Scanf input for char works once then is blank - c

Question

I want to reference the input of name throughout the code I am writing, but for some reason after I use it successfully in the first printf(), the second printf() does not print the name.
int main()
{
char name[50];
char q1[1];
printf( " What is your name?\n");
scanf("%s", name);
printf( " Hi %s, Do you want to have some fun? [Y/N]\n",name);
scanf("%s",q1);
if(strcmp(q1,"Y") == 0||strcmp(q1,"y")==0)
{
printf("Awesome, let's play!\n");
}
else
{
printf("Fine"); goto endgame;
}
printf( "So %s, it's time to get started\n", name);
endgame:
getchar();
return 0;
}
The output for the entry 'Nick' is:
Hi Nick, do you want to have some fun? Awesome let's play So ,
let's get started
I would expect it to say:
So Nick, let's get started.
but for some reason char name is blank after it is used correctly the first time.

Answer 1

The problem, (as I assumed correctly) is with char q1[1]; and then using it like
scanf("%s",q1);
It is causing memory boundary overrun, because, a one-char array is not sufficient to hold a string of having only one element, as it lacks the space for the null-terminator required for the string. This invokes undefined behaviour.
Instead,
change char q1[1]; to char q1;
change scanf("%s",q1); to scanf(" %c", &q1);
change if(strcmp(q1,"Y") == 0||strcmp(q1,"y")==0) to if((q1 =='Y') || q1 == 'y')
That said, as a note,
The recommended signature of main() is int main(void).
To avoid the possibility of buffer overflow by longer input(s), it's better to limit the length of the input with scanf() by writing like
scanf("%49s", name);

Answer 2

Expand the size of your q1 buffer. scanf("%s", q1) doesn't have enough room to store the input. Remember that C uses a null character '\0' to terminate strings. If you don't account for that, the buffer could overrun into other memory causing undefined behavior. In this instance, it's probably overwriting memory allocated to name, so name ends up pointing to "\0ick". This causes printf(%s), which looks for '\0' to know when to stop printing, to think that the string is shorter than it really is.
The code works perfectly if you expand the buffer:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(void)
{
char name[50];
char q1[50];
printf( " What is your name?\n");
scanf("%49s", name);
printf( " Hi %s, Do you want to have some fun? [Y/N]\n",name);
scanf("%49s",q1);
if(strcmp(q1,"Y") == 0||strcmp(q1,"y")==0)
{
printf("Awesome, let's play!\n");
}
else
{
printf("Fine");
}
printf( "So %s, it's time to get started\n", name);
getchar();
return 0;
}
Output:
What is your name?
Nick
Hi Nick, Do you want to have some fun? [Y/N]
y
Awesome, let's play!
So Nick, it's time to get started
Note that I've added the qualifier %49s to avoid buffer overruns like this.
You could also circumvent the need for another string entirely by changing char q1[50] and scanf("%49s") to simply char q1 and scanf("%c%*c", &q1) (note the "address of" operator because q1 is no longer a pointer).
You'll probably even get a performance gain from this (albeit small), because strings are notorious memory hoggers. Comparing a single character is usually preferred over comparing strings.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(void)
{
char name[50];
char q1;
printf( " What is your name?\n");
scanf("%49s%*c", name);
printf( " Hi %s, Do you want to have some fun? [Y/N]\n",name);
scanf("%c%*c",&q1);
if(q1 == 'Y' || q1 == 'y')
{
printf("Awesome, let's play!\n");
}
else
{
printf("Fine");
}
printf( "So %s, it's time to get started\n", name);
getchar();
return 0;
}
if(q1 == 'Y' || q1 == 'y')
{
printf("Awesome, let's play!\n");
}
else
{
printf("Fine");
}
printf( "So %s, it's time to get started\n", name);
getchar();
return 0;
}
If you go this route, you have to ignore the enter key using the format specifier %*c because pressing enter sends a key to the stream as well.