C程序返回数组 [英] C program returning array

问题描述

我工作的一个非常基本的程序，我想长度为2的整数数组回到我的主块。我不能让它工作，虽然，我被告知我可能需要指针做到这一点。如何指针工作，我怎么能在我的程序中使用呢？
这是我目前的code：

  INT [] RETURN2（）;
诠释主（）{  诠释一个[2];  一个=请求（）;
  的printf（％D \\ n，一个[0]，A [1]）;
  返回（0）;
}INT []请求（）
{
  诠释一个[2];  一个[0] = -1;
  一个[1] = 8;  返回;
}
 

解决方案

• 您不能声明函数返回一个数组。

  
    

  ISO / IEC 9899：1999

  
    
    

  §6.9.1函数定义

  
    
    

  ¶3函数的返回类型应为无效或比数组类型以外的对象类型。

  
  

  C2011会说，本质上是一回事。




• 您不应该永远返回一个指针（非静态）局部变量从一个功能，因为它已不再是在范围（因此无效）尽快返回完成

您可以返回一个指向数组的开始如果阵列是静态分配的，或者如果它通过动态分配的malloc（）等

 为int *功能1（无效）
{
    静态int类型的[2] = {-1，+ 1};
    返回;
}静态INT B〔2] = {-1，+ 1};为int *函数2（无效）
{
    返回b;
}/ *调用者必须释放由function3（）返回的指针* /
为int * function3（无效）
{
    INT * C =的malloc（2 * sizeof的（* C））;
    C [0] = 1;
    C [1] = + 1;
    返回℃;
}
 

或者，如果你喜欢冒险，你可以返回一个指向数组：

  / *调用者必须释放由function4（）返回的指针* /
INT（* function4（无效））[2]
{
    INT（* D）[2] =的malloc（sizeof的（* D））;
    （* D）[0] = 1;
    （* D）[1] = + 1;
    返回D组;
}
 

小心那个函数声明！它并不需要太多的变化完全改变它的意义：

  INT（* function4（无效））[2]; //函数返回指向两个int数组
INT（* function5 [2]）（无效）; //两个指针数组来返回int功能
INT（* function6（无效）[2]）; //非法的：函数返回两个指针数组为int
为int *因数7（无效）[2]; //非法的：函数返回两个指针数组为int
 

I am working on a very basic program where I want to return an integer array of length 2 to my main block. I can't get it to work though, and I was told that I may need pointers to do this. How do pointers work, and how can I use this in my program? Here is my current code:

int[] return2();


int main() {

  int a[2];

  a = request();
  printf("%d%d\n", a[0], a[1]);


  return(0);
}

int[] request ()
{
  int a[2];

  a[0] = -1;
  a[1] = 8;

  return a;
}

解决方案

• You can't declare a function returning an array.

  ISO/IEC 9899:1999

  §6.9.1 Function definitions

  ¶3 The return type of a function shall be void or an object type other than array type.

  C2011 will say essentially the same thing.

• You shouldn't ever return a pointer to a (non-static) local variable from a function as it is no longer in scope (and therefore invalid) as soon as the return completes.

You can return a pointer to the start of an array if the array is statically allocated, or if it is dynamically allocated via malloc() et al.

int *function1(void)
{
    static int a[2] = { -1, +1 };
    return a;
}

static int b[2] = { -1, +1 };

int *function2(void)
{
    return b;
}

/* The caller must free the pointer returned by function3() */
int *function3(void)
{
    int *c = malloc(2 * sizeof(*c));
    c[0] = -1;
    c[1] = +1;
    return c;
}

Or, if you are feeling adventurous, you can return a pointer to an array:

/* The caller must free the pointer returned by function4() */
int (*function4(void))[2]
{
    int (*d)[2] = malloc(sizeof(*d));
    (*d)[0] = -1;
    (*d)[1] = +1;
    return d;
}

Be careful with that function declaration! It doesn't take much change to change its meaning entirely:

int (*function4(void))[2]; // Function returning pointer to array of two int
int (*function5[2])(void); // Array of two pointers to functions returning int
int (*function6(void)[2]); // Illegal: function returning array of two pointers to int
int  *function7(void)[2];  // Illegal: function returning array of two pointers to int

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