Common misunderstandings caused by a time reporting assistant of Blue Bridge Cup basic questions

List of articles

• Based on practice —— Time Reporting Assistant
• Common misconceptions about : Returns the address of a local variable in a function
• Summary

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One 、 Time Reporting Assistant

Problem description

        Given the current time , Please read it in English .
　　 Time spent h Sum points m Express , In English pronunciation , The way to read a time is ：
　　 If m by 0, Read out the tense , And then add “o'clock”, Such as 3:00 pronounce as “three o'clock”.
　　 If m Not for 0, Read out the tense , And then read it out , Such as 5:30 pronounce as “five thirty”.
　　 The pronunciation of the hour and the minute is in English number , among 0~20 pronounce as ：
　　0:zero, 1: one, 2:two, 3:three, 4:four, 5:five, 6:six, 7:seven, 8:eight, 9:nine, 10:ten, 11:eleven, 12:twelve, 13:thirteen, 14:fourteen, 15:fifteen, 16:sixteen, 17:seventeen, 18:eighteen, 19:nineteen, 20:twenty.
　　30 pronounce as thirty,40 pronounce as forty,50 pronounce as fifty.
　　 For greater than 20 Less than 60 The number of , First read the whole number of ten , And then add a single digit . Such as 31 Read first 30 add 1 How to read it , pronounce as “thirty one”.
　　 According to the above rules 21:54 pronounce as “twenty one fifty four”,9:07 pronounce as “nine seven”,0:15 pronounce as “zero fifteen”. 

Input format

　　 The input contains two nonnegative integers h and m, The hours and minutes of time . Non zero numbers have no leading 0.h Less than 24,m Less than 60.

Output format

　　 Output time and time in English .

The sample input

      0 15

Sample output

     zero fifteen

Their thinking ：

    At first glance, the text of this topic is very complicated , In fact, if you read it carefully twice, you will find that the topic is very simple , It is to realize a simple English time reporting function . The complexity lies in 24 There are many kinds of hours , We should solve it one by one according to the situation .24 The time of hour system mainly has the following forms ：

1. The minute bit is 0, The hour is not 0, That is, the whole point . Such as ：21:00    Output ：twenty one o'clock

2. The hour is 0, Minutes are not zero , Such as ：00:15    Output ：zero fifteen

3. The hour is not 0, The minute digit is a whole ten digit , Such as ：20:30  Output ：twenty thirty 

4. Hours is not 0, The minute bits have several dozen forms , Such as ：21:37  Output ： twenty one  thirty seven 

Therefore, the general functions of functions can be divided into two , Process the number on the hour , Handle minute digits , Then return a string to the main function , Output... In the main function . Function mainly uses switch Multiple branch statements to complete the matching between numbers and English strings , At the same time C In language string Library function , Complete the copy and splicing of strings .

The complete code is as follows ： 

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#include<stdio.h>
#include<string.h>
// Set three global string array variables  
char cha[20];
char s3[20];
char minute[20];
// Handle "  Hours  " position 
char* function1(int h){
	//char cha[20];
	switch(h){
		case 0:strcpy(cha,"zero");break;
		case 1:strcpy(cha,"one");break;
		case 2:strcpy(cha,"two");break;
		case 3:strcpy(cha,"three");break;
		case 4:strcpy(cha,"four");break;
		case 5:strcpy(cha,"five");break;
		case 6:strcpy(cha,"six");break;
		case 7:strcpy(cha,"seven");break;
		case 8:strcpy(cha,"eight");break;
		case 9:strcpy(cha,"nine");break;
		case 10:strcpy(cha,"ten");break;
		case 11:strcpy(cha,"eleven");break;
		case 12:strcpy(cha,"twelve");break;
		case 13:strcpy(cha,"thirteen");break;
		case 14:strcpy(cha,"fourteen");break;
		case 15:strcpy(cha,"fifteen");break;
		case 16:strcpy(cha,"sixteen");break;
		case 17:strcpy(cha,"seventeen");break;
		case 18:strcpy(cha,"eighteen");break;
		case 19:strcpy(cha,"nineteen");break;
		case 20:strcpy(cha,"twenty");break;
		case 21:strcpy(cha,"twenty one");break;
		case 22:strcpy(cha,"twenty two");break;
		case 23:strcpy(cha,"twenty three");break;
		case 24:strcpy(cha,"twenty four");break;
		case 30:strcpy(cha,"thirty");break;
		case 40:strcpy(cha,"forty");break;
		case 50:strcpy(cha,"fifty");break;
	}
	return cha;
}
// Handle ten bits of minutes  
char *function3(int x){
	//char s3[20];
	switch(x){
		case 1:strcpy(s3,"ten ");break;
		case 2:strcpy(s3,"twenty ");break;
		case 3:strcpy(s3,"thirty ");break;
	    case 4:strcpy(s3,"forty ");break;
	    case 5:strcpy(s3,"fifty ");break;
	}
	return s3;
}
// Handle " minute " position   
char *function2(int m){
	//char minute[20];
	if(m%10!=0){
		if(m<10){
			strcpy(minute,function1(m));
		}
		else{
			// Because only 1~20 Only the numbers between have a specific English match , Others need to be disassembled  
			if(m>20&&m<60){
				strcpy(minute,function3(m/10));// The parameter passed in is ten digits of the number of minutes  
				strcat(minute,function1(m%10));// The parameter passed in is one bit of the minute digit  
			}
			// Figures that do not need to be disassembled  
			else{
				strcpy(minute,function1(m));
			}
		}
		
	} 
	else{
		strcpy(minute,function1(m));//  When m When the single digit of is zero , It's a whole ten  
	}
	return minute;
}

int main(){
	int h,m;
	char hour[20],minute[20];
	scanf("%d%d",&h,&m);
	if(m!=0){
		strcpy(hour,function1(h));
		strcpy(minute,function2(m));
		printf("%s %s",hour,minute);
	}
	else{
		strcpy(hour,function1(h));
		printf("%s o'clock",hour);
	}
	return 0;
}

​

Running results ：

 

 

Two 、 Common misconceptions about

1. You cannot return the address of a local variable in a function

The wrong sample ：

The compiler gives a warning ：

2. Error cause analysis

      When using character arrays , A new backup will be created in the stack , It will be created as the function is called , It will be automatically destroyed as the function ends . Even if it returns an address , But at the end of this function , The contents of the memory space pointed to by this address have been released , Therefore, the content corresponding to that address can no longer be found through the returned address , So it is wrong to return the address of a local variable in a function .

3. Solution

（1） Use global variables

    Because global variables have static storage , Only when the whole program is finished , The data stored in memory will be released .

（2） Use character pointers

    When using character pointers , The string is stored in static storage （ Heap area ）, So the returned address points to the content （ In the pile area ） Will not be released , So it can be found .（ But this way of writing has one drawback , That is, after defining the string pointer , You can't change the string it points to , Unexpected errors may occur after the change ）.

（3） Use static keyword

    Use static Keyword declares it as a static variable , With the method 1 The same principle .

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3、 ... and 、 Summary

1.static Static variables

   static When used for function definition or variable declaration outside the code block , Used to modify the link properties of identifiers , From external links to internal links , But the storage type and scope are not affected , Functions or variables declared in this way can only be accessed in their source files .

    When it is used for variable declaration inside a code block , Used to modify the storage type of variables , Change from automatic variable to static variable , But the link properties and scope are not affected .

2. Function can return local variables , But you cannot return the address of a local variable （ The pointer ）

  The scope of the local variable is inside the function , When the function returns , Local variable memory has been freed . therefore , If the function returns the value of a local variable , Address not involved , The program doesn't go wrong . So the function can return local variables . But if the function returns the address of a local variable ( The pointer ) Words , Error will be reported when the program is running . Because it's just a pointer （ Address ） After copying, it returns , And the memory pointed by the pointer （ Stored content ） Has been released , In this way, the pointer has no permission to access the memory , An error will be reported after calling . To be exact , A function cannot return a pointer to stack memory by ！ Note that this refers to the stack , It is possible to return a pointer to heap memory , Because the contents of heap memory are not released until the end of the whole program .