C语言 将地址传递给 C 中的函数
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Passing addresses to functions in C
提问by Amit
I'm new to C and I have a function that calculates a few variables. But for now let's simplify things. What I want is to have a function that "returns" multiple variables. Though as I understand it, you can only return one variable in C. So I was told you can pass the address of a variable and do it that way. This is how far I got and I was wondering I could have a hand. I'm getting a fair bit of errors regarding C90 forbidden stuff etc. I'm almost positive it's my syntax.
我是 C 的新手，我有一个计算几个变量的函数。但是现在让我们简化一下。我想要的是有一个“返回”多个变量的函数。虽然据我所知，你只能在 C 中返回一个变量。所以我被告知你可以传递一个变量的地址并这样做。这是我走了多远，我想知道我能伸出援手。我在 C90 禁止的东西等方面遇到了相当多的错误。我几乎肯定这是我的语法。
Say this is my main function:
说这是我的主要功能：
void func(int*, int*);
int main()
{
int x, y;
func(&x, &y);
printf("Value of x is: %d\n", x);
printf("Value of y is: %d\n", y);
return 0;
}
void func(int* x, int* y)
{
x = 5;
y = 5;
}
This is essentially the structure that I'm working with. Could anyone give me a hand here?
这基本上是我正在使用的结构。有人可以帮我吗？
回答by Mehrdad Afshari
You should use *variableto refer to what a pointer points to:
您应该使用*variable来指代指针指向的内容：
*x = 5;
*y = 5;
What you are currently doing is to set the pointer to address 5. You may get away with crappy old compilers, but a good compiler will detect a type mismatch in assigning an intto an int*variable and will not let you do it without an explicit cast.
您当前正在做的是将指针设置为地址 5。您可能会使用蹩脚的旧编译器，但是一个好的编译器会在将 赋值int给int*变量时检测到类型不匹配，并且不会让您在没有显式转换的情况下执行此操作。
回答by Amit
void function(int *x, int* y) {
*x = 5;
*y = 5;
}
would change the values of the parameters.
会改变参数的值。
回答by Ingo
You can't forward declare func(int,int) when in reality it is func(int*, int*). Moreover, what should the return type of func be? Since it doesn't use return, I'd suggest using void func(int*, int*).
当实际上它是 func(int*, int*) 时，您不能转发声明 func(int,int)。此外，func的返回类型应该是什么？由于它不使用 return，我建议使用 void func(int*, int*)。
回答by pmg
You can return a single variable of a struct type.
您可以返回结构类型的单个变量。
#include <stdio.h>
#include <string.h>
struct Multi {
int anint;
double adouble;
char astring[200];
};
struct Multi fxfoo(int parm) {
struct Multi retval = {0};
if (parm != 0) {
retval.anint = parm;
retval.adouble = parm;
retval.astring[0] = parm;
}
return retval;
}
int main(void) {
struct Multi xx;
if (fxfoo(0).adouble <= 0) printf("ok\n");
xx = fxfoo(42);
if (strcmp(xx.astring, "\x2a") == 0) printf("ok\n");
return 0;
}
回答by jonsca
In addition to the changes that the other posters have suggested for your function body, change your prototype to void func(int *,int *), and change your function definition (beneath main) to reflect void as well. When you don't specify a return type, the compiler thinks you are trying to imply an int return.
除了其他海报为您的函数体建议的更改之外，将您的原型更改为void func(int *,int *)，并更改您的函数定义（在 main 之下）以反映 void。当您不指定返回类型时，编译器认为您试图暗示 int 返回。
