如何调用 function 作为 strcpy 中的第二个字符串？

Question

I need to call a function from the main to determine a string value. However, as far as I know, string cannot be in the form string_name = call_function() . The assignment of string has to be in form of strcpy(str1, str2) right? So I am wondering what am I doing wrong with this strcpy with str1 as the variable name and the str2 is a return string value from another function.

So far, this is what I have.

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

char *get_name(int num);

char *get_name(int num) {
    char real_name[30];
    
    if (num == 1)
        strcpy(real_name, "Jake Peralta");
    
    return real_name;
}

void main() {
    char name[30];
    int num;
    
    num = 1;
    
    strcpy(name, *get_name(num));
    printf("%s", name);
}

It is not printing anything on my output screen.

What I have tried:

• get_name(num) without using pointers (still does not work)

p/s: This is not the actual code. This is just an example of what I am trying to do, the actual code is longer and I have identified that this part is where the error is coming from.

Answer 1

You are correct, a string cannot be returned as an array via the return statement. When you pass or return an array, only a pointer to its first element is used. Hence the function get_name() returns a pointer to the array defined locally with automatic storage (aka on the stack ). This is incorrect as this array is discarded as soon as it goes out of scope, ie: when the function returns.

There are several ways for get_name() to provide the name to its caller:

• you can pass a destination array and its length: and let the function fill the name in the array, carefully avoiding to write beyond the end of the array but making sure it has a null terminator:

   char *get_name(char *dest, size_t size, int num) { if (num == 1) { snprintf(dest, size, "Jake Peralta"); } else { snprintf(dest, size, "John Doe"); } // return the destination pointer for convenience. return dest; } int main() { char name[30]; int num = 1; get_name(name, sizeof name, num); printf("%s\n", name); return 0; }

• you can allocate memory in get_name() and return a pointer to the allocated array where you copy the string. It will be the caller's responsibility to free this object with free() when it is no longer used.

   char *get_name(int num) { if (num == 1) { return strdup("Jake Peralta"); } else { return strdup("John Doe"); } } int main() { int num = 1; char *name = get_name(num); printf("%s\n", name); free(name); return 0; }

• you can return a constant string, but you can only do this if all names are known at compile time.

   const char *get_name(int num) { if (num == 1) { "Jake Peralta"; } else { "John Doe"; } } int main() { int num = 1; const char *name = get_name(num); printf("%s\n", name); return 0; }

Answer 2

You are returning address of real_name from get_name method which will go out of scope after the function returns. Instead allocate the memory of string on heap and return its address. Also, the caller would need to free the string memory allocated on the heap to avoid any memory leaks.

Answer 3

Like you said, "However, as far as I know, string cannot be in the form string_name = call_function() ." To understand the logic behind this, just look at your get_name() function:

char *get_name(int num)
{
    char real_name[30];
    
    if (num==1)
        strcpy(real_name,"Jake Peralta");
    
    return real_name;
}

Here, you try to return the starting address of real_name , but real_name is destroyed when it goes out of scope (in this case, when the function returns). There are two ways I think you could fix this. One is to add the string that should hold the return value as a parameter. In your case, it would be:

void get_name(int num, char *dest)
{
    char real_name[30];
    
    if (num==1)
    {
        strcpy(real_name,"Jake Peralta");
        strcpy(dest, real_name);
    }
}

Or, just avoid using real_name at all now to make the function shorter:

void get_name(int num, char *dest)
{
    if (num==1)
        strcpy(dest, "Jake Peralta");
}

The other way is to allocate the return value on the heap, but I wouldn't recommend this; you would have to keep track of each allocated string and eventually free all of them. Nevertheless, here is how it would look like in your case:

char *get_name(int num, char *dest)
{
    char *real_name = calloc(30, sizeof(char)); // Using calloc to avoid returning an uninitialized string if num is not 1
    if (num==1)
        strcpy(real_name, "Jake Peralta";
    return real_name;
}

By the way, I kept your strcpy here, but might want to avoid using it in the future, as it can cause buffer overruns. Here's a post with useful answers as to why it's bad . In fact, even strncpy isn't fully safe (See here (credit to @chqrlie for providing the link)). @chqrlie provides a clean and safe alternative using snprintf in his own answer, I'd suggest you use that.