not been mentioned yet, that is the pack instruction of the compiler. It is used to adjust the structure alignment mode. Different compiler names and usages are slightly different. In vc6, the # pragma pack is used for implementation, and you can also directly modify the/ZP compilation switch. # Basic usage of The Pragma Pack: # pragma pack (N), where N is the number of bytes alignment. The default value is 8, if this value is smaller than the sizeof value of the struct member, the offset of the struct member should take the minimum value of the two. The formula is as follows:

Offsetof (item) = min (n, sizeof (item ))

Let's look at the example:

# Pragma pack (push) // Save the current pack setting pressure stack # pragma pack (2) // struct S1 {char C; int I;} must be used before struct definition ;}; struct S3 {char C1; S1 s; char C2}; # pragma pack (POP) // restore previous pack settings

When sizeof (S1) is calculated, the value of min (2, sizeof (I) is 2, so the offset of I is 2, and the value of sizeof (I) is 6, it can be divisible by 2, so the size of the entire S1 is 6. Similarly, for sizeof (S3), the offset of S is 2, the offset of C2 is 8, and the addition of sizeof (C2) is equal to 9, cannot be divisible by 2, add a fill byte, so sizeof (S3) is equal to 10. Now, friends can breathe a sigh of relief ,:)
Note that the size of the "Empty struct" (excluding data members) is not 0, but 1. Imagine how a variable with "no space occupied" can be separated by an address and two different "Empty struct" variables. Therefore, the "null struct" variable must be stored, so that the compiler can allocate only one byte of space for occupying space. As follows:

Struct S5 {}; sizeof (S5); // The result is 1.



8. sizeof containing bitfield struct
As mentioned above, bitfield members cannot be individually sizeof values. Here we will discuss the sizeof of struct containing the bitfield, but specifically list it in consideration of its particularity. C99 specifies that int, unsigned int, and bool can be used as bit domain types, but almost all compilers have extended these types to allow the existence of other types. The main purpose of bit domains is to compress storage. The general rules are as follows:
1) if the types of adjacent fields are the same, and the sum of the bit widths is smaller than the sizeof size of the type, the subsequent fields will be stored next to the previous field until they cannot be accommodated;
2) If the Field Types of adjacent bit fields are the same, but the sum of Bit Width is greater than the sizeof size of the type, the subsequent fields start from the new storage unit, its offset is an integer multiple of its type;
3) if the types of adjacent bit field are different, the implementations of each compiler are different. vc6 adopts the non-compression mode and Dev-C ++ adopts the compression mode;
4) do not compress fields that are interspersed with non-bit fields;
5) the total size of the entire struct is an integer multiple of the size of the widest basic type.

Let's take a look at the example.
Example 1:

struct BF1{char f1 : 3;char f2 : 4;char f3 : 5;};


Its memory layout is:
| _ F1 __|__ F2 __| _ | ____ F3 ___ | ____ |
| _ |
0 3 7 8 1316
The bit field type is Char, and the 1st bytes can only accommodate F1 and F2, so F2 is compressed to 1st bytes, while F3 can only start from the next byte. Therefore, the result of sizeof (bf1) is 2.
Example 2:

struct BF2{char f1 : 3;short f2 : 4;char f3 : 5;};


Because the adjacent bit fields have different types, in vc6, sizeof is 6, and in Dev-C ++ is 2.
Example 3:

struct BF3{char f1 : 3;char f2;char f3 : 5;};


The non-bit field is interspersed in it and will not produce compression. The size obtained in vc6 and Dev-C ++ is 3.



9. sizeof of the Consortium
The struct structure is sequential in the memory structure, while the consortium is overlapping. Each member shares a piece of memory, so the sizeof of the entire consortium is the maximum value of sizeof each member. Struct members can also be composite. Here, composite Members are considered as a whole. Therefore, in the following example, the sizeof value of U is equal to sizeof (s ).

union U{int i;char c;S1 s;};

10. Pen questions

1.
Void * P = malloc (100 );
Sizeof (p) = 4;
--------------------- The number of bytes of the pointer, rather than the content capacity pointed to by the pointer
2.
Void func (char STR [1, 100])
{Sizeof (STR) = 4 ;}
-------------------- When an array is passed as a function parameter, the array is automatically degraded to a pointer of the same type.
3.
Char STR [100];
Sizeof (STR) = 100;
-------------------- When it is not a function parameter
4.
Char STR [] = "hello ";
Char * P = STR;
Int n = 10;
Sizeof (STR) = 6; sizeof (p) = 4; sizeof (n) = 4

 

Reference:

Http://zhidao.baidu.com/question/19171773.html

Http://blog.csdn.net/hjhcs121/article/details/7627359

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