Beginner's Problem

I'm a novice C programmer using the Ubuntu terminal to compile. As in, I just started today. The compiler keeps coming up with this:

Variable1.c: In function 'main':
Variable1.c:9:46: error: expected ')' before string constant

I checked all over for mismatched parentheses. Here is the code.

Code:

#include <stdio.h>

int main()
{
  int this_is_a_number;

  printf ( "Please enter a number: " );
  scanf ( "%d", &this_is_a_number );
  printf ( "You entered %d", this_is_a_number"\n" );
  getchar();
  if ( this_is_a_number==1 ){
    printf ( "Correct! \n");
  }
  else {
    printf ( "Incorrect.  Press enter to be exited from the program. \n");
  }
  getchar();
  return 0;
}

Any ideas on what is wrong? I am a beginner, so it's probably something fundamental that I missed.

By the way, this is just experimental. There is absolutely no purpose to this program than a learning experience.

Originally Posted by Gigabitten

I'm a novice C programmer using the Ubuntu terminal to compile. As in, I just started today. The compiler keeps coming up with this:

Variable1.c: In function 'main':
Variable1.c:9:46: error: expected ')' before string constant

I checked all over for mismatched parentheses. Here is the code.

Code:

#include <stdio.h>

int main()
{
  int this_is_a_number;

  printf ( "Please enter a number: " );
  scanf ( "%d", &this_is_a_number );
  printf ( "You entered %d", this_is_a_number"\n" );
  getchar();
  if ( this_is_a_number==1 ){
    printf ( "Correct! \n");
  }
  else {
    printf ( "Incorrect.  Press enter to be exited from the program. \n");
  }
  getchar();
  return 0;
}

Any ideas on what is wrong? I am a beginner, so it's probably something fundamental that I missed.

the problem is in this statement->

printf ( "You entered %d", this_is_a_number"\n" );

Originally Posted by Gigabitten

By the way, this is just experimental. There is absolutely no purpose to this program than a learning experience.

replace line 9 with

Code:

printf( "You entered %d \n", this_is_a_number);

hey giga i am a beginner too just been into c for about a month now i think the problem with the code is the fact that you declared and integer and used 'getchar' which is for characters to collect an integer.
i have not ran the code yet i may be try scanf and see what happens

The problem was what ragarshi stated....include the '\n' in the quotation marks after the %d and it'd work.

Originally Posted by Nyah Check

hey giga i am a beginner too just been into c for about a month now i think the problem with the code is the fact that you declared and integer and used 'getchar' which is for characters to collect an integer.
i have not ran the code yet i may be try scanf and see what happens

You need to go and look up the return value from getchar() ....

Such a simple syntax error...thanks, guys! I was amazed at how fast the answer came!

Understand what the error means though. The "\n" is a string constant, and the compiler can only work with 1 variable in that part of the printf() code. There is no format in the format section, for another variable of any kind - it expects the closing parenthesis: ) (and semi-colon)