C + + interview-memory alignment is a different data type storage space

Below is a list of the dev-c++ of the base type and the range of values:

Basic type number of placeholder value range Input Example output character example

----Char 8-2^7 ~ 2^7-1%c %c,%d,%u

Signed--Char 8-2^7 ~ 2^7-1%c %c,%d,%u

Unsigned--Char 8 0 ~ 2^8-1%c %c,%d,%u

[Signed] short [int] 16-2^15 ~ 2^15-1%HD %hd

unsigned short [int] 0 ~ 2^16-1%hu%hu ,%ho,%HX

[Signed]--int 32-2^31 ~ 2^31-1%d

unsigned--[int] 0 ~ 2^32-1%u,%o,%x

[Signed] long [int] 32-2^31 ~ 2^31-1%ld

unsigned long [int] 0 ~ 2^32-1%lu,%lo,%LX

[Signed] long long [int] 64-2^63 ~ 2^63-1%i64d

unsigned long long [int] 0 ~ 2^64-1%i64u,%i64o,%i64x

----float +/-3.40282e+038%f,%e,%g

----double +/-1.79769e+308%lf,%le,%lg%f,%e,%g

--long double + +/-1.79769e+308%lf,%le,%LG

Note: int data is not the same as the number of digits in a system of different digits some compilers have different effects

16 2*8 Bit

32 4*8 Bit

64 8*8 Bit

1 byte=8 BIT

Memory alignment can be summed up in a sentence:

"Data items can only be stored in memory locations where the address is an integer multiple of the data item size"

For example, the int type occupies 4 bytes, and the address can only be located in 0,4,8.

The double type occupies 8 bytes, and the address can only be placed above the 0,8,16.

If there is a special provision in the code, the alignment is program-based

#pragma pack (n)

If the above statement appears in the code

Then the alignment mechanism is the address only in the 0,n,2*n,3*n position

1#include <iostream>2#include <cstdlib>3#include <pthread.h>4#include <windows.h>5 using namespacestd;6 structX17 {8     intI//4 bytes9     CharC1;//1 bytesTen     CharC2;//1 bytes One     LongA//4 bytes A     floatb; - } x1; - structX2 the { -     CharC1;//1 bytes -     intI//4 bytes -     CharC2;//1 bytes +     LongA//4 bytes -     floatb; + } x2; A structX3 at { -     CharC1;//1 bytes -     CharC2;//1 bytes -     intI//4 bytes -     LongA//4 bytes -     floatb; in } x3; - structX4 to { +     intI//4 bytes -     CharC1;//1 bytes the     CharC2;//1 bytes *     LongA//4 bytes $     floatb;Panax Notoginseng } x4; - intMain () the { +cout<<"Long"<<sizeof(Long) <<"\ n"; Acout<<"float"<<sizeof(float) <<"\ n"; thecout<<"int"<<sizeof(int) <<"\ n"; +cout<<"Char"<<sizeof(Char) <<"\ n"; -cout<<"size of the X1"<<sizeof(x1) <<"\ n"; $cout<<"size of the X2"<<sizeof(x2) <<"\ n"; $cout<<"size of the X3"<<sizeof(x3) <<"\ n"; -cout<<"size of the X4"<<sizeof(x4) <<"\ n"; -     return 0; the } - Operation Result:Wuyi Long 4 the float 4 - int 4 Wu Char 1 -Size of the X1 - AboutSize of the X2 - $Size of the X3 - -Size of the X4 -

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The data in the three struct structures is the same, but the structure takes up a different space. This is how the alignment mechanism affects storage

C + + interview-memory alignment is a different data type storage space