如何返回变量？

我正在编写一个程序，将字母翻译成摩尔斯电码，然后将它们传输到 LED 并闪烁。我无法返回值“.- -...”

#include <stdio.h>
#include <string.h>

char *encode()
{
  char *name = "AB";
  char name_m[100] = "";

  for (int i = 0; i < strlen(name); i++)
  {
    if (name[i] == 65)
    {
      strcat(name_m, ".- ");
    }
    else if (name[i] == 66)
    {
      strcat(name_m, "-...");
    }
  };
  printf("%s", name_m);
  
  return name_m;
}

int main()
{
    char *name_m;
    name_m = encode();
    printf("\n%s", name_m);
}

main.c:22:10: warning: function returns address of local variable [-Wreturn-local-addr]
   22 |   return name_m;
      |          ^~~~~~
.- -...
(null)

在 C 中，您无法返回对局部变量的引用（指针），因为当函数返回时它不再存在。

您需要将 te 缓冲区传递给该函数：

char *encode(char *name_m, const char *name)
{
  for (int i = 0; i < strlen(name); i++)
  {
    if (name[i] == 'A')
    {
      strcat(name_m, ".- ");
    }
    else if (name[i] == 'B')
    {
      strcat(name_m, "-...");
    }
  }
  printf("%s", name_m);
  
  return name_m;
}

动态分配它：

char *encode( const char *name)
{
  char *name_m = malloc(100);

  if(name_m)
  {
    for (int i = 0; i < strlen(name); i++)
    {
        if (name[i] == 'A')
        {
        strcat(name_m, ".- ");
        }
        else if (name[i] == 'B')
        {
        strcat(name_m, "-...");
        }
    }
    printf("%s", name_m);
  }
  
  return name_m;
}

或最糟糕的解决方案 - 将其定义为静态

char *encode( const char *name)
{
  static char name_m[100];

  if(name_m)
  {
    for (int i = 0; i < strlen(name); i++)
    {
        if (name[i] == 'A')
        {
        strcat(name_m, ".- ");
        }
        else if (name[i] == 'B')
        {
        strcat(name_m, "-...");
        }
    }
    printf("%s", name_m);
  }
  
  return name_m;
}