Using long int as a parameter for malloc - c

Question

int main() {
int n;
long u=0,d=0,count=0,i=0;
char *p=(char *)malloc(sizeof(char)*n);
scanf("%ld",&n);
scanf("%s",p);
for(i=0;i<n;i++){
if(p[i]=='U'){
u=u+1;
}
if(p[i]=='D'){
d=d+1;
}
if((d-u)==0 && p[i]=='U'){
count=count+1;}
}
printf("%ld",count);
return 0;
}
In this standard syntax for implicit memory allocation, if i replace "int n;" with "long int n;"
An error pops up saying:
GDB trace:
Reading symbols from solution...done.
[New LWP 10056]
Core was generated by `solution'.
Program terminated with signal SIGSEGV, Segmentation fault.
I have searched everywhere for a solution, rather i quite dont know what to search for,
i would be greatful if anyone helps me out. Thanks :)
(This was executed on an online compiler)

Answer 1

There are a couple of things that I would like to point out:
First of all, you do not have to declare n as "long int". "long int" and "long" are the same. So,
long int n; //is same as
long n;
malloc() works perfectly fine whether n is an "int" or a "long". However, you don't seem to have initialized n. What is the value of n? C does not perform auto-initialization of variables and n might have a garbage value (even negative) which might cause your program to crash. So please give a value to n.
long n = 10; //example
or use a scanf() to input a value.
Now in your code, what is scanf() doing "after" malloc? I presume that you intended to read a value for n and then pass it to malloc. So please change the order of code to this:
scanf("%ld",&n);
char *p=(char *)malloc(sizeof(char)*n);
I ran your program with these changes on my system and it works fine (no segmentation fault)
malloc() limits: We know that malloc allocates from a heap. But I really don't see malloc returning NULL on current platforms (which are generally 64 bit). However, if you do try to allocate a very large chunk of memory, malloc might return NULL which will cause your program to crash.
So it's good to check the return value for malloc() and if that's NULL then take appropriate actions (such as retry or exit the program)
Having a check like the one below will always help:
if (p == NULL) {
printf("Malloc error");
exit(1);
}

Answer 2

Extracting the relevant parts of your code:
int n;
char *p=(char *)malloc(sizeof(char)*n);
The parameter to malloc is of type size_t, which is an unsigned type. If you pass an argument of any other integer type, it will be implicitly converted to size_t.
You report that with int n; you don't see a problem, but with long int n; your program dies with a segmentation fault.
In either case, you're passing an uninitialized value to malloc(). Just referring to the value of an uninitialized object has undefined behavior.
It may be that the arbitrary long int value you're passing to malloc() happens to cause it to fail and return a null pointer, causing a segmentation fault later when you try to dereference the pointer; the arbitrary int value might just happen to cause malloc to succeed. Checking whether malloc succeeded or failed would likely avoid the segmentation fault.
Passing an uninitialized value to malloc() is a completely useless thing to do. The fact that it behaves differently depending on whether that uninitialized value is an int or a long int is not particularly significant.
If you're curious, you might add a line to print the value of n before calling malloc(), and you definitely should check whether malloc() reported failure by returning a null pointer. Beyond that, you know the code is incorrect. Don't waste too much time figuring out the details of how it fails (or, worse, why it sometimes doesn't fail). Just fix the code by initializing n to the number of bytes you actually want to allocate. (And define n as an object of type size_t.)
Some more points:
The code in your question is missing several required #include directives. If they're missing in your actual code, you should add them. If they're present in your actual code, you should have included them in your question. Don't make assumptions about what you can safely leave out.
int main() should be int main(void). (This is a minor point that probably doesn't make any practical difference.)
scanf("%s",p);
This is inherently dangerous. It reads a blank-delimited string that can be arbitrarily long. If the user enters more characters than the buffer p points to can hold, you have undefined behavior.
u=u+1;
Not incorrect, but more idiomatically written as u ++;.
(d-u)==0 is more clearly and safely written as d == u. (For extreme values of d and u the subtraction can overflow; an equality comparison doesn't have that problem.)