I/O - Manipulation of C Strings - c

Question

I'd like to iterate through a string (entered by the user), returning the inputted string with an added space after each character (i.e. "Hello" --> "H e l l o ".
If I preset the value for str (i.e. char str[] = "Hello";) then the desired result is printed ("H e l l o "), but not so with user input (i.e. If the user inputs "Hello" the output is "H"). How does one successfully extract and manipulate a C string based on user input?
#include <stdio.h>
#include <string.h>
int main()
{
char str[] = "";
printf("\nEnter a string: ");
scanf("%s", &str);
printf("\nYou typed: %s \n", str);
int i = 0;
char newstr[150] = "";
for (i = 0; i < strlen(str); i++)
{
newstr[2*i] = str[i];
newstr[2*i+1] = ' ';
}
newstr[2 * strlen(str)] = '\0';
printf("\nExpanded String: ");
printf("%s", newstr);
return 0;
}

Answer 1

Here:
char str[] = "";
the size of str is inferred from the initializer, which is in this case one byte large. Thus str cannot hold a string larger than one byte, and since the zero-terminator is one byte large, there is no more space for a payload. A fix is to specify a size:
char str[1024] = "";
Now str has enough space for a kilobyte of data, or 1023 characters in addition to the terminator. The size is deliberately chosen to be much larger than the input you expect.
In addition to this, it would be a good idea to prevent scanf from writing past the end of the buffer by including the size in the format string. That is
scanf("%1023s", str); // now scanf will not read more than 1023 bytes plus sentinel.
...and in turn, it would be a good idea to increase the size of newstr accordingly (to twice that of str), i.e.
char newstr[2047]; // 2 * 1023 + terminator
...or, you know, make str smaller, depending on how long a string you want to support.
Thanks to Cool Guy for catching the superfluous & and newstr size implications.

Answer 2

"How does one successfully extract and manipulate a C string based on user input?"
You can use getchar() instead.
For example, you can store the user input in an array first. Then the problem becomes the same as if you did your 'char str[] = "Hello":
int index = 0
while((temp1 = getchar())!= '\n'){
str[index++] = temp1;
}

Answer 3

the following code
-complies cleanly
-checks and handles errors
-does the job
-doesn't use unneeded memory
(well actually) the logic could be a loop
that reads one char, outputs char, outputs space
or something similar if a trailing space is a problem
then the input buffer could be a single character
#include <stdio.h>
#include <stdlib.h> // exit, EXIT_FAILURE
#include <string.h>
int main()
{
// char str[] = "";
// there actually has to be room for the string
char str[100] = {'\0'};
printf("\nEnter a string: ");
if( 1 != scanf("%s", str) ) // arrays degenerate to pointer so no extra '&' needed
{ // then scanf failed
perror( "scanf failed" );
exit( EXIT_FAILURE );
}
// implied else, scanf successful
printf("\nYou typed: %s \n", str);
// there is no need to keep the modified string in memory
// when all that will be done is print it
int i = 0; // loop counter
printf("\nExpanded String: ");
for (i = 0; i < strlen(str); i++)
{
printf("%c", str[i]);
if( i < (strlen(str)-1) )
{ // then another char will follow
printf( " " );
}
else
{
printf( "\n" );
} // end if
} // end for
return 0;
} // end function: main