C language - deep understanding of arrays

Why is the first subscript of an array from 0 Start ？

Because the array address must be calculated from 1 At the beginning, we have to take one more step , from 0 Start , It will be more efficient

#include <stdio.h>
#include <math.h>

void foo(int ary[])
{
    
    ary++;
}
int main(int argc,char* argv[])
{
    
    
    // Consistency and continuity 
    // Array name ary It's No 0 Address constants for elements 
    int ary[5] = {
    1,2,3,4,5};
    
    foo(ary); //foo(0x18ff34)
    printf("%08x\r\n",ary);//ary It's a constant , So we can't ary++
    /*  Arrays are random access data structures  type ary[N] = ...; int n = ...; ary[n] address is: (int)ary + sizeof(type)*n */
    
    /* printf("%d\r\n",&ary[2]);// Subscript access operator  printf("%d\r\n",(int)ary + sizeof(int)*2); printf("%d\r\n",&ary[argc]); printf("%d\r\n",(int)ary + sizeof(int)*argc); printf("%d\r\n",&ary[argc + 2]); printf("%d\r\n",(int)ary + sizeof(int)*( argc + 2)); */
    printf("%08x\r\n",&ary[0]);//0x0018ff34
    printf("%08x\r\n",&ary[8]);//0x0018ff54
    printf("%08x\r\n",&3[ary]);//0x0018ff40
    //0x00400000 -> 0x00905a4d ary[(0x00400000-(int)ary)/sizeof(int)] 0x00905a4d
    // (00400000 - 0x0018ff3c )/4 + 2 = 9C033
    printf("%08x\r\n",ary[(0x00400000-(int)ary)/sizeof(int)]);//0x00400000
    system("pause");
    return 0;
}

Function exchange numeric code

#include <stdio.h>
#include <math.h>

// Indirect access within a function will affect the actual parameters 
// * [] ->  These three operators generate indirect access 

void foo(int x,int y)
{
    
	int nTmp = x;
	x = y;
	y = nTmp;
	
}
int main(int argc,char* argv[])
{
    


		int x = 8;
		int y = 9;
	foo(x,y); //foo(0x18ff34)
	printf("%d\t%d\r\n",x,y);

	system("pause");
	return 0;
}

Array exchange numeric code ：

#include <stdio.h>
#include <math.h>
// Indirect access within a function will affect the actual parameters 
// * [] ->  These three operators generate indirect access 

void foo(int ary[])
{
    
	int nTmp = ary[0];
	ary[0] = ary[1];
	aty[1] = nTmp;
	
}
int main(int argc,char* argv[])
{
    

	// Consistency and continuity 
	// Array name ary It's No 0 Address constants for elements 
	int ary[5] = {
    1,2,3,4,5};

	foo(ary); //foo(0x18ff34)
	printf("%d\t%d\r\n",ary[0],ary[1]);

	system("pause");
	return 0;
}

main Stack structure of ：

Use to see argc Value , The bottom of the caller stack can be pushed

The bottom and top of the caller stack will be initialized as local variables 0xcccccccc

The array is then initialized to 1,2,3,4,5

Press F11 Step into debugging

callee Stack structure of ：

guess

Continue to press F11

Final value conversion

Conclusion ：

Indirect access within a function will affect the actual parameters

• [] -> These three operators generate indirect access