i just started studying programming at school, and i am so lost already. Here's the task:

Answer all 3 tasks in a separate file and return it. What is the result if the format string (control character) of the printf function is %.3f and the number to be printed is 456.87654321 0.17023 443.14159

How do i even do this? My code is this but it's obviously wrong.

#include <stdio.h>

int main() {
int num1, num2, num3;

printf("Give a number 1\n");
scanf("%i", &num1);

printf("Answer is on %.3f", &num1);

return 0;

}

It gave me 0 as answers or 0.000 what ever. Only 0's.

I don't know what to do really, my teacher is already on another subject and has no time helping me much.

CodePudding user response：

This source code:

#include <stdio.h>


int main(void)
{
    printf("%.3f\n", 456.87654321);
    printf("%.3f\n", 0.17023);
    printf("%.3f\n", 443.14159);
}

produces this output:

456.877
0.170
443.142

CodePudding user response：

You declared num1 to be an int (integer... whole numbers only).

Then you read that number from the keyboard.
I'm guessing you're entering 456.87654321.
(hint, that is not a whole number. it will not fit in an int)

Then you try to print it out with:

printf("Answer is on %.3f", &num1);

This has several problems.

• %.3f is for printing out doubles, not ints.
• You passed the "address" (& sign) of the number you wanted to print.
  Just pass the variable directly

Fixing up your code, I get:

#include <stdio.h>

int main() {
    double num1;                       // Declare DOUBLE, not INT

    printf("Give a number 1\n");
    scanf("%f", &num1);                // Get a DOUBLE from input, not an INT

    printf("Answer is on %.3f", num1); // Remove the & (address)

    return 0;
}