#include <stdio.h>
#include <conio.h>
int main()
{
int a[3][5] = {
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15}
};
printf("%d", (a + 1));
printf("\n%d", *(a + 1));
_getch();
}
The result of the first printf and the result of the second printf are the same in the above code
I don't understand why it looks the same.
*(a + 1) means "Read the value in the street number pointed to by the pointer (a + 1)
The result (as a result of reading the value contained in the address) was a[1].
That is, the value included in the address indicated by the pointer a+1 is the address value of a[1].
In other words, (a + 1) means that it has a street value of 'something with a street value of a[1]'.
Doesn't that make sense? A +1 to A means that it went to A by 4*5 bytes
What do you mean there's a street value of something with a street value of a[1].
I'm sure I misunderstood something, but I can't tell even if I read the blog.
(Sorry for confusing the sentence. But I think I can't help expressing my question myself.)
A one-dimensional array would have a different value. However, a above is a two-dimensional array.
So if you look at the next one,
#include <stdio.h>
int main(){
int a[3][5] = {
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15}
};
printf("%d", a[1][2]); // output as follows:
a[1] //So next is the pointer.
The a[1]=*(a+1) //[] operator is as follows:
This is also a pointer.
So this is also a pointer.
}
And since both values point to 1*3 away from the beginning, they have the same value.
In a two-dimensional array, you have to write two *operators.
I think I'm stammering. You understand, right?
© 2024 OneMinuteCode. All rights reserved.