They created a two-dimensional array of chars, and they also created a pointer to it. I want the pointer to point to a[0][1], so I did (*pa)++ (I know pa is the pointer to a, *pa is the pointer to the first row of a, and **pa is the pointer to the first value of the first row). I don't know why. Please help me ㅠ<
#include <stdio.h>
int main()
{
char a[100][100];
char (*pa)[100];
pa=a;
a[0][0]='a';
a[0][1]='e';
a[1][0]='g';
(*pa)++;
printf("%c\n",*(*pa+1));
}
The potential/post increment operator is only available for changeable lvalue.
++b is equivalent to b = b+1.
b++ is equivalent to temp = b; b = b + 1; temp
The value is eventually substituted for b whether it is a potential or a posterior.
In order to substitute a value for b, b must be a variable.
If you look at (*pa)++, *pa is an array of characters with 100 elements.
The array is also converted to the address of the first element, so (*pa) becomes the address value of a[0].
The address value is not a variable, so it is not a substitution operation. Potential/postal increment operator is not available because it is not a substitution operation.
Simply put, (*pa)++ is not allowed, such as why (1)++ is not allowed.
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