函数，字符串作为C中的返回类型

Question

I needed a function to return a string. I used the following code to declare the function:

const char* serv_con(char app_data[50])
{
    char send_data[1024],recv_data[1024];
    //i am avoiding code segments irrelevant to the issue.
    return recv_data;
}

and then called the function in the main like this:

int main()
{
    char ser_data[50], app_data[50];
    ser_data[0] = '\0';
    app_data[0] = '\0';
    //avoiding code segments irrelevant to the issue.
    app_data = serv_con(ser_data); //function call
}

On compiling it gives the error:

connect.c:109: error: incompatible types when assigning to type ‘char[50]’ from type ‘const char *’

Then i replaced the const char in the declaration with std::string. The declaration now is as follows:

std::string serv_con(char app_data[50])
{
    char send_data[1024],recv_data[1024];
    //avoiding code segments irrelevant to the issue.
    return recv_data;
}

And called it in the same way as mentioned above. But still it gives the following error on compiling:

connect.c:13: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘:’ token

Please tell me how I can give a string as a return type from a function. The platform I work on is linux. Thanks in advance.

Answer 1

const char* serv_con(char app_data[50])
{
  char send_data[1024],recv_data[1024];
  //i am avoiding code segments irrelevant to the issue.
  return recv_data;
}

this cannot work, since you're returning a pointer to a local variable, which is invalid after returning. You need to allocate recv_data on the heap in order to use it after returning

char* serv_con(char app_data[50])
{
  char send_data[1024];
  char *recv_data = malloc(1024);
  if (!recv_data)
      return NULL;

  // ...
  return recv_data;
 }

then change the main function to be something like

int main()
{
 char ser_data[50];
 char *app_data;
 ser_data[0] = '\0';
 //avoiding code segments irrelevant to the issue.
 app_data = serv_con(ser_data); //function call
 if (!app_data) {
   // error
 }
}

Answer 2

What you are doing is a really bad idea. The serv_con function allocates some space on the stack for the recv_data array and then returns a pointer to that location. Of course, as soon as you call another function that data will be wiped out, leading to a bug that is difficult to diagnose. If you must return a block of memory from a function, allocate it using malloc . Basically, never return pointers to objects on the stack.

When you get the pointer out of serv_con you assign it to app_data , which already has some space allocated. There is no reason for doing this: simply declare app_data as a char * that will point to the storage allocated in serv_con .

Answer 3

Returning pointer to local array or the array itself is a bad idea.

Even if you do malloc inside the function, the caller needs to free that memory too which is again not a very good practice.

I suggest the best way is to pass an out parameter ie pass the array in which you want the result to be copied as a parameter.

int serv_con(char app_data[50], char recv_data[50])
{
    char send_data[1024];
     //recv_data[1024];
    //i am avoiding code segments irrelevant to the issue.
    return 1; // return status success or failure which can be tested in main
}

and then called the function in the main like this:

int main()
{
    char ser_data[50], app_data[50];
    ser_data[0] = '\0';
    app_data[0] = '\0';
    //avoiding code segments irrelevant to the issue.

    serv_con(ser_data, app_data); //function call
}

I found and corrected a slight mistake. The app_data you are passing in main is of size 50 where as the one you were trying to return from the function serv_con was 1024. This should be consistent. I have used both of size 50 in code above.

Answer 4

From your question I can be pretty sure that you are a beginner. Please go through a good text book ie The C Programming Language .

Along with the error you must be getting this.
warning: function returns address of local variable.

Look at the warnings always. The question is already answered by Greg .