C语言 如何从 C 中的函数返回字符数组？
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How can I return a character array from a function in C?
提问by Jonathon Reinhart
is that even possible? Let's say that I want to return an array of two characters
这甚至可能吗？假设我想返回一个包含两个字符的数组
char arr[2];
arr[0] = 'c';
arr[1] = 'a';
from a function. What type do I even use for the function? Is my only choice to use pointers and void the function? So far I've tried having a char* function or a char[]. Apparently you can only have functions of char(*[]). The only reason I want to avoid using pointers is the fact that the function has to end when it encounters a "return something;" because the value of "something" is a character array (not a string!) that might change size depending on the values I pass into the function through the main function. Thanks to anyone who responds in advance.
从一个函数。我什至使用什么类型的功能？我唯一的选择是使用指针并使函数无效吗？到目前为止，我已经尝试过使用 char* 函数或 char[]。显然你只能有 char(*[]) 的函数。我想避免使用指针的唯一原因是函数在遇到“返回内容”时必须结束；因为“某物”的值是一个字符数组（不是字符串！），它可能会根据我通过主函数传递给函数的值改变大小。感谢任何提前回复的人。
回答by Jonathon Reinhart
You've got several options:
你有几个选择：
1)Allocate your array on the heapusing malloc(), and return a pointer to it. You'll also need to keep track of the length yourself:
1)使用在堆上分配数组malloc()，并返回指向它的指针。您还需要自己跟踪长度：
void give_me_some_chars(char **arr, size_t *arr_len)
{
/* This function knows the array will be of length 2 */
char *result = malloc(2);
if (result) {
result[0] = 'c';
result[1] = 'a';
}
/* Set output parameters */
*arr = result;
*arr_len = 2;
}
void test(void)
{
char *ar;
size_t ar_len;
int i;
give_me_some_chars(&ar, &ar_len);
if (ar) {
printf("Array:\n");
for (i=0; i<ar_len; i++) {
printf(" [%d] = %c\n", i, ar[i]);
}
free(ar);
}
}
2)Allocate space for the array on the stackof the caller, and let the called function populate it:
2)在调用者的堆栈上为数组分配空间，并让被调用的函数填充它：
#define ARRAY_LEN(x) (sizeof(x) / sizeof(x[0]))
/* Returns the number of items populated, or -1 if not enough space */
int give_me_some_chars(char *arr, int arr_len)
{
if (arr_len < 2)
return -1;
arr[0] = 'c';
arr[1] = 'a';
return 2;
}
void test(void)
{
char ar[2];
int num_items;
num_items = give_me_some_chars(ar, ARRAY_LEN(ar));
printf("Array:\n");
for (i=0; i<num_items; i++) {
printf(" [%d] = %c\n", i, ar[i]);
}
}
DO NOT TRY TO DO THIS
不要尝试这样做
char* bad_bad_bad_bad(void)
{
char result[2]; /* This is allocated on the stack of this function
and is no longer valid after this function returns */
result[0] = 'c';
result[1] = 'a';
return result; /* BAD! */
}
void test(void)
{
char *arr = bad_bad_bad_bad();
/* arr is an invalid pointer! */
}
回答by Iharob Al Asimi
You can pass the array to the function and let the function modify it, like this
您可以将数组传递给函数并让函数修改它，就像这样
void function(char *array)
{
array[0] = 'c';
array[1] = 'a';
}
and then
进而
char array[2];
function(array);
printf("%c%c\n", array[0], array[1]);
If you want it as a return value, you should use dynamic memroy allocation,
如果你想要它作为返回值，你应该使用动态内存分配，
char *function(void)
{
char *array;
array = malloc(2);
if (array == NULL)
return NULL;
array[0] = 'c';
array[1] = 'a';
return array;
}
then
然后
char *array = function();
printf("%c%c\n", array[0], array[1]);
/* done using `array' so free it because you `malloc'ed it*/
free(array);
Important Note:
重要说明：
You should be aware of the fact that the array as filled above is not a string, so you can't for instance do this
您应该知道上面填充的数组不是字符串，因此您不能这样做
printf("%s\n", array);
because the "%s"expects a matching string to be passed, and in c an array is not a string unless it's last character is '\0', so for a 2 character string you need to allocate space for 3 characters and set the last one to '\0'.
因为"%s"期望传递匹配的字符串，并且在 c 中数组不是字符串，除非它的最后一个字符是'\0'，所以对于 2 个字符的字符串，您需要为 3 个字符分配空间并将最后一个设置为'\0'.
回答by DSLima90
You can return a pointer for the array from a function, however you can't return pointers to local arrays, the reference will be lost.
您可以从函数返回数组的指针，但不能返回指向本地数组的指针，引用将丢失。
So you have 3 options:
所以你有3个选择：
Use a global variable:
char arr[2]; char * my_func(void){ arr[0] = 'c'; arr[1] = 'a'; return arr; }Use dinamic allocation (the caller will have the responsability to free the pointer after using it, make it clear in your documentation)
char * my_func(void){ char *arr; arr = malloc(2); arr[0] = 'c'; arr[1] = 'a'; return arr; }Make the caller allocate the array and use as a reference (my recomendation)
void my_func(char * arr){ arr[0] = 'c'; arr[1] = 'a'; }
使用全局变量：
char arr[2]; char * my_func(void){ arr[0] = 'c'; arr[1] = 'a'; return arr; }使用动态分配（调用者将有责任在使用后释放指针，请在文档中明确说明）
char * my_func(void){ char *arr; arr = malloc(2); arr[0] = 'c'; arr[1] = 'a'; return arr; }让调用者分配数组并用作参考（我的推荐）
void my_func(char * arr){ arr[0] = 'c'; arr[1] = 'a'; }
if you realy need the function to return the array, you can return the same reference as:
如果您确实需要该函数返回数组，则可以返回与以下相同的引用：
char * my_func(char * arr){
arr[0] = 'c';
arr[1] = 'a';
return arr;
}
回答by this
Since you have a predetermined size of you array you can in-fact return the array if you wrap it with a struct:
由于您有预先确定的数组大小，如果您用结构包装它，实际上可以返回数组：
struct wrap
{
char a[2] ;
} ;
struct wrap Get( void )
{
struct wrap w = { 0 } ;
w.a[0] = 'c';
w.a[1] = 'a';
return w ;
}
回答by Dan Korn
char* getCharArray()
{
return "ca";
}
回答by Elephant
This works perfecly:
这完美地工作：
int comm_read_data(int file_i2c, unsigned char** buffer)
{
*buffer = malloc(BUFFER_SIZE);
if (i2c_read_bytes(file_i2c, *buffer, BUFFER_SIZE) != 0)
{
return -1;
}
return BUFFER_SIZE;
}
And then call the function:
然后调用函数：
unsigned char* buffer;
int length = comm_read_data(file_i2c, &buffer);
/* parse the buffer here */
free(buffer);
