Change the value of the argument by address through malloc and pointer

#include"stdio.h"
#include"stdlib.h"
#include"string.h"

void getMem(int* p)
{
	
	p=(int*)malloc(sizeof(int));
	
	*p=5;
	
}

int main(void)
{
	
	int a=1;
	int* p1=&a;
	getMem(p1);
	printf("a=%d\n",a);
		
	return 0;
}
 The original meaning of the code is ： Defining pointer variables p1,p1 Point to main Variables in functions a, By calling getMem Function to variable a Change arguments by addressing a Value , Change it to 5, But the output result is 1, Explanatory variable a The value of . The reason lies in the wrong use of malloc.

【 change 】：
#include"stdio.h"
#include"stdlib.h"
#include"string.h"

void getMem(int* p)
{
	
//	p=(int*)malloc(sizeof(int));
	 You need to block the above sentence , Because after passing in the parameters here , What comes in is main Function a The address of , But this statement makes the pointer variable p It points to a newly opened space ,p No longer point to a 了 , So below *p=5 It only plays a role in malloc The memory space is assigned as 5, But this time p Stored in is no longer a The address of the .
	*p=5;
	
}

int main(void)
{
	
	int a=1;
	int* p1=&a;
	getMem(p1);
	printf("a=%d\n",a);
		
	return 0;
}

Let's do another example , It involves address and value transmission , as well as malloc, The secondary pointer

#include"stdio.h"
#include"stdlib.h"
#include"string.h"

void getMemory(char* *p)
{
	*p=(char*)malloc(15);
	
	
}

int main(void)
{
	char* str =NULL;
	
	getMemory(&str); 
 Change the argument by calling a function that does not return a value str Value , Should be str Pointer to as a function parameter , This function parameter must be of pointer type （ Several levels of pointer are still ignored ）

 By calling a function that does not return a value , You need to change the value of the argument , Address should be used .

 because str It's a pointer in itself , therefore *str It's a secondary pointer .
	
	strcpy(str,"hello");
	printf("%s",str);
	
	
	
	return 0;
}