我有一个关于 C 中这个简单程序的问题，用于练习指针和数组。
代码：
#include <stdio.h>
fun(int *p, int *v) {
*p++;
*(v+2) = p[3];
*v++ = p[0];
v[0] = *(p+4);
}
int main() {
int v[] = {90, 89, 8, 78, 60, 120};
int k[] = {0, 0, 0, 0, 0, 0};
int *ptr = k;
fun(v, ptr);
for (ptr = k; ptr < k + 6; ptr++)
printf("%d\n", *ptr);
}
所以我的问题是它为什么会打印89 120 60 0 0 0。
It's useful to draw out (even by hand on paper!) what's going on with pointers and arrays. For the following function:
void fun(int *p, int *v) {
*p++;
*(v+2) = p[3];
*v++ = p[0];
v[0] = *(p+4);
}
You have passed these arrays:
[90] [89] [8] [78] [60] [120]
|
p
[0] [0] [0] [0] [0] [0]
|
v
What happens in fun?
*p++
This could be read one of two ways: Dereference p and (post) increment that value, or post-increment p and dereference the old value of p. Which is it? Our handy table of operator precedence says ++ comes before *, so it's the second one. Notice that we deference a pointer, but don't do anything the value. That means we could have written p++ and gotten the same result.
After the increment, our pointer has changed:
[90] [89] [8] [78] [60] [120]
|
p
The next line uses both p and v:
*(v+2) = p[3];
You can read this as "assign the 3rd value to the right of p to the 2nd spot to the right of v." The third spot to the right of p has 60, and it goes here:
[0] [0] [60] [0] [0] [0]
|
v
Next up:
*v++ = p[0];
From before, we know that *v++ means (post) increment v. The difference between ++v (pre-increment) and v++ (post-increment) is what value gets returned - in this case, we'll modify v but use its old value, assiging what p is pointing at. Notice that v has been incremented:
[89] [0] [60] [0] [0] [0]
|
v
And finally:
v[0] = *(p+4);
This can be read as "assign th e 4th value to the right of p to the 0th spot to the right of v." The forth spot to the right of p has 120, and it goes here:
[89] [120] [60] [0] [0] [0]
|
v
So at the end of the function, this is what our pointers and memory look like:
[90] [89] [8] [78] [60] [120]
| |
| p
main-v
[89] [120] [60] [0] [0] [0]
| |
| v
main-k
main-ptr