从函数获取const char *数组

Question

Need help in getting const char* array from function so elements can be printed in main.

main:

const char* values[3];
strings_to_array();

printf("%s\n", values[1]);
printf("%s\n", values[2]);

function:

const char* strings_to_array()
{
    char one_str[16];
    char two_str[16];
    char three_str[16];

    strcpy(one_str, "one");
    strcpy(two_str, "two");
    strcpy(three_str, "three");

    const char* values[] = {one_str, two_str, three_str};
    return values;
}

What is incorrect here and how to get values to main?

Answer 1

The main problem with this code is that: functions in C should not return pointers to local variables as they are stored on the stack, which means they are not available once the function returns.

So this line:

const char* values[] = {one_str, two_str, three_str};

Can be replaced with:

const char** values = malloc(3*sizeof(char *));
values[0] = strdup(one_str);
values[1] = strdup(two_str);
values[2] = strdup(three_str);

The full working code of the above example:

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

const char** strings_to_array()
{
    char one_str[16];
    char two_str[16];
    char three_str[16];

    strcpy(one_str, "one");
    strcpy(two_str, "two");
    strcpy(three_str, "three");

    const char** values = malloc(3*sizeof(char *));
    values[0] = strdup(one_str);
    values[1] = strdup(two_str);
    values[2] = strdup(three_str);
    return values;
}

int main() {
  const char** values = strings_to_array();

  printf("%s\n", values[1]);
  printf("%s\n", values[2]);

  free((void *)values[0]);
  free((void *)values[1]);
  free((void *)values[2]);
  free(values);      

  return 0;
}

Answer 2

1. It is wrong syntactically because you declared return type as const char* and you are trying to return const char** .

2. It is wrong semantically because you are trying to return a pointer to array allocated on stack.

Answer 3

Nothing initializes the array of character pointers named "values" in main. What is that function supposed to do? It does nothing, and its return value is ignored in main.

Answer 4

The variables one_str, two_str, three_str, and values are local variables of the 'strings_to_array' function. They have no valid existence outside this function.

By returning 'values' (ie: a pointer to a local variable), you return a pointer to an invalid memory location.

Answer 5

One way is to declare your variables as static since you are going to access them outside of that function. It just means that memory for those are not lost once you exit the function. Also, you should be returning a char ** instead of char * .

const char** strings_to_array()
{
    static char one_str[16];
    static char two_str[16];
    static char three_str[16];

    strcpy(one_str, "one");
    strcpy(two_str, "two");
    strcpy(three_str, "three");

    static const char* values[] = {one_str, two_str, three_str};
    return values;
}

Answer 6

As you are dealing with literals ( "one" , ...) you could simply assign their addresses.

main:

void string_to_array(const char **);

const char * values[3] = {NULL};
strings_to_array(values);

printf("%s\n", values[0]);    
printf("%s\n", values[1]);
printf("%s\n", values[2]);

function:

void strings_to_array(const char ** values)
{
  values[0] = "one";
  values[1] = "two";
  values[2] = "three";
}

And if you really need it as a function returning a reference to the array do it this way:

const char ** strings_to_array2(const char ** values)
{
  values[0] = "one";
  values[1] = "two";
  values[2] = "three";

  return values;
}

This keeps you from doing the ugly call to free() on const char * s.