Question

I am very new to C Programming and have a doubt… I’ve been asked to find errors in certain segments of C code… and this segment has me a bit confused so would appreciate the help…

int main(void)     
{
    int myInt = 5;
    printf("myInt = %d");
    return 0;
}

As far as i understand there is nothing wrong in this code. What i wanna know is why is this statement printing out a random number ??

The output i get is

myInt = 1252057154

Would appreciate the help… Thanks

Answer 1

You should read more about C programming.

And you should enable all warnings and debugging when compiling. With GCC, this means gcc -Wall -Wextra -g (on Linux at least).

When compiling with

gcc -Wall -Wextra -g john.c -o john

I am getting the following warnings:

john.c: In function ‘main’:
john.c:4:5: warning: implicit declaration of function ‘printf’ [-Wimplicit-function-declaration]
john.c:4:5: warning: incompatible implicit declaration of built-in function ‘printf’ [enabled by default]
john.c:4:5: warning: format ‘%d’ expects a matching ‘int’ argument [-Wformat]
john.c:3:9: warning: unused variable ‘myInt’ [-Wunused-variable]

So the correction is simple:

/* file john.c */
#include <stdio.h>
int main(void)     
{
  int myInt = 5;
  printf("myInt = %d\n", myInt);
  return 0;
}

which gets compiled without warnings.

Notice the \n at the end of printf format string. It is important.

Always enable all the warnings the compiler can give you and trust the compiler, so correct your code till no warnings is given.

And learn to use the debugger (e.g. gdb on Linux).

The behavior you observed is undefined behavior; anything could happen with a standard conforming implementation of C (even an explosion).

Happy hacking.