C programming learning notes [edited by Mr. Tan Haoqiang] (Chapter III sequence programming) 03 operators and expressions

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One 、 Operators and Expressions

Two 、 Common arithmetic operators ：

Be careful ：
1. Divide two real numbers , The result is a double precision real number
2. The result of dividing two integers is an integer
3.% For integers only

3、 ... and 、 Self increasing （++） Self reduction （–） Operator

++i,–i In the use of ｉ Before , First of all ｉ The value of the add / reduce 1
i++,i-- In the use of ｉ after , send ｉ The value of the add / reduce 1

++i It's to execute first i=i+1, Reuse i Value ; and i++ Use first i Value , Re execution i=i+1.

int i=3,j;
j=++i;	//i The value of becomes 4,  Reassign to ｊ,j The value of is ４

int i=3;
printf(“%d”,++i);	// Output ４

int i=3,j;
j=i++;	// First the  i Value 3 Assign to ｊ,ｊ The value of is ３, then ｉ Turn into ４

int i=3;
printf(“%d”,i++);	// Output 3

Four 、 Type conversion （ a key ）

Automatic type conversion
1.+、-、*、/ One of the two numbers in the operation is float or double type , The result is double type , Because the system will all float Type data are converted to double type , And then you do the math .
2. If int Type and float or double Type of data , The first int The type and float Type data is converted to double type , And then you do the math , The result is double type .
3. character (char) Type data and integer data , That is to put the character ASCII Code and integer data . If character data and real data are operated , Then the ASCII Code conversion to double Type data , And then you do the math .

give an example 1：

int i=3;
float f=2.5;
double d=7.5;
printf("%lf",10+'a'+i*f-d/3);

10+‘a’+if-d/3
① Conduct 10+′a′ Arithmetic ,′a′ The value of is an integer 97, The result of operation is 107.
② because “” Than “+” High priority , to if Arithmetic . First the i And f All turned into double type , The result of operation is 7.5,double type .
③ Integers 107 And ｉｆ Add the products of . Integer first 107 Convert to double , The sum is 114.5,double type .
④ Conduct d/3 Arithmetic , First the 3 convert to double type ,d/3 The result is 2.5,double type .
⑤ take 10+′a′+i*f Result 114.5 And d/3 The business of 2.5 Subtracting the , The result is 112.0,double type .

give an example 2：

Given a capital letter , Output in lowercase letters is required .

#include <stdio.h>
int main()
{
    
	char c1,c2;
	c1='A'; 			// The character ′A′ Of ASCII Put the code in c1 variable 
	c2=c1+32;			// Get the characters ′a′ Of ASCII Code , Put it in c2 variable  
	printf("%c\n",c2);	// Output c2 Value , Is a character 
	printf("%d\n",c2);	// Output c2 Value , Is the character ′a′ Of ASCII Code 
	return 0;
}

Cast ： When automatic type conversion cannot achieve its purpose , You can use cast .
Cast operators

( Type name )( expression )

(double)a		 take ａ convert to double type 
(int)(x+y)		 take x+y The value of is converted to int type 
(float)(5%3)		 take 5%3 The value of is converted to float type 
(int)x+y		 Only will x Convert to integer , Then with y Add up 

int a; float x,y;double b;
a=(int)x
 Perform forced type operation (int)x And then I got a int Temporary value of type , Its value is equal to ｘ The integral part of , Give it to a, Be careful x The value and type of have not changed , Still float type . The temporary value no longer exists after assignment .