C语言 如何将函数输出分配给数组类型？在 C?

How to assign function output to an array type? In C?

提问by Sother

I try to assign to assign the output of a function to an array, but upon compiling, it doesn't seem to work.

我尝试分配将函数的输出分配给数组，但在编译时，它似乎不起作用。

The function takeInputsshould return an array. And, as I have it, I thought that holdInputsis an array. However, it doesn't seem to compile. What is the error here?

该函数takeInputs应该返回一个数组。而且，据我所知，我认为这holdInputs是一个数组。但是，它似乎无法编译。这里的错误是什么？

// Holds the expression that the user enters
struct Inputs  
{
    char word[10]; 
};

// Declare these: 
struct Inputs* takeInputs(struct Inputs *userInputs, int numInputs);  // must return a pointer to a pointer because returning array 
struct Inputs* printInputs(struct Inputs *userInputs);

struct Inputs* takeInputs(struct Inputs *userInputs,int numInputs){
    /*Inputs:
        userInputs: an array of struct Inputs, each of which contain a string called "word" 
        numInputs:  integer from user 
    */

    int i;
    for (i=0;i<numInputs;i++){
        printf("please input the word");        
        fgets(userInputs[i].word,10,stdin);
    }


}

int main(int argc, char const *argv[]){

    //user Input should look like this:  ./takes_Input.exe 7
    if (argc!=2){
        error("user Input should look like this:  ./takes_Input.exe 7");
    }

    // represents the number of words the user is entering 
    int numInputs = atoi(argv[2]);

    struct Inputs allInputs[numInputs];
    struct Inputs holdInputs[numInputs];

    holdInputs = takeInputs(allInputs,numInputs);
    // printInputs(holdInputs);
    return 0;
}

Error output:

错误输出：

takes_Input.c: In function ‘main':
takes_Input.c:53:13: error: assignment to expression with array type
  holdInputs = takeInputs(allInputs,numInputs);

But I thought I initialized holdInputs as an array?? Thanks.

但我以为我将holdInputs 初始化为一个数组？？谢谢。

采纳答案by autistic

The function takeInputsshould return an array. And, as I have it, I thought that holdInputsis an array. However, it doesn't seem to compile. What is the error here?

该函数takeInputs应该返回一个数组。而且，据我所知，我认为这holdInputs是一个数组。但是，它似乎无法编译。这里的错误是什么？

The error in this explanation is that a function can't return an array, as Jonathan Leffler pointed out in the comments.

这个解释中的错误是函数不能返回数组，正如 Jonathan Leffler 在评论中指出的那样。

---

---

takes_Input.c:53:13: error: assignment to expression with array type holdInputs = takeInputs(allInputs,numInputs);

But I thought I initialized holdInputs as an array??

take_Input.c:53:13: 错误：赋值给数组类型为holdInputs = takeInputs(allInputs,numInputs)的表达式；

但我以为我将holdInputs 初始化为一个数组？？

You did indeed declare holdInputsas an array, and it is an array. Though you haven't initialised it, that shouldn't be a problem. The error message is telling you that you can't assign to an array. For example,

你确实声明holdInputs为一个数组，它是一个数组。虽然你还没有初始化它，但这应该不是问题。错误消息告诉您无法分配给数组。例如，

char foo[4];
foo = "bar";        // This is an error
strcpy(foo, "bar"); // This is fine and dandy, like sour candy...

---

---

Or to take an example from your code, you're assigning to the array here:

或者以您的代码为例，您在此处分配给数组：

holdInputs = takeInputs(allInputs,numInputs);

Perhaps you meant:

也许你的意思是：

memcpy(holdInputs,takeInputs(allInputs,numInputs),numInputs*sizeof *allInputs);

回答by Andre

Try this,

尝试这个，

// Holds the expression that the user enters
#include <stdlib.h>
#include <stdio.h>
#include <string.h>

struct Inputs  { char word[10]; };

// Declare these: 
__typeof__(struct Inputs (*)[]) takeInputs(struct Inputs (*userInputs)[], 
int numInputs);  // must return a pointer to a pointer because returning array 
void printInputs(struct Inputs (*userInputs)[]);

void printInputs(struct Inputs (*userInputs)[]){ 
        struct Inputs * ptr = *userInputs; 
        for(;*ptr->word != 0; ptr++){ 
                        printf("%s", ptr->word);
        }
}

__typeof__(struct Inputs (*)[]) takeInputs(struct Inputs (*userInputs)[], int numInputs){
/*Inputs:
    userInputs: an array of struct Inputs, each of which contain a string called "word" 
    numInputs:  integer from user 
*/

int i;
for (i=0;i<numInputs;i++){
               struct Inputs * ptr = (*userInputs+i);
               printf("please input the word\t");        
               fgets(ptr->word, 10, stdin);
      }
    return userInputs;
}

int main(int argc, char const *argv[]){

//user Input should look like this:  ./takes_Input.exe 7
//if (argc!=2){
//    printf("user Input should look like this:  ./takes_Input.exe 7");
//    exit(1);
//}

// represents the number of words the user is entering 
//const char num[] = {4};
int numInputs = 7;//atoi(num);//atoi(argv[2]);

struct Inputs allInputs[numInputs];

/* this array pointer points to allInputs[] */
struct Inputs (*holdInputs)[numInputs];


holdInputs = takeInputs(&allInputs, numInputs);

printInputs(holdInputs);
return 0;
}