C language function review (pass value and address [binary search], recursion [factorial, Hanoi Tower, etc.))

function

Function call by value

The formal and actual parameters of a function occupy different code blocks , Modifying the formal parameter does not change the value of the argument

Address calling of function

Addressing call refers to a call method that passes the memory address created outside the function to the function parameter

This parameter transfer method can operate variables outside the function inside the function

practice

Two points search

#include<stdio.h>

int binary_2(int arr[] , int k,int len);

int main() {
    
	int arr[ ] = {
     1,2,3,4,5,6,7,8,9,10 };
	int len = sizeof(arr) / sizeof(arr[0]);
	int element ;
	printf(" Please enter the element you want to find ");
	scanf("%d", &element);
	int ret;
	ret = binary_2(arr ,element,len);
	if (ret == -1) {
    
		printf(" The specified element was not found \n");
	}
	else {
    
		printf(" eureka , The subscript of the element is %d\n",ret);
	}

	return 0;
}
int binary_2(int arr[] , int k,int len) {
    
	//int len = sizeof(arr) / sizeof(arr[0]);
	int left = 0;
	int right = len - 1;
	
	while (left <= right) {
    
		int mid = (left + right) / 2;
		if (arr[mid] < k) {
    
			left = mid + 1;
		}
		else if (arr[mid] > k) {
    
			right = mid - 1;
		}
		else {
    
			return mid;
		}
		
	}
	return -1;
}

Be careful ：

1） One is function len To be in main Function to calculate internally , Because if you calculate inside a function ,size（arr） The value of is actually 4 , Why? , Because what is passed in is a pointer variable , Points to the address of the first element , The calculated size is int* size

#include<stdio.h>
int main() {
    
 The subscript of a prime is %d\n",ret);
	printf("%d\n", sizeof(arr));//40
	printf("%d\n", sizeof(arr[0]));//4

	return 0;
}

2） The second thing to note is , The parameter passed in int arr[] Why pointer variables , int arr[] Equivalent to int * arr, The two are the same

Modular use of functions

From a source file A To another source file B in , Want to add the source file A Use source files in B The function in ,

1） First, in the source file A Add #include “ The function name you want to use ”

2. Create a new and... In the header file B Header file with the same name （.h） ending

For example ：

Source file A, The name is str.c ：

#include<stdio.h>
#include"add.h"

int main2() {
    
	int x = 1, y = 4;
	int sum = Add(x, y);
	printf("%d", sum);
	return 0;
}

Source file B, The name is add.c ：

#include<stdio.h>

int Add(int x, int y) {
    
	return x + y;
}

Header files to be actively added C, The name is add.h( This is generally the same as the header file ) :

#ifndef __ADD_H_// if not define __XX_X_
#define __ADD_H_

int Add(int x, int y);// Here is the declaration of the function 

#endif

This is the format , To understand the ： The first line finds that there is no definition , If defined , You don't have to define , Avoid repeated definitions

Recursion of a function

What is recursion

A programming skill of function call itself (recursion)

The main thinking ： Turn big into small , Layer by layer dialing

I think it can be imagined as an inverted pyramid , The bottom floor is open , The upper layers have permissions and cannot be opened , The process of recursion is the process of unlocking upward

practice 1：

Write functions to print out numbers one by one “1234” Medium 1 , 2, 3,4

#include<stdio.h>

void print(int i) {
    
	if (i > 9)
	{
    
		print(i/10);
	}
	printf("%d\n", i%10);
}

int main() {
    
	unsigned int i = 1234;
	// Customize by calling print Function to complete 1,2,3,4 Output 
	/*  For the first time 1234 Divide into “123” and 4  Continue to divide “12" and ”3“”4“  Go on ”1“2”3“4”  This is the programming idea of "from big to small"  */
	print(i);


	return 0;
}

Their thinking ：

From big to small , Set the permissions , Layer by layer recursion ！

practice 2：

Written by myself my_strlen Play a strlen The role of , Number of output strings , In two ways

Tips ： One is to create temporary variables , One does not create

#include<stdio.h>

int my_strlen1(char * );
int my_strlen2(char *);
int main() {
    
	char arr[] = "superzkx";
	int len1 = my_strlen1(arr);// This is a function obtained through temporary variables 
	int len2 = my_strlen2(arr);// This is the second way 
	printf("%d\n", len1);
	printf("%d\n", len2);

	return 0;
}

int my_strlen1(char* arr) {
    // The idea is to save a temporary variable 
	int coust = 0;
	while (*arr != '\0') {
    
		*arr++;
		coust++;
	}
	return coust;
}

int my_strlen2(char* arr) {
    // This is the pointer pointing back one by one , The first step turns into 1+"uperzkx"...
	if (*arr != '\0') {
    
		return 1 + my_strlen2(arr + 1);
	}
	return 0;
}

thus it can be seen , Recursion converts the original while Instead of if, Among them return In fact, it plays a role while The role of

practice 3：

Function calculation n! Value

#include<stdio.h>

int main() {
    
	int n;
	int sum1, sum2;
	printf(" Please enter the number of factorials to be calculated ");
	scanf("%d", &n);
	sum1 = factorial_1(n);
	sum2 = factorial_2(n);
	printf("sum1==%d,sum2==%d", sum1, sum2);
	return 0;
}

int factorial_1(int n) {
    // Circular approach 
	int i;
	int sum = 1;
	for (i = 1; i <= n; i++) {
    
		sum = sum * i;
	}
	return sum;
}
int factorial_2(int n) {
    // Recursive method 
	if (n == 1)
		return 1;
	else if (n > 1) {
    
		return n * factorial_2(n - 1);

	}
}

It can be seen that loop and recursive In fact, they can express each other

practice 4：

Calculating Fibonacci series ( Not stack overflow )

#include<stdio.h>

int Fibo1(int n);
int Fibo2(int n);

int main() {
    
	int n;
	int sum1, sum2;
	printf(" Please enter to calculate Fibo Number of numbers ");
	scanf("%d", &n);
	sum1 = Fibo1(n);
	sum2 = Fibo2(n);
	printf("sum1== %d,sum2== %d", sum1, sum2);
	return 0;
}

// Ordinary recursive method 
int Fibo1(int n) {
    
	if (n == 1)
		return 1;
	else if (n == 2)
		return 1;
	else
		return Fibo1(n - 1) + Fibo2(n - 2);
}

int Fibo2(int n) {
    
	int a = 1;
	int b = 1;
	int c = 0;
	if (n == 1) {
    
		return a;
	}
	if (n == 2) {
    
		return b;
	}
	while (n > 2) {
    
		c = a + b;
		a = b;
		b = c;
		n--;
	}
	return c;
	//1 1 2 3 5 8 13 21 34
	//a b c 
	// a b c
}

Classical recursion problem ：

Hanoi

#include<stdio.h>

void Hanoi(int n, char A, char B, char C);

int main() {
    
	int n =0;
	char A = 'A';
	char B = 'B';
	char C = 'C';
	printf(" Please enter the number of floors of Hanoi tower ( The goal is to A The column moves to C column )\n");
	scanf("%d", &n);
	
	Hanoi(n,A,B,C);
	return 0;
}

void Hanoi(int n,char A, char B ,char C ) {
    
	if (n == 1) {
    
		printf(" take %c-->%c\n",A,C);
	}
	else {
    
		Hanoi(n - 1, A, C, B);// Before the n-1 individual   from  A  adopt C  All moved to B
		printf("%c-->%c\n",A,C);// then A The last big one arrives C
		Hanoi(n - 1, B, A, C);// Last B adopt A To C

	}
}

The main idea ：n>=2 when （n>=3) More obvious when , The former n-1 Doughnuts as a whole , In fact, the repetition is n=2 Time operation

Namely ！

Before the n-1 individual from A adopt C All moved to B

then A The last big one arrives C

Last B adopt A To C

A simple KO!