Why is strlen causing a segmentation fault in C? - c

Question

(Warning) Yes this is a part of an assignment I am working on, but I am completely desperate at this point and NO I am not looking for you guys to solve it for me, but any hint would be much appreciated!(/Warning)
I am pretty much trying to make an interactive menu, the user is meant to input an expression (For example "5 3 +") and the program should detect that it's in postfix notation, unfortunately I have been getting segmentation fault errors and I suspect they have something to do with the use of the strlen function.
EDIT: I was able to make it work, first the char expression[25] = {NULL}; line
becomes char expression[25] = {'\0'};
And when calling the determine_notation function I removed the [25] from the array I am passing like so:
determine_notation(expression, expr_length);
Also the input[length] part I changed to input[length-2] since like mentioned in a previous comment, input[length] == '\0' and input[length--] == '\n'.
All in all thanks for all the help!
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int determine_notation(char input[25], int length);
int main(void)
{
char expression[25] = {NULL}; // Initializing character array to NULL
int notation;
int expr_length;
printf("Please enter your expression to detect and convert it's notation: ");
fgets( expression, 25, stdin );
expr_length = strlen(expression[25]); // Determining size of array input until the NULL terminator
notation = determine_notation( expression[25], expr_length );
printf("%d\n", notation);
}
int determine_notation(char input[25], int length) // Determines notation
{
if(isdigit(input[0]) == 0)
{
printf("This is a prefix expression\n");
return 0;
}
else if(isdigit(input[length]) == 0)
{
printf("This is a postfix expression\n");
return 1;
}
else
{
printf("This is an infix expression\n");
return 2;
}
}

Answer 1

You probably got a warning explaining that you are converting a char to a pointer in this call:
expr_length = strlen(expression[25]);
// ^^^^
This is the problem - your code is referencing a non-existent element past the end of the array (an undefined behavior) and tries to pass it to strlen.
Since strlen takes a pointer to the beginning of the string, the call needs to be
expr_length = strlen(expression); // Determining size of array input until the NULL terminator