void main(char *param_1)

{
  long lVar1;
  int iVar2;
  
  lVar1 = strtol(param_1,(char **)0x0,10);
  if (1000 < lVar1 - 1U) {
  }
  iVar2 = fun9(n1,lVar1);
  if (iVar2 == 2) {
    return;
  }
}

Trying to backwards engineer but stuck on this. Some of the variables don't make sense so hard to understand what the output will be. Trying to use test cases but those aren't going through either.

CodePudding user response：

There is no real answer here. The standard specifies main as follows:

The function called at program startup is named main. The implementation declares no prototype for this function. It shall be defined with a return type of int and with no parameters:

    int main(void) { /* ... */ }

or with two parameters (referred to here as argc and argv, though any names may be used, as they are local to the function in which they are declared):

    int main(int argc, char *argv[]) { /* ... */ }

or equivalent;⁹⁾ or in some other implementation-defined manner.

Footnote 9 reads as follows:

9. Thus, int can be replaced by a typedef name defined as int, or the type of argv can be written as char ** argv, and so on.

So, simply by having void main(char *param_1), your behavior falls under that "or in some other implementation-defined manner", so the only definition of the behavior is what's documented by some particular implementation, nothing more and nothing less. Without knowing what implementation is involved, none of us can comment on it meaningfully.

For the moment, let's put that aside, and treat this as if it were some function other than main, so having a void return and one parameter has meaning.

Let's start by simplifying the code a bit.

Let's start with the first if statement:

  if (1000 < lVar1 - 1U) {
  }

There's nothing inside the {}, so execution is the same regardless. We can simply remove this without affecting execution at all.

The second if statement controls a return, but it's right before the end of the function, so it'll return either way. It has no effect either.

With those removed, we're left with:

void main(char *param_1)

{
  long lVar1;
  int iVar2;
  
  lVar1 = strtol(param_1,(char **)0x0,10);
  iVar2 = fun9(n1,lVar1);
}

You haven't shown us fun9 so we can't even begin to guess what it does. The only part we can comment on meaningfully is:

void main(char *param_1) {
    long lVar1 = strtol(param_1,(char **)0x0,10);
}

When cast to a pointer type, the 0x0 will produce a null pointer, so this is more easily interpreted as: long lvar1 = strtol(parm_1, NULL, 10). So, this attempts to interpret whatever string as passed in param_1 as a base-10 integer, and puts its value in lvar1.

But it doesn't return that value or print it out, or anything else, so there's no output--a reasonable optimizing compiler will find that this is dead code and eliminate it.

If there's any output of any sort, it's in the fun9 you haven't shown us.

CodePudding user response：

This is a C library function which is used to converts the initial parts of the string str into long int value according to the given base which must be between 2 and 36

for example #include <stdio.h> #include <stdlib.h>

int main () {
   char str[30] = "2030300 This is test";
   char *ptr;
   long ret;

   ret = strtol(str, &ptr, 10);
   printf("The number(unsigned long integer) is %ld\n", ret);
   printf("String part is |%s|", ptr);

   return(0);
}

Output The number(unsigned long integer) is 2030300 String part is | This is test|