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Today, I would like to share with you the relevant knowledge of how C++ integers are stored in memory. The content is detailed and the logic is clear. I believe most people still know too much about this knowledge, so share this article for your reference. I hope you can get something after reading this article, let's take a look at it.

Storage format:

For integers, the data stored in memory is actually a complement. The reason is that in the computer system, the numerical value is represented and stored by the complement code, because the symbol bit and numerical range can be treated uniformly by using the complement code, and at the same time, the addition and subtraction can also be handled uniformly (cpu only the adder). In addition, the complement code and the original code are converted into each other, the operation process is exactly the same, and no additional hardware circuit is needed.

Introduction to the size end:

Large segment storage mode: the low weight bits of the data are stored at the high address of memory, and the high weight bits are stored at the low address of memory.

Small segment storage mode: refers to the high weight bits of data stored in the high address of memory, low weight bits stored in the low address of memory.

# define _ CRT_SECURE_NO_WARNINGS 1#include # include int Judge1 () / method 1: use the address int to forcibly redirect char judgment. {int I = 1; return * (char *) & I; / / I has four addresses. Take the smallest address and add a strong de-reference to take only the value in the lowest address space. } int Judge2 () / / method II: take advantage of the characteristics of the unio consortium: all members share space. The space size of the consortium is the maximum of the space occupied by the members, and the small space starts from the low address. {union {int a; char i;} U1; U1.a = 1; return u1.i;} int main () {int ret; ret = Judge2 (); if (1 = = ret) {printf ("small end storage mode\ n");} else if (0 = = ret) {printf ("large end storage mode\ n");} system ("pause") Return 0;}

Add and subtract in memory

Question 1:

Char a =-1; / / original code 1000 0001 complement 1111 1111 signed char b =-1; / / original code 1000 0001 complement 1111 1111 unsigned char c =-1 / / original code 1000 0001 complement 1111 1111 / /% d char to int type occurs × × × promotion, signed number occurs × × × lift preceding complement symbol bit, unsigned number complement 0 printf ("aversion% d\ n", a) / / output source code after promotion: 1000 0000 0000 1111 1111 1111 (complement): 1000 0000 0000 0000 00000001 =-1 printf ("baked% d\ n", b); / / printf ("cased% d\ n", c) after promotion; / / output original code: 0000 0000 0000 1111 1111 (complement) after promotion

Question 2:

1 > char a =-128; / the original code 1 1000 0000 symbol bit overflow truncated into 1000 0000 complement: 1 1000 0000 truncated 1000 0000 printf ("% u\ n", a); / / × × upgrade: 1111 1111 1111 1000000 (complement) because it is unsigned × × output, so output the decimal number corresponding to the binary. 2 > char a = 128; / / original code: 0 1000 0000 symbol bit overflow truncated to 1000 0000 complement: 1 1000 0000 truncated 1000 0000 printf ("% u\ n", a) again; / / it can be seen that the answer is the same as above

Question 3:

Unsigned char i = 0; / / indicates the range of values: 0,255 for (I = 0; I)

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