从C中的函数返回多个参数 [英] return more than one parameter from a function in C

问题描述

我一直在试图找出如何在那里我可以从一个函数用C返回多个值。

i have been trying to figure out ways where i can return more than one value from a function in C.

例如说我有

int compute(int a, int b)
{
int c;
// do some computation on a,b,c
// return a,b,c
}

现在我将这些值在数组中说了，我试图做到这一点，我然后我意识到，数组需要是动态的，因此它可以在被引用的main（）也是如此。

now i can return these values say in an array, and i was trying to do that and i then i realized that the array needs to be dynamic so it can be referenced in main() as well.

另一种方式可能是我创建了一个结构，然后只返回结构变量。

Another way could be that i create a structure and then just return the struct variable.

是否有这样做的任何简单而强大的方式，除了上述工作周围的其他？我必须回到它们是相互独立的，从方法大约25不同的计算值，我在我的code已经有很多结构。

Is there any simple and robust way of doing this, other than the above work around ? i have to return about 25 different computation values which are independent of each other from a method, and i have a lot structs in my code already.

感谢您。

推荐答案

如果该值都逻辑上相关的，那么它是有道理的把它们放入一个结构。但是，如果你改变这种不紧耦合的只有几个值，就可以将它们作为这样的指针：

If the values are all logically related, then it makes sense to put them into a structure. But if you're changing only a few values that aren't tightly coupled, you can pass them as pointers like this:

int swap(int *a, int *b) {
  int tmp;

  if (a == b) { // Pointers are equal, so there's nothing to do!
    return -1;  // Indicate the values haven't changed.
  }

  tmp = *a;
  *a = *b;
  *b = tmp;
  return 0;  // Indicate the swap was successful.
}

void main(...) {
  int first = 12;
  int second = 34;

  if (swap(&first, &second) == -1) {
    printf("Didn't swap: %d, %\n", first, second);
  } else {
    printf("Swapped: %d, %d\n", first, second);
  }
}

这是将数据放入该函数的参数，以及有函数返回一个值，以指示成功/失败，或一些其他条件相当标准的技术。\\

It's a fairly standard technique to put data into the function arguments, and have the function return a value to indicate success/failure, or some other condition.\

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