C printf printing random number - c

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.

Answer 2

printf (and similarly scanf) works like this:
Let's say you issue a call to printf
printf("%d some text %f %u %hu some more text\n", arg1, arg2, arg3, arg4)
printf goes over the format string and replaces %? with an argument, based on ?
%d some text %f %u %hu some more text
| | | |
arg1 arg2 | arg4
arg3
Now the thing with functions taking variable number of arguments is that, they don't know if the arguments exist, that's why they just take data from a specific part of the stack based on the format string. If you write:
printf("%d %f %u\n");
it reads three non existing data from the stack, most probably get the values stored when calling the function (values that should be hidden from you)

Answer 3

Well, if it's printing something wrong, the problem is in the printf call:
printf("myInt = %d");
Which arguments you're expected to pass?

Answer 4

It should be this:
printf("myInt = %d",myInt);
If you don't include the variable, then it basically pulls a random chunk of memory in. In addition, it might cause more wierd stuff to happen if you do this with a larger chunk of code. Always make sure that you have as many variables from a printf statement as you need, or bad stuff will result.

Answer 5

printf is a function that receives one or more arguments.
In your case it received only one argument (which is legal) but the argument containes %d which tells printf to take the second argument instead of the %d.
printf takes the "second argument" from the stack, and because only one argument was pushed to the stack (the string), it uses the return address as the second argument.

Answer 6

int main(void)
{
int myInt = 5;
printf("myInt = %d",myInt);
return 0;
}
The only change in the code is I added myInt variable in the printf statement if you see at the end.Whenever a variable is assigned some value it can be displayed only by passing it to the printf() function with the corresponding type specifier.Its a rule in C.