Question

C language practice

On line 25 (see below) it says that "unknown conversion type character 'y' in format", why is that?? and also, I don't think 'x/y' is not working right, so I used another variable z to represent it but I guess it doesn't work either (it keeps showing x=0)

Answer 1

#include <stdio.h>
#include<math.h>
#include<stdlib.h>

int main( int argc, char ** argv ) {
int x;
int y;
int z;
z= x / y;

printf("Enter x value: \n");
fflush(stdout);
scanf("%d",&x);

printf("Enter y value: \n");
fflush(stdout);
scanf("%d",&y);

printf("Results: \n");
printf("x value\t |y value|expressions \t | results \n");
printf("%d \t | %d \t | x=y+3 \t | x= %d\n", x, y, y+3);
printf("%d \t | %d \t | x=y-2 \t | x= %d\n", x, y, y-2);
printf("%d \t | %d \t | x=y*5 \t | x= %d\n", x, y, y*5);
printf("%d \t | %d \t | x=x/y \t | x= %d\n", x, y, z);
printf("%d \t | %d \t | x=x % y \t | x= %d\n", x, y, x % y);

return 0;
}

Answer 2

line 25 is printf("%d \t | %d \t | x=x % y \t | x= %d\n", x, y, x % y);

Answer 3

Is there any other way to write this in a shorter way?

Answer 4

That is already pretty short.

Answer 5

The percent sign is a meta-character in printf format strings. To use a percent sign as a regular character, it must be escaped. The escape character in printf format strings is the percent sign, so:

printf("%d \t | %d \t | x=x %% y \t | x= %d\n", x, y, x % y);

Answer 6

Aah I see!
Thank you again rsmith :D

Thanks also e.mccormick for letting me know :)