Getting no output why is that?

Question

I', learning C and I'm getting no output for some reason, probably I don't return as I should but how I should? (described the problem in the comments below :D) Any help is appreciated!

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

char *makeUpperCase (char *string);

int main()
{
    printf(makeUpperCase("hello")); //Here there is no output, and when I'm trying with the format %s it returns null
    return 0;
}

char *makeUpperCase(char *string)
{
    char str_out[strlen(string) + 1];
    for (int i = 0; i < strlen(string); ++i)
        str_out[i] = toupper(string[i]);

    printf(str_out); //Here I get the output.
    return str_out;
}

Answer 1

You declared within the function a local variable length array that will not be alive after exiting the function

char str_out[strlen(string) + 1];

So your program has undefined behavior.

If the function parameter declared without the qualifier const then it means that the function changes the passed string in place. Such a function can be defined the following way

char * makeUpperCase( char *string )
{
    for ( char *p = string; *p != '\0'; ++p )
    {
        *p = toupper( ( unsigned char )*p );
    }

    return string;
}

Otherwise you need to allocate dynamically a new string. For example

char * makeUpperCase( const char *string )
{
    char *str_out = malloc( strlen( string ) + 1 );

    if ( str_out != NULL )
    { 
        char *p = str_out;

        for ( ; *string != '\0'; ++string )
        {
            *p++ = toupper( ( unsigned char )*string );
        }

        *p = '\0';
    }

    return str_out;
}

Here is a demonstration program.

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

char *makeUpperCase( const char *string )
{
    char *str_out = malloc( strlen( string ) + 1 );

    if (str_out != NULL)
    {
        char *p = str_out;

        for (; *string != '\0'; ++string)
        {
            *p++ = toupper( ( unsigned char )*string );
        }

        *p = '\0';
    }

    return str_out;
}

int main( void )
{
    char *p = makeUpperCase( "hello" );

    puts( p );

    free( p );
}

The program output is

HELLO

Answer 2

The problem is that printf() is buffering output based on a bit complex mechanism. When you are outputting to a terminal, printf() just buffers everything until the buffer fills (which is not going to happen with just the string "hello", or until it receives a '\n' character (which you have not used in your statement)

So, to force a buffer flush, just add the following statement

    fflush(stdout);

after your printf() call.