Segmentation fault on certain inputs and not others - c

Question

Heres a function I wrote that has some debugging elements in it already. When i enter either a "y" or a "Y" as the input I get a segmentation fault during runtime. When I enter any other value the code runs. The seg fault kicks out after it scans and gives me the response but before the "scan worked" line is output. DOn't know why it would act like this only on these values. If anyone needs the function call I have that as well.
query_user(char *response [10])
{
printf("response after query call before clear=%s\n",response);
strcpy(response,"");
printf("response after clearing before scan=%s\n",response);
printf("Enter another person into the line? y or n\n");
scanf("%s", response);
printf("response after scan=%s\n",response);
printf("scan worked");
}
main()
{
char response [10];
strcpy(response,"y");
printf("response=%s\n",response);
printf("When finished with program type \"done\" to exit\n");
while (strcmp(response,"done") != 0)
{
printf("response after while loop and before query call=%s\n",response);
query_user(&response);
}
}
output on error:
response after query call before clear=y
response after clearing before scan=
Enter another person into the line? y or n
y
response after scan=y
Segmentation Fault (core dumped)
output on non-error:
response after query call before clear=y
response after clearing before scan=
Enter another person into the line? y or n
n
response after scan=n
scan worked
Cycle number 0
(program continues to run outside this function)

Answer 1

Your declaration of the parameter for query_user is wrong. You've declared an array of pointers to char. You need a simple char buffer. Like this:
query_user(char response[])
or
query_user(char* response)
Use whichever you prefer.
When you call the function, you can do it like this:
query_user(response);
Also I would point out that your declaration for main is incorrect. You should use
int main(void)

Answer 2

What you are doing is not totally secure.
What will happen if the user tape more than size of your tab ? (Segfault probably).
You really should use something like scanf("%9s", response) (You will have to empty the buffer after it !) :
(Example for tab size = 10) :
query_user(char *response)
{
printf("response after query call before clear=%s\n",response);
strcpy(response,"");
printf("response after clearing before scan=%s\n",response);
printf("Enter another person into the line? y or n\n");
scanf("%9s", response);
clean_buffer();
printf("response after scan=%s\n",response);
printf("scan worked");
}
void clean_buffer()
{
int c;
while ((c = getchar ()) != '\n' && c != EOF);
}
Hope it helps !

Answer 3

First, the code you posted doesn't reproduce the problem. I tested on Linux with Gcc 4.6.2 and on Windows with Visual Studio 2010. In both cases with your exact code my output is:
response=y
When finished with program type "done" to exit
response after while loop and before query call=y
response after query call before clear=y
response after clearing before scan=
Enter another person into the line? y or n
y
response after scan=y
scan workedresponse after while loop and before query call=y
response after query call before clear=y
response after clearing before scan=
Enter another person into the line? y or n
So to better diagnose we need a full working set of code which shows this problem. There are a lot of warnings with the posted code (not sure if those are in your base as well) and sometimes ignoring warnings can get you in a bad place:
main should return int: int main()
Put a return in your main at the end return 0;
query_user needs a void return type: void query_user()
strcpy takes a char * you're giving it a char **
For this last point you can fix it by passing the array to your function such as:
query_user(response); // remove the & operator
And your header for query_user() needs to change such as:
query_user(char response [10]) // remove the * operator
Also note that scanf will leave a '\n' character on stdin, to make sure you don't consume that you can put a single space before the %s:
scanf(" %s", response);
I don't think that any of this will make a big difference, but make sure you make the changes. Please post an update to your code that actually shows the error.