I am making a basic program, and decided to use functions, and pointers, in the first input I have the choice of typing either, "kitchen", or "upstairs", however when I use fgets()
I am getting a segmentation fault, and have no idea why. I tried to print out the string to see if I was getting the correct output, of course due to the segmentation fault it's not happening.
Here's the code:
#include <stdio.h>
#include <string.h>
char one(char *oOne[]);
void two();
void three();
int main() {
char *oOne[10], oTwo[10], oThree[10]; // Options one, two, and three.
puts("WELCOME TO ASAD'S ADVENTURE");
puts("");
one(oOne);
return 0;
}
char one(char *oOne[]) {
puts("You are in a creepy house! Would you like to go \"upstairs\", or into the \"kitchen\"?");
printf("\n> ");
if (fgets(*oOne, sizeof *oOne, stdin) == NULL) { // Receiving the input from the user, and removing the \n caused by fgets().
puts("EOF Occurred");
return 1;
}
*oOne[strcspn(*oOne, "\n")] = 0;
printf("%s\n", *oOne);
// puts(*oOne);
}
Here is the output I am receiving:
WELCOME TO ASAD'S ADVENTURE
You are in a creepy house! Would you like to go "upstairs", or into the "kitchen"?
> kitchen
Segmentation fault
How can I fix the segmentation fault?
A character string in C is of type pointer-to-character, e.g.,:
char *oOne;
Or an array-of-character, if you want to statically allocate your memory:
char oOne[10];
But not a pointer-to-array-of-character, which is what you have:
char *oOne[10];
You need:
char oOne[10];
And then:
one(oOne);
And then modify your one
function with proper types:
char one(char *oOne)
{
if (fgets(oOne, sizeof oOne, stdin) == NULL)
{
puts("EOF Occurred");
return 1;
}
*(oOne + strcspn(oOne, "\n")) = 0;
printf("%s\n", oOne);
}
Although I would pass in an explicit length, rather than using
sizeof
, because this won't work with dynamically allocated strings:
car one(char *oOne, int len)
{
if (fgets(oOne, len, stdin) == NULL)
{
puts("EOF Occurred");
return 1;
}
// etc...
}