please help me with this files program

Please give me code for the following :
The program should take one file as input which contains integers and should take it into a int variable. lets say, print that variable on the screen.

I wrote the following code but its showing errors..

Code:

# include <stdio.h>

int main (int argc, int argv[1] )
{
FILE *file_num ;
int num ;

file_num = fopen (argv[1],"r");
  if (!file_num) 
  printf ("File not found\n");
fscanf (file_num,"%d",&num);
printf("%d",num);
return ;
}

The programming is functioning when I replace these int types with char type but I need integers not characters.

What error?
Check argc > 1.
check fscanf() return value.

main should be either

Code:

int main(void)
int main(int ,char * [ ]);

main 2nd argument is array of pointer to char.
You are using 2nd argument of your main function as file name.
myprogram test.txt

argv[0] = "myprogram"
argv[1] = "test.txt"

This is what I am getting at terminal when I execute the code.
I even removed 1 from main (int argc, int argv[1], still this is what I 'm getting

test.c: In function ‘main’:
test.c:10: warning: passing argument 1 of ‘fopen’ makes pointer from integer without a cast
/usr/include/stdio.h:249: note: expected ‘const char * __restrict__’ but argument is of type ‘int’

Hello...

Code:

int main (int argc, int argv[1] )

Try...

Code:

int main (int argc, char *argv[])

> I even removed 1 from main (int argc, int argv[1], still this is what I 'm getting
But you didn't change the int to char did you?

Read your C book again to see how main should be declared.

Yes i did changes even that @salem

Then post your latest code and latest error message(s), if you're still stuck.

@salem

Code:

test.c: In function ‘main’:
test.c:10: warning: passing argument 1 of ‘fopen’ makes pointer from integer without a cast
/usr/include/stdio.h:249: note: expected ‘const char * __restrict__’ but argument is of type ‘char’

> Then post your latest code and latest error message(s), if you're still stuck.
If this is too hard for you, try something other than programming.

I'm just curious to see how many times he has to be told it's ... char *argv[] ... before he actually changes it.

I'm betting that the fopen call is now something like

Code:

fopen( argv[1][0], "r");

Originally Posted by Salem

I'm betting that the fopen call is now something like

Code:

fopen( argv[1][0], "r");

Code:

fopen (*&argv*[1][0],"%r++");

Dear Programmers!
I'm very sorry, first I didn't go through all your posts clearly. so, I didn't catch what you are saying. ..
I have changed to what Mr. Salem has said char *argv[].

@ Salem, @ CommonTater : Thanks for your great comments

Code:

fopen( argv[1][0], "r");
fopen (*&argv*[1][0],"%r++");

Its working!