Use Pointers to do addition operations without a plus sign

We usually add with something like this

#include<stdio.h>
int main()
{
 int a=2,b=3;
 printf("%d\n",a+b);
 return 0;
}

The result is 5, adding a and b

But how can we add without the plus sign? So let's look at one of the array operations

#include<stdio.h>
int main()
{
 int a[]={1,2};
 printf("%d\n",a[0]+a[1]);
 return 0;
}

The result is 3, which is the same as the following code

#include<stdio.h>
int main()
{
 int a[]={1,2};
 printf("%d\n",*(a+0)+*(a+1));
 return 0;
}

Because a [i] = * (a + i)

So let's just do 1 pointer

#include<stdio.h>
int main()
{
 int a=2,b=3;
 int *p=a,*q=b;
 printf("%d\n",*p+*q);
 return 0;
}

p and q point to the addresses of a and b respectively, calculate the sum of a and b, and the result is 5

Now use the pointer and just said a[i]=*(a+i) and variable location 1 combination

#include<stdio.h>
int main()
{
 int a=2,b=3;
 int *p;
 printf("%d\t%d\n",&a,&b);
 p=&a;
 printf("%d\n",(int)&((char *)a)[b]);
 return 0;
}

The output is

[

1638212 1638208
5

]

In front of the two values are respectively a and b address, because is int type and in the 32-bit compiler I use is stored in 4 bytes, however char type is according to 1 byte stored in one byte, use Pointers p variable a, see (char *) a first, it is a casting, converting a this variable char * type of address, ((char*)a)[b] moves the address back by b bytes, but ((char*)a)[b] is equal to *(a+b), we know & The alpha and alpha cancel each other out, so we have alpha & ((char*)a)[b], but it is still of type char*, and the preceding (int) casts its bit int for output.

When a and b, as defined by long at the time, were cast to int, long was 8 bytes and int was 4 bytes, then data would be lost.

Finally, as I just learned, you can test whether the compiler is 64-bit or 32-bit with Pointers. The 64-bit compiler pointer variable is 8 bytes, and the 32-bit compiler pointer variable is 4 bytes.

conclusion