02B-5

解答例1 †

#include <stdio.h>

int main(void){
  int x = 30, y = 4;
  double z = (double) x / y;
  printf("%f\n", z);
  return 0;
}

解答例2 †

#include <stdio.h>

int main(void){
  int x = 30, y = 4;
  double z = x /(double) y;
  printf("%f\n", z);
  return 0;
}

解答例3 †

#include <stdio.h>

int main(void){
  int x = 30, y = 4;
  double z = (double) x /(double) y;
  printf("%f\n", z);
  return 0;
}

解答例4 †

#include <stdio.h>

int main(void){
  printf("%f\n", (double) 30 / 4);
  return 0;
}

解答例5 †

#include <stdio.h>

int main(void){
  printf("%f\n", 30 / (double) 4);
  return 0;
}

解答例6 †

#include <stdio.h>

int main(void){
  printf("%f\n", (double) 30 / (double) 4);
  return 0;
}