The function realizes the exchange of two numbers (indirect access to variables outside the function through pointers)

Function to exchange two numbers

First of all, let's see if we can't use functions, how can we exchange two numbers?

#include <stdio.h>
int main()
{
    
    
	int a = 50;
	int b = 100;
	int temp = a;
	a = b;
	b = temp;
	printf("a=%d b=%d\n", a, b);
	system("pause");
	return 0;
}

Result: a and b successfully realized the exchange.

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So how to use the function?
Many people think that just put the part of the above exchange numbers into the function, then let's try:

#include <stdio.h>
void swap(int x, int y){
    
    
	int temp = x;
	x = y;
	y = temp;
}
int main()
{
    
    
	int a = 50; 
	int b = 100;
	swap(a, b);
	printf("a=%d b=%d\n", a, b);
	system("pause");
	return 0;
}

result:

Analysis: We can see that in this code, there is no exchange between a and b.
So did x and y swap?
We can add code to output x and y values ​​in the function to take a look.

Code:

void swap(int x, int y){
    
    
	int temp = x;
	x = y;
	y = temp;
	printf("x=%d y=%d\n", x, y);
}

Result:

Analysis: In this code, we found that a and b did not exchange. What is exchanged inside the function is x and y, x and y are equivalent to new local variables created, but initialized by a and b. Exchanging x and y does not affect a and b.

The reason is: x and y accessed inside the function are not the ontology of a and b .

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  To solve this problem, you need to find a way to make the variables outside the function accessible inside the function. In C language, we cannot directly access local variables outside the function, but we can use pointers to indirectly access external variables .

Code:

void swap(int* x, int* y){
    
    
	int temp = *x;
	*x = *y;
	*y = temp;
}

The swap function here is equivalent to the following code:

    int* x = &a;
	int* y = &b;
	
	int temp = *x;
	*x = *y;
	*y = temp;

	int temp = a;
	a = b;
	b = temp;

    printf("a = %d, b = %d\n", a, b);

In this code, first define two pointer variables x and y, where x holds the address of a and y holds the address of b. Represents the dereference of the pointer and indicates the target pointed to by the pointer. So x is equivalent to a, *y is equivalent to, and then exchange a, b through the temporary variable temp.

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The final code implementation:

#include <stdio.h>
void swap(int* x, int* y){
    
    
	int temp = *x;
	*x = *y;
	*y = temp;
}

int main()
{
    
    
	int a = 50;
	int b = 100;
	swap(&a, &b);
	printf("a=%d b=%d\n", a, b);
	system("pause");
	return 0;
}