我如何使用返回 void [...] 的函数更改作为参数传递给函数的变量 [...]
指针是做到这一点的唯一方法吗?
Yes.
如何执行此操作的示例:
#include <stdio.h> /* for printf() */
struct S
{
int i;
char c;
};
void foo(struct S * ps)
{
ps->i = 42;
ps->c = 'x';
}
int main(void)
{
struct S s = {1, 'a'}; /* In fact the same as:
struct S s;
s.i = 1;
s.c = 'a'
*/
printf(s.i = %d, s.d = %c\n", s.i, s.c);
foo(&s);
printf(s.i = %d, s.d = %c\n", s.i, s.c);
}
Prints:
s.i = 1, s.d = a
s.i = 42, s.d = x
另一个例子是(取自/基于Bruno https://stackoverflow.com/users/2458991/bruno's 删除的答案 https://stackoverflow.com/revisions/55990582/2):
void f(int * v1, float * v2)
{
*v1 = 123; // output variable, the previous value is not used
*v2 += 1.2; // input-output variable
}
int main(void)
{
int i = 1;
float f = 1.;
f(&i, &f);
// now i values 123 and f 2.2
return 0;
}