当尝试使用函数交换 2 个指针的值来反转数组时，核心不断转储
#include<stdio.h>
int main() {
/* read integer array */
int n, i;
scanf("%d", &n);
int *a = (int *)malloc(n * sizeof(int));
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
/* reverse the array */
for (i = 0; i < n / 2; i++) {
exchange(a[i], a[n - i - 1]);
}
/* print the array */
for (i = 0; i < n; i++) {
printf("%d ", a[i]);
}
/* free the memory */
free(a);
return;
}
/* write a function that takes two pointers of two intergers and then exchanges the values
in those locations */
void exchange(int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
这是代码，函数交换尝试交换这两个指针指向的整数的值。 不断地倾倒核心。
#include<stdio.h>
int main() {
/* read integer array */
int n, i;
scanf("%d", &n);
int *a = (int *)malloc(n * sizeof(int));
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
/* reverse the array */
for (i = 0; i < n / 2; i++) {
exchange(a[i], a[n - i - 1]);
}
/* print the array */
for (i = 0; i < n; i++) {
printf("%d ", a[i]);
}
/* free the memory */
free(a);
return;
}
/* write a function that takes two pointers of two intergers and then exchanges the values
in those locations */
void exchange(int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
This is the code, the function exchange tries to exchange the values of the integers which these two pointers are pointing to.
Keeps dumping the core.
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评论（3）
在此函数交换调用中，
两个参数的类型均为
int，而函数需要的参数类型为int *。您需要像这样调用该函数
In this call of the function exchange
the both arguments have the type
intwhile the function expects arguments of the typeint *.You need to call the function like
您需要发送交换函数的地址值，因为它的参数被声明为指针。因此，您需要将此行更改
为此行，因为正如我提到的，它需要地址。
You need to send address values for exchange functions since arguments of it are declared as pointers. Therefore, you need to change this line
to this line because as I mentioned it expects addresses.
更好的方法是将整个数组和索引传递给交换。
像这样：
在线尝试A nicer way would be passing the whole array and indexes to swap.
Like this:
TRY IT ONLINE