如何从一个函数返回不同的类型 [英] How to return different types from a single function

问题描述

我有以下c代码：

  #include  
 #include< stdlib.h> 
 
 void * func（int a）{
 if（a == 3）{
 int a_int = 5; 
 int * ptr_int =& a_int; 
 return（void *）ptr_int; 
} 
 else if（a == 4）{
 char a_char ='b'; 
 char * ptr_char =& a_char; 
 return（void *）ptr_char; 
} 
 else {
 fprintf（stderr，返回值为NULL）; 
返回NULL; 
 
 
 $ b int main（int argc，char * argv []）{
 int * ptr_int =（int *）func（3）; 
 char * ptr_char =（char *）func（4）; 
 fprintf（stdout，int value =％d\\\
，* ptr_int）; 
 fprintf（stdout，char value =％c\\\
，* ptr_char）; 
返回0; 
} 
  

但是当我使用gcc来测试这段代码，我有以下结果：

  int value = 98 
 char值= 
 root @ coupure：/ home / bohao / Desktop / test1375＃gcc test.c -o test $ b $ root root @ coupure：/ home / bohao / Desktop / test1375＃./测试
 int值= 98 
 char值= 
 root @ coupure：/ home / bohao / Desktop / test1375＃gcc test.c -o test $ b $ root @ coupure：/ home / bohao / Desktop / test1375＃./test 
 int value = 98 
 char value = 
 root @ coupure：/ home / bohao / Desktop / test1375＃gcc test.c -o test 
 root @ coupure：/ home / bohao / Desktop / test1375＃./test 
 int value = 98 
 char value = 
 root @ coupure：/ home / bohao / Desktop / test1375 ＃gcc test.c -o test $ b $ root root @ coupure：/ home / bohao / Desktop / test1375＃./test 
 int value = 98 
 char value = 
 root @ coupure ：/ home / bohao / Desktop / test1375＃./test 
 int value = / home / bohao / Desktop / test1375＃gcc test.c -o test $ b $ root root @ coupure： 98 
 char值= g 
 root @ coupure：/ home / bohao / Desktop / test1375＃gcc test.c -o test $ b $ root root @ coupure：/ home / bohao / Desktop / test1375＃ ./test 
 int value = 98 
 char value = 
 root @ coupure：/ home / bohao / Desktop / test1375＃gcc test.c -o test 
 root @ coupure： / home / bohao / Desktop / test1375＃./test 
 int value = 98 
 char value =！ 
  

为什么我要 98 对于ptr_int和ptr_char的随机值？

我想要的是通用函数，它们可以返回不同类型的值而不是使用两个函数。那可能吗 ？如果是这样，该怎么办？

解决方案

问题就在我看到的时候，您正试图 return 函数（范围）中的局部变量的地址，并试图从调用者访问返回的内存。在调用者中，内存无效，任何用法都会导致未定义的行为。
$ b

解决方案：您需要为指针（ malloc（） / calloc（）），您希望返回来自函数。这将解决这个问题，因为动态分配的内存的生命周期是手动或直到程序终止，直到更早的时候，直到 free（） -d。

话虽如此，这种方法并不好。如果你想返回多种类型中的一种，请为包含所有类型成员的 struct >使用标志来标记类型。填充相应的成员变量，设置标志和 return 结构变量。为了更好，您可以实际使用 union 作为结构的成员，因为您一次只需要一种类型。有关工作代码，请参阅 @pmg 的其他答案。

I have the following c code :

#include <stdio.h>
#include <stdlib.h>

void *func(int a) { 
    if (a==3) {
        int a_int = 5;
        int *ptr_int = &a_int;
        return (void *)ptr_int;
    } 
    else if (a==4) {
        char a_char = 'b';
        char *ptr_char = &a_char;
        return (void *)ptr_char;
    }
    else {
        fprintf(stderr, "return value is NULL");
        return NULL;
    }
}

int main (int argc, char *argv[]) {
    int *ptr_int = (int *)func(3);
    char *ptr_char = (char *)func(4);
    fprintf(stdout, "int value = %d\n", *ptr_int);
    fprintf(stdout, "char value = %c\n", *ptr_char);
    return 0; 
}

But when I use gcc to test this code, I have the following results :

int value = 98
char value = �
root@coupure:/home/bohao/Desktop/test1375# gcc test.c -o test
root@coupure:/home/bohao/Desktop/test1375# ./test 
int value = 98
char value = 
root@coupure:/home/bohao/Desktop/test1375# gcc test.c -o test
root@coupure:/home/bohao/Desktop/test1375# ./test 
int value = 98
char value = 
root@coupure:/home/bohao/Desktop/test1375# gcc test.c -o test
root@coupure:/home/bohao/Desktop/test1375# ./test 
int value = 98
char value = 
root@coupure:/home/bohao/Desktop/test1375# gcc test.c -o test
root@coupure:/home/bohao/Desktop/test1375# ./test 
int value = 98
char value = 
root@coupure:/home/bohao/Desktop/test1375# gcc test.c -o test
root@coupure:/home/bohao/Desktop/test1375# ./test 
int value = 98
char value = g
root@coupure:/home/bohao/Desktop/test1375# gcc test.c -o test
root@coupure:/home/bohao/Desktop/test1375# ./test 
int value = 98
char value = 
root@coupure:/home/bohao/Desktop/test1375# gcc test.c -o test
root@coupure:/home/bohao/Desktop/test1375# ./test 
int value = 98
char value = !

Why do I have the 98 for ptr_int and the random value of ptr_char ?
What I want is to have a general function who can return different types of values instead of using two functions. Is that possible ? If so, how to do it ?

解决方案

The problem as I see it, you're trying to return the address of a local variable from a function (scope) and trying to access the returned memory from the caller. In the caller, the memory is invalid and any usage will lead to undefined behavior.

Solution: You need to use dynamic memory allocation for the pointer (malloc()/ calloc()) which you want to return from the function. This will overcome the issue here, as the lifetime of the dynamically allocated memory is untill free()-d manually or till program termination, whichever is earlier.

Having said that, this approach is not a good one. If all you want to return one of multiple types, go for a struct containing members for all types and use a flag to mark the type. Fill the corresponding member variable, set the flag and return the structure variable.

For better, you can actually use an union as a member of the structure, as you only need one type at a time. For a working code, please refer to the other answer by @pmg.

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