How do I pass a string array from a function to main

Question

I'm trying to Initialize a string in Initialize then pass it to int main() for screen output, but it seems that the strings that are initialized have become corrupted.

Headers

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

Initialize

void
Initialize(char* STRINGs)
{
    STRINGs = malloc(sizeof(char)*5);
    STRINGs = "hello" ;

    printf("1: %s\n",STRING);
}

Main

int
main (char* STRINGs)
{
    Initialize(STRINGs);

//The program stops working when it reaches this section
    printf("2: %s",STRINGs);
    return 0;
}

Answer 1

First, you have wrong prototype for int main (char* STRINGs), which must be either:

• int main(), or
• int main( int argc, char *argv[] )

How do I pass a string array from a function to main

As it stands, you can create a string inside your Initialize() then return a pointer to that string.

There are several issues in your Initialize() though.

Here's a suggestion to change:

char * 
Initialize()
{
    char *STRINGs = malloc(strlen("hello") + 1);  // <-- malloc must include an additional space for the NULL terminator.
    strcpy( STRINGs, "hello" );  // <-- NEVER use assignment for string type.
    printf("1: %s\n",STRINGs);
    return STRINGs;
}

Then your main() can be like this:

int main()
{
    char *str = Initialize();
    printf( "str = %s\n", str );
    return 0;
}

NOTE: do not forget to add #include <string.h>

Answer 2

You can use this code to initialize the string variable

  char * Initialize()
  {
    char* STRINGs="HELLO";

     printf("1: %s\n",STRINGs);

     return STRINGs;
  }

int main ()
{

 char *strings =Initialize();

 //The program stops working when it reaches this section
 printf("2: %s",strings);
 return 0;
}

Answer 3

Here's an answer. First, when allocating memory for any variable, it must be freed or you'll get some nasty system errors at some point or at the very least, a memory leak.

In the int main(), the declaration should ideally be int main(int argc, char* argv[]).

I also recommend allocating at least one more byte of memory just in case you create a string and a function you use later on requires a null character appended to it.

I fixed your code to make it work at its bare minimum.

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

char* Initialize(){
  char* string = malloc(sizeof(char)*6);
  strcpy(string,"hello");
  printf("1: %s\n",string);
  return string;
}

int main (int argc, char* argv[]){
  char *strings=Initialize();
  printf("2: %s\n",strings);
  free(strings);
  return 0;
}

For a shorter version of your code, I suggest this:

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

int main (int argc, char* argv[]){
  char* strings=malloc(6);
  strcpy(string,"hello");
  printf("2: %s\n",strings);
  free(strings);
  return 0;
}

Answer 4

For the sake of understanding, let us suppose STRINGs of main function is a.

And STRINGs of initialize function is b.

At first, in main a is pointing to some unknown location say u. When you pass this to the initialize function then b also starts pointing to the location u.

But, after the allocation of memory, b starts pointing to some other memory that was allocated by malloc say m.

Now you change the contents of memory m by use of b. But a is still pointing to the unknown location u.

So both the pointers are now pointing towards two different memory locations. So when you print contents where b is pointing it works perfectly and then you printf contents of a which has no specific location or may be null.

So, because of this your problem came.

And also, there is another error in your program that is in the printf of initialize function, it has given a parameter STRING which is undeclared....make it STRINGs.

Hope you will like the explanation. Its a little bit tricky.

Thanks :-)

Answer 5

You can use:

void Initialize(char** STRING)

Instead:

void Initialize(char* STRINGs)

because you want to change the the address to which STRING points

Also you have wrong prototype of main

Try:

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

void Initialize(char** STRING)
{
    *STRING = malloc(6);
    strcpy(*STRING, "hello");

    printf("1: %s\n",*STRING);
}


int main (int argc, char *argv[])
{
    char* STRING;

    Initialize(&STRING);

    printf("2: %s\n",STRING);

    free(STRING);
    return 0;
}