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parse error at..., parse error before...

This is a general-purpose syntax error. It is frequently caused by a missing semicolon. For example, the following code:

#include <stdio.h>

/* To shorten example, not using argp */
int main()
{
  printf ("Hello, world!\n")
  return 0;
}

generates the following error:

semicolon.c: In function `main':
semicolon.c:6: parse error before `return'

Adding a semicolon (;) at the end of the line printf ("Hello, world!") will get rid of this error.

Notice that the error refers to line 6, but the error is actually on the previous line. This is quite common. Since C compilers are lenient about where you place whitespace, the compiler treats line 5 and line 6 as a single line that reads as follows:

printf ("Hello, world!\n") return 0;

Of course this code makes no sense, and that is why the compiler complains.

Often a missing curly bracket will cause one of these errors. For example, the following code:

#include <stdio.h>

/* To shorten example, not using argp */
int main()
{
  if (1==1)
    {
      printf ("Hello, world!\n");


  return 0;
}

generates the following error:

brackets.c: In function `main':
brackets.c:11: parse error at end of input

Because there is no closing curly bracket for the if statement, the compiler thinks the curly bracket that terminates the main function actually terminates the if statement. When it does not find a curly bracket on line 11 of the program to terminate the main function, it complains. One way to avoid this problem is to type both members of a matching pair of brackets before you fill them in.