Equivalent of C++ cin.tie() for C stdio? [duplicate] - c++

Question

I'm using Eclipse to code in C/C++ and I'm struggling with what might be something pretty easy. In my code below I use printf() and after scanf(). Althougth printf is written before scanf() the output differs. I was able to find out something about similar issue here. But I wasn't able to solve it. Any ideas?
Code:
#include <stdio.h>
int main()
{
int myvariable;
printf("Enter a number:");
scanf("%d", &myvariable);
printf("%d", myvariable);
return 0;
}
Expected output:
Enter a number:1
1
Instead I get:
1
Enter a number:1

Answer 1

Your output is being buffered.
You have 4 options:
explicit flush
fflush after each write to profit from the buffer and still enforce the desiredbehavior/display explicitly.
fflush( stdout );
have the buffer only buffer lines-wise
useful for when you know that it is enough to print only complete lines
setlinebuf(stdout);
disable the buffer
setbuf(stdout, NULL);
disable buffering in your console through what ever options menu it provides
Examples:
Here is your code with option 1:
#include <stdio.h>
int main() {
int myvariable;
printf("Enter a number:");
fflush( stdout );
scanf("%d", &myvariable);
printf("%d", myvariable);
fflush( stdout );
return 0;
}
Here is 2:
#include <stdio.h>
int main() {
int myvariable;
setlinebuf(stdout);
printf("Enter a number:");
scanf("%d", &myvariable);
printf("%d", myvariable);
return 0;
}
and 3:
#include <stdio.h>
int main() {
int myvariable;
setbuf(stdout, NULL);
printf("Enter a number:");
scanf("%d", &myvariable);
printf("%d", myvariable);
return 0;
}

Answer 2

Ok, so finally I used something similar to what #zsawyer wrote as an option labelled 3.
In my code I inserted this line:
setvbuf(stdout, NULL, _IONBF, 0);
As a first line in main():
#include <stdio.h>
int main()
{
setvbuf(stdout, NULL, _IONBF, 0);
int myvariable;
printf("Enter a number:");
scanf("%d", &myvariable);
printf("%d", myvariable);
return 0;
}
I got it from here.