How to write to pointer directly using gets() function? - c

Question

I got these test code and just curious, why passing pointer to gets() results in a runtime error?
void main()
{
char *value="gogo";
puts(value);
value="11";
puts(value);
gets(value);
}

Answer 1

Because char *value="gogo"; is more likely than not allocated to READ ONLY MEMORY!
Better:
#include <stdio.h>
#include <string.h>
#define MAX_LINE 80
int main()
{
char value[MAX_LINE] ="gogo";
puts(value);
strcpy (value, "11");
puts(value);
fgets(value, MAX_LINE, stdin);
puts(value);
return 0;
}
Here is a good link with more details: Storage for Strings in C
PS:
gets() is Evil. Avoid it if at all possible: Why gets() is bad

Answer 2

The pointer value points to (the static array associated with) a string literal.
Attempting to modify a string literal has undefined behavior. In this case, your compiler stores the string "gogo" in memory marked as read-only by the operating system, and attempting to modify it causes your program to crash.
If you declare a pointer to a string literal, you should define it as const:
const char *value = "gogo";
so the compiler will diagnose any attempt to modify it. Or, if you really want to modify the string, define it as an array:
char value[] = "gogo";
which the means that you can't assign a value to value, but you can use strcpy to update it.
Some more problems:
void main() is wrong [*]; the correct definition is int main(void). If you're using a book that told you to use void main(), please get a better one; its author does not know C very well.
Never use the gets function; it is inherently unsafe, and has been removed from the language. (It cannot guard against input longer than the array into which the value is stored.) You can use fgets instead; it's a bit more complicated to use, but it can be used safely.
You need to add
#include <stdio.h>
to the top of your source file to make these functions visible. If your compiler didn't complain about calls to undeclared functions, find out how to increase its warning level.
[*] Saying that void main() is wrong slightly overstates the case. A conforming compiler may permit it, and no compiler is required to complain about it, but there is no good reason to take advantage of that. int main(void) is always correct. Any C book or tutorial that advocates using void main() was almost certainly written by someone who does not know C well enough to be writing books or tutorials about it.

Answer 3

Here, your pointer points to a string literal ("gogo"). String literals are not guaranteed to be writable. You need to allocate your own memory:
char value[50] = "gogo";
...
gets(value);
However, this is not safe, as gets does not take the size of the buffer, and thus might overflow your buffer. (Which could also lead to a runtime error). NEVER use gets, as the manpage states:
BUGS
Never use gets(). Because it is impossible to tell without knowing the data in advance how many characters gets() will read, and because gets() will continue to store characters past the end of the buffer, it is extremely dangerous to use. It has been used to break computer security. Use fgets() instead.
Much better to allocate your own memory:
char user_input[200];
fgets(user_input, 200, stdin);
You might need to check user_input, to see if it ends with a newline. If it does, then fgets read a whole line. If it doesn't, then the the user typed more than ~200 characters into the line, and you'll need to read more to get the whole line. I've used 200 here. Choose a size that makes sense for your data. You can also use malloc to allocate memory on-the-fly, and put fgets into a loop in order to read an entire line into a buffer.