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It has to be clear on the whole :

1) There are three kinds of program structure : Sequential structure 、 Selection structure ( Branching structure )、 Loop structure .

2) Read the program from main() entrance , Then read down from the top ( Run into a cycle and make a cycle , Meet the choice to make a choice ), Yes and only — individual main function .

3) The data of the computer is stored in the form of binary . Where the data is stored is His address .

4)bit Yes Refers to as 0 perhaps 1. byte It means byte , A byte = Eight .

Concepts are often tested ：

1、 Compile preprocessing is not C Part of the language , No running time , Don't semicolon .C A program compiled in a language is called a source program , It uses ASCII Values are stored in text files .

2、#define PI 3.1415926; This writing is wrong , There must be no semicolons .

define a 1+2 define a (1+2)

a=a*a=1+2*1+2=5a=a*a=3*3=9

3、 Every C Language program main There is only one function .

4、 You can no longer define a function in a function .

5、 Algorithm ： No input , But there must be output .

6、break It can be used for cyclic structure and switch sentence .

7、 The comma operator has the lowest level , The assigned level is the next to last .

Chapter one C Basic knowledge of language

Section 1 、 Yes C Basic knowledge of language

1、C A program written in a language is called a source program , Also called compilation unit .

2、C The form of writing is free , You can write more than one statement per line , Can write many lines .

3、 One C There is and only one language program main function , Is the starting point of program operation .

In the second quarter 、 be familiar with vc++

1、VC It's software , Used to run write C Language program .

2、 Every C When the language program is finished , They all compile first , Post link , Last run .（.c---à.obj---à.exe） Pay attention to this process .c and .obj File cannot be run , Only .exe File to run .（ Regular examination ！）

In the third quarter 、 identifier

1、 identifier （ What must be tested ）：

Legal requirements are made by the letters , Numbers , Underline composition . There are other elements that are wrong .

And the first one must be a letter or an underline . The first number is wrong

2、 Identifiers are divided into keywords 、 Predefined identifiers 、 User identifier .

keyword : It can not be used as a user identification symbol .main define scanf printf None of them are keywords . A place to confuse you lf It can be used as a user identifier . because lf The first letter in is capitalized , So it's not keywords .

Predefined identifiers ： recite define scanf printf include. Remember that predefined identifiers can be used as user identifiers .

User identifier ： Basically every year , Please refer to the exercises in the book for details .

The fourth quarter, ： Base conversion

Convert decimal to binary 、 octal 、 Hexadecimal .

Binary system 、 octal 、 Convert hex to decimal .

Section V number

1）C The language is only eight 、 Ten 、 Hexadecimal , No binary . But at runtime , All the bases must be converted to binary for processing .（ Twice in the exam ）

a、C The octal rules in language should be 0 start .018 The value of is illegal , Octal is not 8 Of , Meet 8 Into the 1.

b、C The hexadecimal rules in a language are as follows 0x start .

2) The legal writing of decimals ：C If there is a zero on both sides of the decimal point of the language , You don't have to write .

1.0 stay C It can be written as 1.

0.1 stay C Language can be written as .1.

3） The legal form of real data ：

a、2.333e-1 It's legal , And the data is 2.333×10-1.

b、 Test formula ：e front e There must be a few in the future ,e Must be an integer after . Please combine the examples in the book .

4） The integer is usually 4 Bytes , The character type is 1 Bytes , Double precision is generally 8 Bytes ：

long int x; Express x It's a long one .

unsigned int x; Express x It's an unsigned integer .

The sixth 、 Seven quarter ： Arithmetic expressions and assignment expressions

The core ： The expression must have a number ！

1、 Arithmetic expressions ：+,-,*,/,%

Be sure to pay attention to ：“/” If it's full size on both sides , The result is an integer . 3/2 The result is that 1.

“/” If one side is a decimal , So the result is a decimal . 3/2.0 The result is that 1.5

“%” Please note that the sign is the remainder , The examination is the easiest to count as a divisor .)% Both sides of the symbol are required to be integers . It's not a whole number .[ Be careful !!!]

2、 Assignment expression ： The expression value is the leftmost value ,a=b=5; The expression is 5, Constant cannot be assigned .

1、int x=y=10: Wrong , Definition time , You can't assign values consecutively .

2、​​​​​​​int x,y:

x=y=10; Yes , After definition , Can be assigned continuously .

3、 The left side of the assignment can only be a variable .

4、int x=7.7; Yes ,x Namely 7

5、float x=7; Yes ,x Namely 7.0

3、 Compound assignment expression ：

int a = 2;

a*=2+3; After completion of operation ,a The value of is 12.

Be sure to pay attention to , In the first 2+3 Put brackets on the top of . become （2+3） Re operation .

Compound statements must be used ｛｝;

4、 Self adding expression ：

Self adding 、 Self reducing expression ： hypothesis a=5,++a（ Is for 6）, a++（ by 5）;

The mechanism of operation ：++a First add the value of the variable to 1, And then put the resulting value in the variable a in , Then use this ++a The value of the expression is 6, and a++ First, the value of the expression is 5, Then take it. a The value of plus 1 by 6,

And then put it in variables a in . the ++a and a++ After that, it will be used in the following program to a It's all variables a Medium 6 了 .

Test formula ：++ Add before use ,++ Use first and then add .

5、 Comma expression ：

The lowest priority . The value of the expression the value of the expression to the far right of the comma .

（2,3,4） The value of the expression of is 4.

z=（2,3,4）( The whole assignment expression ) This is the time z The value of is 4.（ It's a little difficult ！）

z= 2,3,4 （ The whole thing is a comma expression ） This is the time z The value of is 2.

Add ：

1、 Empty statements cannot be executed at will , Will lead to logical errors .

2、 Notes are the key points of exams in recent years , The note is not C Language , No running time , There's no semicolon . You can't nest ！

3、 Cast ：

It must be （int）a No int（a）, Note that there must be brackets on the type .

Be careful （int）（a+b） and （int）a+b The difference between . The front is a a+b The transformation of , After that is the a Transformation plus b.

4、 Three cases of rounding and deciphering ：

１、int a =1.6;

２、(int)a;

３、1/2; 3/2;

4． Don't lose decimals , Prefix the corresponding format .2 Retain 2 position , rounding

Section 8 、 character

1） The legal form of character data :：

'1' It is the character that accounts for — Bytes ,"1" It's a string that takes up two bytes ( Contains an end symbol ).

'0' Of ASCII The value is expressed as 48,'a' Of ASCII Values are 97,'A' Of ASCII Values are 65.

The general test represents the form of a single character error ：'65' "1"

Characters can be used for arithmetic operations , remember ：　'0'-0=48

How to convert capital letters and lowercase letters ：　'A'+32='a' They are usually different from each other 32.

2） Escape character ：

Escape characters are divided into general escape characters 、 Octal escape character 、 Hexadecimal escape character .

General escape characters ： recite \0、 \n、 \’、 \”、 \\.

Octal escape character ： ‘\141’ It's legal. , Leading 0 It can't be written .

Hexadecimal escape character ：’\\x6d’ It's legal , Leading 0 Can't write , also x It's lowercase .

3、 Character types and integers are close relatives ： The two have great similarities

char a = 65 ;

printf(“%c”,a); The output is ：a

printf(“%d”, a);　 The output is ：65

Chapter nine 、 An operation

1） The examination of bit operation ： There will be one or two questions .

The general treatment ： Almost all the problems of bit operation should be handled according to this process （ Change decimal system into binary system and then decimal system ）.

example 1：　char a = 6,b;

b = a<<2; The calculation of this kind of problem is to put a Decimal system of 6 Into binary , And then do bit arithmetic .

example 2: Remember that , Bitwise sign of XOR ”^”.0 Exclusive or 1 obtain 1.

0 Exclusive or 0 obtain 0. Two girls can't be born .

Test memory method : A man (1)— Woman (0) To have a baby (1).

example 3: When the data is not discarded ,<< Move one bit to the left to multiply by 2;>> Move one bit to the right to divide by 2

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